Chapter 03 Reading Questions Due: 11:59pm on Friday, May 23, 2014 You will receive no credit for items you complete after the assignment is due. Grading Policy Chapter 3 Reading Quiz Question 1 Part A Isotopes of an element differ from each other by the _____. ANSWER: Correct Chapter 3 Reading Quiz Question 2 Part A Which one of the following statements about pH is correct? ANSWER: Correct Lemon juice is an acid. Chapter 3 Reading Quiz Question 17 Part A In which form are water molecules most closely bonded to each other? ANSWER: number of electrons number of neutrons types of electrons number of protons Stomach acid has more OH- ions than H+ ions. Baking soda has more H+ ions than OH- ions. Lemon juice has more H+ ions than OH- ions. Seawater is slightly acidic. Chapter 03 Reading Questions http://session.masteringenvironmentalscience.com/myct/assignmentPrintV… 1 of 9 5/21/2014 7:58 PM Correct Chapter 3 Reading Quiz Question 16 Part A Which one of the following is a molecule but NOT a compound? ANSWER: Correct Oxygen is a molecule made up of just one element. Therefore, it is not a compound. Chapter 3 Reading Quiz Question 3 Part A Which one of the following is a carbohydrate and one of Earth’s most abundant organic molecule? ANSWER: Correct equally closely bonded in water vapor and ice solid ice forming part of an Antarctic sheet liquid water a few degrees above the freezing point water vapor above a boiling pot of water CH4 O2 CO2 H2O oil protein cellulose DNA Chapter 03 Reading Questions http://session.masteringenvironmentalscience.com/myct/assignmentPrintV… 2 of 9 5/21/2014 7:58 PM Chapter 3 Reading Quiz Question 4 Part A Which one of the following is a protein that functions as a catalyst? ANSWER: Correct Chapter 3 Reading Quiz Question 18 Part A The process of translation involves the use of _____. ANSWER: Chapter 3 Reading Quiz Question 5 Part A The cooling effect of sweating best represents _____. ANSWER: glucose cellulose enzyme RNA proteins to make lipids lipids to make carbohydrates carbohydrates to make proteins nucleic acids to make proteins latent heat transfer conduction radiation convection Chapter 03 Reading Questions http://session.masteringenvironmentalscience.com/myct/assignmentPrintV… 3 of 9 5/21/2014 7:58 PM Correct Chapter 3 Reading Quiz Question 6 Part A When plants use sunlight in photosynthesis, the plants are using a form of _____. ANSWER: Correct Chapter 3 Reading Quiz Question 8 Part A Which of the following converts mass to energy? ANSWER: Correct Chapter 3 Reading Quiz Question 19 Part A When a windmill turns to generate electricity, the amount of kinetic energy input _____. ANSWER: chemical energy in sunlight nuclear fission electromagnetic radiation conduction conduction the breaking of chemical bonds nuclear fission photosynthesis Chapter 03 Reading Questions http://session.masteringenvironmentalscience.com/myct/assignmentPrintV… 4 of 9 5/21/2014 7:58 PM Correct Chapter 3 Reading Quiz Question 20 Part A Which of the following best represents kinetic energy? ANSWER: Correct Chapter 3 Reading Quiz Question 21 Part A Which of the following processes reduces entropy? ANSWER: Correct Chapter 3 Reading Quiz Question 9 is unrelated to the amount of electrical energy produced is more than the amount of electrical energy produced equals the amount of electrical energy produced is less than the amount of electrical energy produced a charged battery gunpowder in a bullet the energy in the wax molecules of a candle a hot burner on a stove burning gasoline in an automobile engine photosynthesis in a leaf a person walking up a flight of stairs cell respiration in a leaf Chapter 03 Reading Questions http://session.masteringenvironmentalscience.com/myct/assignmentPrintV… 5 of 9 5/21/2014 7:58 PM Part A Which one of the following planets is a gas giant? ANSWER: Correct Chapter 3 Reading Quiz Question 10 Part A What is the main driving force that causes Earth’s tectonic plates to drift? ANSWER: Correct Chapter 3 Reading Quiz Question 23 Part A In which of the following locations would you expect to find large quantities of young rocks? ANSWER: Venus Jupiter Mars Mercury Heat from Earth’s core causes the mantle rock to circulate. The weight of the tectonic plates causes them to sink and melt. Currents of magma from the core of Earth circulate just beneath the tectonic plates. Electromagnetic radiation from the sun heats the tectonic plates, causing them to expand. the Appalachian Mountains the Himalayas deep in the central parts of India the Mid-Atlantic Ridge Chapter 03 Reading Questions http://session.masteringenvironmentalscience.com/myct/assignmentPrintV… 6 of 9 5/21/2014 7:58 PM Chapter 3 Reading Quiz Question 12 Part A The oxygen-rich atmosphere of Earth is mainly the result of _____. ANSWER: Correct Chapter 3 Reading Quiz Question 13 Part A A scientist working on the chemical reactions in the ozone layer is studying the _____. ANSWER: Correct Chapter 3 Reading Quiz Question 24 Part A The total amount of moisture in the air is highest when relative humidity is _____. ANSWER: volcanic activity chemical reactions between the early Earth atmosphere and iron photosynthetic organisms erosion of rocks into soil troposphere thermosphere stratosphere mesosphere Chapter 03 Reading Questions http://session.masteringenvironmentalscience.com/myct/assignmentPrintV… 7 of 9 5/21/2014 7:58 PM Chapter 3 Reading Quiz Question 15 Part A You are enjoying a spring day but expect a storm to arrive soon . As the storm arrives and the rain begins to fall, you notice that the temperature drops dramatically. Most likely, you have just experienced the arrival of a _____. ANSWER: Correct Chapter 3 Reading Quiz Question 25 Part A Every day tremendous amounts of the sun’s energy strikes Earth. Why doesn’t Earth overheat? ANSWER: Correct Earth’s energy budget is balanced. Over the course of a year, the energy input is equal to the energy output. Chapter 3 Reading Quiz Question 7 low and temperatures are low high and temperatures are high high and temperatures are low low and temperatures are high cold front Hadley cell intertropical convergence stratospheric event The energy is ultimately radiated back to space. Much of the heat melts rocks, forming lava deep inside of Earth. Most of the energy is used in photosynthesis to help plants grow and survive. The energy mostly is absorbed in various weather systems. Chapter 03 Reading Questions http://session.masteringenvironmentalscience.com/myct/assignmentPrintV… 8 of 9 5/21/2014 7:58 PM Part A How many calories are required to heat up 1,000 grams of liquid water (about 1 liter) from 20 °C to 70 °C? ANSWER: Correct Chapter 3 Reading Quiz Question 14 Part A Hadley cells near the Equator consist of _____. ANSWER: Correct Score Summary: Your score on this assignment is 85.5%. You received 19.67 out of a possible total of 23 points. 100 1,000 5,000 50,000 rising dry air associated with deserts and falling moist air that produces precipitation and rainforests rising moist air that produces precipitation and rainforests, and falling dry air associated with deserts warm, moist air rising up the sides of mountains and cool, dry air descending on the leeward sides cool, dry air rising up the sides of mountains and warm, moist air descending on the leeward sides Chapter 03 Reading Questions http://session.masteringenvironmentalscience.com/myct/assignmentPrintV… 9 of 9 5/21/2014 7:58 PM

Chapter 03 Reading Questions Due: 11:59pm on Friday, May 23, 2014 You will receive no credit for items you complete after the assignment is due. Grading Policy Chapter 3 Reading Quiz Question 1 Part A Isotopes of an element differ from each other by the _____. ANSWER: Correct Chapter 3 Reading Quiz Question 2 Part A Which one of the following statements about pH is correct? ANSWER: Correct Lemon juice is an acid. Chapter 3 Reading Quiz Question 17 Part A In which form are water molecules most closely bonded to each other? ANSWER: number of electrons number of neutrons types of electrons number of protons Stomach acid has more OH- ions than H+ ions. Baking soda has more H+ ions than OH- ions. Lemon juice has more H+ ions than OH- ions. Seawater is slightly acidic. Chapter 03 Reading Questions http://session.masteringenvironmentalscience.com/myct/assignmentPrintV… 1 of 9 5/21/2014 7:58 PM Correct Chapter 3 Reading Quiz Question 16 Part A Which one of the following is a molecule but NOT a compound? ANSWER: Correct Oxygen is a molecule made up of just one element. Therefore, it is not a compound. Chapter 3 Reading Quiz Question 3 Part A Which one of the following is a carbohydrate and one of Earth’s most abundant organic molecule? ANSWER: Correct equally closely bonded in water vapor and ice solid ice forming part of an Antarctic sheet liquid water a few degrees above the freezing point water vapor above a boiling pot of water CH4 O2 CO2 H2O oil protein cellulose DNA Chapter 03 Reading Questions http://session.masteringenvironmentalscience.com/myct/assignmentPrintV… 2 of 9 5/21/2014 7:58 PM Chapter 3 Reading Quiz Question 4 Part A Which one of the following is a protein that functions as a catalyst? ANSWER: Correct Chapter 3 Reading Quiz Question 18 Part A The process of translation involves the use of _____. ANSWER: Chapter 3 Reading Quiz Question 5 Part A The cooling effect of sweating best represents _____. ANSWER: glucose cellulose enzyme RNA proteins to make lipids lipids to make carbohydrates carbohydrates to make proteins nucleic acids to make proteins latent heat transfer conduction radiation convection Chapter 03 Reading Questions http://session.masteringenvironmentalscience.com/myct/assignmentPrintV… 3 of 9 5/21/2014 7:58 PM Correct Chapter 3 Reading Quiz Question 6 Part A When plants use sunlight in photosynthesis, the plants are using a form of _____. ANSWER: Correct Chapter 3 Reading Quiz Question 8 Part A Which of the following converts mass to energy? ANSWER: Correct Chapter 3 Reading Quiz Question 19 Part A When a windmill turns to generate electricity, the amount of kinetic energy input _____. ANSWER: chemical energy in sunlight nuclear fission electromagnetic radiation conduction conduction the breaking of chemical bonds nuclear fission photosynthesis Chapter 03 Reading Questions http://session.masteringenvironmentalscience.com/myct/assignmentPrintV… 4 of 9 5/21/2014 7:58 PM Correct Chapter 3 Reading Quiz Question 20 Part A Which of the following best represents kinetic energy? ANSWER: Correct Chapter 3 Reading Quiz Question 21 Part A Which of the following processes reduces entropy? ANSWER: Correct Chapter 3 Reading Quiz Question 9 is unrelated to the amount of electrical energy produced is more than the amount of electrical energy produced equals the amount of electrical energy produced is less than the amount of electrical energy produced a charged battery gunpowder in a bullet the energy in the wax molecules of a candle a hot burner on a stove burning gasoline in an automobile engine photosynthesis in a leaf a person walking up a flight of stairs cell respiration in a leaf Chapter 03 Reading Questions http://session.masteringenvironmentalscience.com/myct/assignmentPrintV… 5 of 9 5/21/2014 7:58 PM Part A Which one of the following planets is a gas giant? ANSWER: Correct Chapter 3 Reading Quiz Question 10 Part A What is the main driving force that causes Earth’s tectonic plates to drift? ANSWER: Correct Chapter 3 Reading Quiz Question 23 Part A In which of the following locations would you expect to find large quantities of young rocks? ANSWER: Venus Jupiter Mars Mercury Heat from Earth’s core causes the mantle rock to circulate. The weight of the tectonic plates causes them to sink and melt. Currents of magma from the core of Earth circulate just beneath the tectonic plates. Electromagnetic radiation from the sun heats the tectonic plates, causing them to expand. the Appalachian Mountains the Himalayas deep in the central parts of India the Mid-Atlantic Ridge Chapter 03 Reading Questions http://session.masteringenvironmentalscience.com/myct/assignmentPrintV… 6 of 9 5/21/2014 7:58 PM Chapter 3 Reading Quiz Question 12 Part A The oxygen-rich atmosphere of Earth is mainly the result of _____. ANSWER: Correct Chapter 3 Reading Quiz Question 13 Part A A scientist working on the chemical reactions in the ozone layer is studying the _____. ANSWER: Correct Chapter 3 Reading Quiz Question 24 Part A The total amount of moisture in the air is highest when relative humidity is _____. ANSWER: volcanic activity chemical reactions between the early Earth atmosphere and iron photosynthetic organisms erosion of rocks into soil troposphere thermosphere stratosphere mesosphere Chapter 03 Reading Questions http://session.masteringenvironmentalscience.com/myct/assignmentPrintV… 7 of 9 5/21/2014 7:58 PM Chapter 3 Reading Quiz Question 15 Part A You are enjoying a spring day but expect a storm to arrive soon . As the storm arrives and the rain begins to fall, you notice that the temperature drops dramatically. Most likely, you have just experienced the arrival of a _____. ANSWER: Correct Chapter 3 Reading Quiz Question 25 Part A Every day tremendous amounts of the sun’s energy strikes Earth. Why doesn’t Earth overheat? ANSWER: Correct Earth’s energy budget is balanced. Over the course of a year, the energy input is equal to the energy output. Chapter 3 Reading Quiz Question 7 low and temperatures are low high and temperatures are high high and temperatures are low low and temperatures are high cold front Hadley cell intertropical convergence stratospheric event The energy is ultimately radiated back to space. Much of the heat melts rocks, forming lava deep inside of Earth. Most of the energy is used in photosynthesis to help plants grow and survive. The energy mostly is absorbed in various weather systems. Chapter 03 Reading Questions http://session.masteringenvironmentalscience.com/myct/assignmentPrintV… 8 of 9 5/21/2014 7:58 PM Part A How many calories are required to heat up 1,000 grams of liquid water (about 1 liter) from 20 °C to 70 °C? ANSWER: Correct Chapter 3 Reading Quiz Question 14 Part A Hadley cells near the Equator consist of _____. ANSWER: Correct Score Summary: Your score on this assignment is 85.5%. You received 19.67 out of a possible total of 23 points. 100 1,000 5,000 50,000 rising dry air associated with deserts and falling moist air that produces precipitation and rainforests rising moist air that produces precipitation and rainforests, and falling dry air associated with deserts warm, moist air rising up the sides of mountains and cool, dry air descending on the leeward sides cool, dry air rising up the sides of mountains and warm, moist air descending on the leeward sides Chapter 03 Reading Questions http://session.masteringenvironmentalscience.com/myct/assignmentPrintV… 9 of 9 5/21/2014 7:58 PM

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During deployment processing, soldiers undergo medical screening.After a wait, each soldier see same dicaltechnician who reviews his or her records.If the records review shows no problems—the soldier is physically qualified to deploy—then the soldier departs to the next step in deployment processing.It he records do reveal a potential medical issue,the soldier instead sees a doctor,who assesses the soldier’s condition and determines both deployability and for those who are medically disqualified, treatment needs.In rare cases ,the doctor in itiate a medical board to evaluate the soldier for retention in the military. Currently, 80 soldiers per hour arrive for deployment screening,and 80% of them pass there cords review.On average,20 people are waiting for the medical records review,which takes 6 minutes. When the records review indicates a soldier must see a doctor,the soldier reports to a waiting room,where an average of 8 soldiers are waiting. After a wait,the soldier sees a doctor, who reviews the soldier’s condition and either approves the soldier for deployment(75%of the time) or disapproves deployment and conducts a morein-deptxeam to determine treatment(20%of the time)or the need for a medical board(5%ofthetime).Each doctor’s exam takes,on average,6minutes if the solider is medically able to deploy(the doctor pretty much replicates the records review),15 minutes if the soldier requires some kind of treatment,and 30 minutes in those rare cases that require the doctor to initiate a medical review board. Assume the process is stable;that is,average inflow rate equals average outflow rate.[Finally,thisisNOTaqueuingproblem.] a. On average,how long does a soldier spend in the deployment process?

During deployment processing, soldiers undergo medical screening.After a wait, each soldier see same dicaltechnician who reviews his or her records.If the records review shows no problems—the soldier is physically qualified to deploy—then the soldier departs to the next step in deployment processing.It he records do reveal a potential medical issue,the soldier instead sees a doctor,who assesses the soldier’s condition and determines both deployability and for those who are medically disqualified, treatment needs.In rare cases ,the doctor in itiate a medical board to evaluate the soldier for retention in the military. Currently, 80 soldiers per hour arrive for deployment screening,and 80% of them pass there cords review.On average,20 people are waiting for the medical records review,which takes 6 minutes. When the records review indicates a soldier must see a doctor,the soldier reports to a waiting room,where an average of 8 soldiers are waiting. After a wait,the soldier sees a doctor, who reviews the soldier’s condition and either approves the soldier for deployment(75%of the time) or disapproves deployment and conducts a morein-deptxeam to determine treatment(20%of the time)or the need for a medical board(5%ofthetime).Each doctor’s exam takes,on average,6minutes if the solider is medically able to deploy(the doctor pretty much replicates the records review),15 minutes if the soldier requires some kind of treatment,and 30 minutes in those rare cases that require the doctor to initiate a medical review board. Assume the process is stable;that is,average inflow rate equals average outflow rate.[Finally,thisisNOTaqueuingproblem.] a. On average,how long does a soldier spend in the deployment process?

6+0.75*6+.2*15+.05*30 = 15 miutes     Timeindeploymentsystem:15 (minutes)
Who are the actors in the dialogue and who is more powerful? A. The Lacedaemons and the Melians. The Melians are more powerful. B. The Athenians and the Melians. The Athenians are more powerful. C. The Lacedaemons, the Melians, and the Athenians. The Melians are most powerful. D. The Athenians and the Melians. The Melians are more powerful. E. The Lacedaemons and the Melians. They are equally powerful. The Athenian envoy was dispatched to address the Melians. Who did they end up talking to? A. The people of Lacedaemon. B. A group of Melian magistrates. C. The Melian magistrates and the population. D. Just the general Melian population. The magistrates did not show out of protest. E. The Lacedaemon population and the Melian population. What is the main point in question upon which both parties seem to agree? A. Whether to go to war against Athenian enemies. B. Whether to form an equal alliance with the Athenians. C. Whether there will be war between the Melians and the Athenians or whether the Melians will submit to Athenian power. D. How many ships and soldiers the Melians will give to the Athenian campaign. E. How many ships and soldiers the Athenians will give to the Melian movement for independence. What do the Athenians suggest about right and justice in terms of power? A. Questions about what is “right” only happen among equals in power. B. Justice and right come about when the powerful are wise. C. Power has no relation to right and justice. D. Questions about what is “right” only happen among equals in power AND justice and right come about when the powerful are wise. E. Justice and right come about when the powerful are wise AND power has no relation to right and justice. What is the necessary “law of their nature” that the Athenians are referring to? A. That states, both weak and strong, will seek peace. B. The law of chance and luck operates universally among men. C. The belief that, given the opportunity, men will seek to rule whenever they can. D. Whenever a treaty is created, states will settle those treaties with due consideration to the weaker side. E. That the nature of men is towards justice rather than power.

Who are the actors in the dialogue and who is more powerful? A. The Lacedaemons and the Melians. The Melians are more powerful. B. The Athenians and the Melians. The Athenians are more powerful. C. The Lacedaemons, the Melians, and the Athenians. The Melians are most powerful. D. The Athenians and the Melians. The Melians are more powerful. E. The Lacedaemons and the Melians. They are equally powerful. The Athenian envoy was dispatched to address the Melians. Who did they end up talking to? A. The people of Lacedaemon. B. A group of Melian magistrates. C. The Melian magistrates and the population. D. Just the general Melian population. The magistrates did not show out of protest. E. The Lacedaemon population and the Melian population. What is the main point in question upon which both parties seem to agree? A. Whether to go to war against Athenian enemies. B. Whether to form an equal alliance with the Athenians. C. Whether there will be war between the Melians and the Athenians or whether the Melians will submit to Athenian power. D. How many ships and soldiers the Melians will give to the Athenian campaign. E. How many ships and soldiers the Athenians will give to the Melian movement for independence. What do the Athenians suggest about right and justice in terms of power? A. Questions about what is “right” only happen among equals in power. B. Justice and right come about when the powerful are wise. C. Power has no relation to right and justice. D. Questions about what is “right” only happen among equals in power AND justice and right come about when the powerful are wise. E. Justice and right come about when the powerful are wise AND power has no relation to right and justice. What is the necessary “law of their nature” that the Athenians are referring to? A. That states, both weak and strong, will seek peace. B. The law of chance and luck operates universally among men. C. The belief that, given the opportunity, men will seek to rule whenever they can. D. Whenever a treaty is created, states will settle those treaties with due consideration to the weaker side. E. That the nature of men is towards justice rather than power.

Who are the actors in the dialogue and who is … Read More...
Extra Credit Due: 11:59pm on Thursday, May 15, 2014 You will receive no credit for items you complete after the assignment is due. Grading Policy A Man Running to Catch a Bus A man is running at speed (much less than the speed of light) to catch a bus already at a stop. At , when he is a distance from the door to the bus, the bus starts moving with the positive acceleration . Use a coordinate system with at the door of the stopped bus. Part A What is , the position of the man as a function of time? Answer symbolically in terms of the variables , , and . Hint 1. Which equation should you use for the man’s speed? Because the man’s speed is constant, you may use . ANSWER: c t = 0 b a x = 0 xman(t) b c t x(t) = x(0) + vt xman(t) = −b + ct Correct Part B What is , the position of the bus as a function of time? Answer symbolically in terms of and . Hint 1. Which equation should you use for the bus’s acceleration? Because the bus has constant acceleration, you may use . Recall that . ANSWER: Correct Part C What condition is necessary for the man to catch the bus? Assume he catches it at time . Hint 1. How to approach this problem If the man is to catch the bus, then at some moment in time , the man must arrive at the position of the door of the bus. How would you express this condition mathematically? ANSWER: xbus(t) a t x(t) = x(0) + v(0)t + (1/2)at2 vbus(0) = 0 xbus = 1 a 2 t2 tcatch tcatch Typesetting math: 15% Correct Part D Inserting the formulas you found for and into the condition , you obtain the following: , or . Intuitively, the man will not catch the bus unless he is running fast enough. In mathematical terms, there is a constraint on the man’s speed so that the equation above gives a solution for that is a real positive number. Find , the minimum value of for which the man will catch the bus. Express the minimum value for the man’s speed in terms of and . Hint 1. Consider the discriminant Use the quadratic equation to solve: . What is the discriminant (the part under the radical) of the solution for ? xman(tcatch) > xbus(tcatch) xman(tcatch) = xbus(tcatch) xman(tcatch) < xbus(tcatch) c = a  tcatch xman(t) xbus(t) xman(tcatch) = xbus(tcatch) −b+ct = a catch 1 2 t2 catch 1 a −c +b = 0 2 t2 catch tcatch c tcatch cmin c a b 1 a − c + b = 0 2 t2 catch tcatch tcatch Typesetting math: 15% Hint 1. The quadratic formula Recall: If then ANSWER: Hint 2. What is the constraint? To get a real value for , the discriminant must be greater then or equal to zero. This condition yields a constraint that exceed . ANSWER: Correct Part E Assume that the man misses getting aboard when he first meets up with the bus. Does he get a second chance if he continues to run at the constant speed ? Hint 1. What is the general quadratic equation? The general quadratic equation is , where , \texttip{B}{B}, and \texttip{C}{C} are constants. Depending on the value of the discriminant, \Delta = c^2-2ab, the equation may have Ax2 + Bx + C = 0 x = −B±B2−4AC 2A  = cc − 2ab tcatch c cmin cmin = (2ab) −−−−  c > cmin Ax2 + Bx + C = 0 A Typesetting math: 15% two real valued solutions 1. if \Delta > 0, 2. one real valued solution if \Delta = 0, or 3. two complex valued solutions if \Delta < 0. In this case, every real valued solution corresponds to a time at which the man is at the same position as the door of the bus. ANSWER: Correct Adding and Subtracting Vectors Conceptual Question Six vectors (A to F) have the magnitudes and directions indicated in the figure. Part A No; there is no chance he is going to get aboard. Yes; he will get a second chance Typesetting math: 15% Which two vectors, when added, will have the largest (positive) x component? Hint 1. Largest x component The two vectors with the largest x components will, when combined, give the resultant with the largest x component. Keep in mind that positive x components are larger than negative x components. ANSWER: Correct Part B Which two vectors, when added, will have the largest (positive) y component? Hint 1. Largest y component The two vectors with the largest y components will, when combined, give the resultant with the largest y component. Keep in mind that positive y components are larger than negative y components. ANSWER: C and E E and F A and F C and D B and D Typesetting math: 15% Correct Part C Which two vectors, when subtracted (i.e., when one vector is subtracted from the other), will have the largest magnitude? Hint 1. Subtracting vectors To subtract two vectors, add a vector with the same magnitude but opposite direction of one of the vectors to the other vector. ANSWER: Correct Tactics Box 3.1 Determining the Components of a Vector Learning Goal: C and D A and F E and F A and B E and D A and F A and E D and B C and D E and F Typesetting math: 15% To practice Tactics Box 3.1 Determining the Components of a Vector. When a vector \texttip{\vec{A}}{A_vec} is decomposed into component vectors \texttip{\vec{A}_{\mit x}}{A_vec_x} and \texttip{\vec{A}_{\mit y}}{A_vec_y} parallel to the coordinate axes, we can describe each component vector with a single number (a scalar) called the component. This tactics box describes how to determine the x component and y component of vector \texttip{\vec{A}}{A_vec}, denoted \texttip{A_{\mit x}}{A_x} and \texttip{A_{\mit y}}{A_y}. TACTICS BOX 3.1 Determining the components of a vector The absolute value |A_x| of the x component \texttip{A_{\mit x}}{A_x} is the magnitude of the component vector \texttip{\vec{A}_{\1. mit x}}{A_vec_x}. The sign of \texttip{A_{\mit x}}{A_x} is positive if \texttip{\vec{A}_{\mit x}}{A_vec_x} points in the positive x direction; it is negative if \texttip{\vec{A}_{\mit x}}{A_vec_x} points in the negative x direction. 2. 3. The y component \texttip{A_{\mit y}}{A_y} is determined similarly. Part A What is the magnitude of the component vector \texttip{\vec{A}_{\mit x}}{A_vec_x} shown in the figure? Express your answer in meters to one significant figure. ANSWER: Correct |A_x| = 5 \rm m Typesetting math: 15% Part B What is the sign of the y component \texttip{A_{\mit y}}{A_y} of vector \texttip{\vec{A}}{A_vec} shown in the figure? ANSWER: Correct Part C Now, combine the information given in the tactics box above to find the x and y components, \texttip{B_{\mit x}}{B_x} and \texttip{B_{\mit y}}{B_y}, of vector \texttip{\vec{B}}{B_vec} shown in the figure. Express your answers, separated by a comma, in meters to one significant figure. positive negative Typesetting math: 15% ANSWER: Correct Conceptual Problem about Projectile Motion Learning Goal: To understand projectile motion by considering horizontal constant velocity motion and vertical constant acceleration motion independently. Projectile motion refers to the motion of unpowered objects (called projectiles) such as balls or stones moving near the surface of the earth under the influence of the earth's gravity alone. In this analysis we assume that air resistance can be neglected. An object undergoing projectile motion near the surface of the earth obeys the following rules: An object undergoing projectile motion travels horizontally at a constant rate. That is, the x component of its velocity, \texttip{v_{\mit x}}{1. v_x}, is constant. An object undergoing projectile motion moves vertically with a constant downward acceleration whose magnitude, denoted by \texttip{g}{g}, is equal to 9.80 \rm{m/s^2} near the surface of the earth. Hence, the y component of its velocity, \texttip{v_{\mit y}}{v_y}, changes continuously. 2. An object undergoing projectile motion will undergo the horizontal and vertical motions described above from the instant it is launched until the instant it strikes the ground again. Even though the horizontal and vertical motions can be treated independently, they are related by the fact that they occur for exactly the same amount of time, namely the time \texttip{t}{t} the projectile is in the air. 3. The figure shows the trajectory (i.e., the path) of a ball undergoing projectile motion over level ground. The time t_0 = 0\;\rm{s} corresponds to the moment just after the ball is launched from position x_0 = 0\;\rm{m} and y_0 = 0\;\rm{m}. Its launch velocity, also called the initial velocity, is \texttip{\vec{v}_{\rm 0}}{v_vec_0}. Two other points along the trajectory are indicated in the figure. One is the moment the ball reaches the peak of its trajectory, at time \texttip{t_{\rm 1}}{t_1} with velocity \texttip{\vec{v}_{\rm 1}}{v_1_vec}. Its position at this moment is denoted by (x_1, y_1) or (x_1, y_{\max}) since it is at its maximum \texttip{B_{\mit x}}{B_x}, \texttip{B_{\mit y}}{B_y} = -2,-5 \rm m, \rm m Typesetting math: 15% The other point, at time \texttip{t_{\rm 2}}{t_2} with velocity \texttip{\vec{v}_{\rm 2}}{v_2_vec}, corresponds to the moment just before the ball strikes the ground on the way back down. At this time its position is (x_2, y_2), also known as (x_{\max}, y_2) since it is at its maximum horizontal range. Projectile motion is symmetric about the peak, provided the object lands at the same vertical height from which is was launched, as is the case here. Hence y_2 = y_0 = 0\;\rm{m}. Part A How do the speeds \texttip{v_{\rm 0}}{v_0}, \texttip{v_{\rm 1}}{v_1}, and \texttip{v_{\rm 2}}{v_2} (at times \texttip{t_{\rm 0}}{t_0}, \texttip{t_{\rm 1}}{t_1}, and \texttip{t_{\rm 2}}{t_2}) compare? ANSWER: Correct Here \texttip{v_{\rm 0}}{v_0} equals \texttip{v_{\rm 2}}{v_2} by symmetry and both exceed \texttip{v_{\rm 1}}{v_1}. This is because \texttip{v_{\rm 0}}{v_0} and \texttip{v_{\rm 2}}{v_2} include vertical speed as well as the constant horizontal speed. Consider a diagram of the ball at time \texttip{t_{\rm 0}}{t_0}. Recall that \texttip{t_{\rm 0}}{t_0} refers to the instant just after the ball has been launched, so it is still at ground level (x_0 = y_0= 0\;\rm{m}). However, it is already moving with initial velocity \texttip{\vec{v}_{\rm 0}}{v_0_vec}, whose magnitude is v_0 = 30.0\;{\rm m/s} and direction is \theta = 60.0\;{\rm degrees} counterclockwise from the positive x direction. \texttip{v_{\rm 0}}{v_0} = \texttip{v_{\rm 1}}{v_1} = \texttip{v_{\rm 2}}{v_2} > 0 \texttip{v_{\rm 0}}{v_0} = \texttip{v_{\rm 2}}{v_2} > \texttip{v_{\rm 1}}{v_1} = 0 \texttip{v_{\rm 0}}{v_0} = \texttip{v_{\rm 2}}{v_2} > \texttip{v_{\rm 1}}{v_1} > 0 \texttip{v_{\rm 0}}{v_0} > \texttip{v_{\rm 1}}{v_1} > \texttip{v_{\rm 2}}{v_2} > 0 \texttip{v_{\rm 0}}{v_0} > \texttip{v_{\rm 2}}{v_2} > \texttip{v_{\rm 1}}{v_1} = 0 Typesetting math: 15% Part B What are the values of the intial velocity vector components \texttip{v_{0,x}}{v_0, x} and \texttip{v_{0,y}}{v_0, y} (both in \rm{m/s}) as well as the acceleration vector components \texttip{a_{0,x}}{a_0, x} and \texttip{a_{0,y}}{a_0, y} (both in \rm{m/s^2})? Here the subscript 0 means “at time \texttip{t_{\rm 0}}{t_0}.” Hint 1. Determining components of a vector that is aligned with an axis If a vector points along a single axis direction, such as in the positive x direction, its x component will be its full magnitude, whereas its y component will be zero since the vector is perpendicular to the y direction. If the vector points in the negative x direction, its x component will be the negative of its full magnitude. Hint 2. Calculating the components of the initial velocity Notice that the vector \texttip{\vec{v}_{\rm 0}}{v_0_vec} points up and to the right. Since “up” is the positive y axis direction and “to the right” is the positive x axis direction, \texttip{v_{0,x}}{v_0, x} and \texttip{v_{0,y}}{v_0, y} will both be positive. As shown in the figure, \texttip{v_{0,x}}{v_0, x}, \texttip{v_{0,y}}{v_0, y}, and \texttip{v_{\rm 0}}{v_0} are three sides of a right triangle, one angle of which is \texttip{\theta }{theta}. Thus \texttip{v_{0,x}}{v_0, x} and \texttip{v_{0,y}}{v_0, y} can be found using the definition of the sine and cosine functions given below. Recall that v_0 = 30.0\;\rm{m/s} and \theta = 60.0\;\rm{degrees} and note that \large{\sin(\theta) = \frac{\rm{length\;of\;opposite\;side}}{\rm{length\;of\;hypotenuse}}} \large{= \frac{v_{0, y}}{v_0}}, \large{\cos(\theta) = \frac{\rm{length\;of\;adjacent\;side}}{\rm{length\;of\;hypotenuse}}} \large{= \frac{v_{0, x}}{v_0}.} What are the values of \texttip{v_{0,x}}{v_0, x} and \texttip{v_{0,y}}{v_0, y}? Enter your answers numerically in meters per second separated by a comma. ANSWER: ANSWER: 15.0,26.0 \rm{m/s} Typesetting math: 15% Correct Also notice that at time \texttip{t_{\rm 2}}{t_2}, just before the ball lands, its velocity components are v_{2, x} = 15\;\rm{m/s} (the same as always) and v_{2, y} = – 26.0\;\rm{m/s} (the same size but opposite sign from \texttip{v_{0,y}}{v_0, y} by symmetry). The acceleration at time \texttip{t_{\rm 2}}{t_2} will have components (0, -9.80 \rm{m/s^2}), exactly the same as at \texttip{t_{\rm 0}}{t_0}, as required by Rule 2. The peak of the trajectory occurs at time \texttip{t_{\rm 1}}{t_1}. This is the point where the ball reaches its maximum height \texttip{y_{\rm max}}{y_max}. At the peak the ball switches from moving up to moving down, even as it continues to travel horizontally at a constant rate. Part C What are the values of the velocity vector components \texttip{v_{1,x}}{v_1, x} and \texttip{v_{1,y}}{v_1, y} (both in \rm{m/s}) as well as the acceleration vector components \texttip{a_{1,x}}{a_1, x} and \texttip{a_{1,y}}{a_1, y} (both in \rm{m/s^2})? Here the subscript 1 means that these are all at time \texttip{t_{\rm 1}}{t_1}. ANSWER: 30.0, 0, 0, 0 0, 30.0, 0, 0 15.0, 26.0, 0, 0 30.0, 0, 0, -9.80 0, 30.0, 0, -9.80 15.0, 26.0, 0, -9.80 15.0, 26.0, 0, +9.80 Typesetting math: 15% Correct At the peak of its trajectory the ball continues traveling horizontally at a constant rate. However, at this moment it stops moving up and is about to move back down. This constitutes a downward-directed change in velocity, so the ball is accelerating downward even at the peak. The flight time refers to the total amount of time the ball is in the air, from just after it is launched (\texttip{t_{\rm 0}}{t_0}) until just before it lands (\texttip{t_{\rm 2}}{t_2}). Hence the flight time can be calculated as t_2 – t_0, or just \texttip{t_{\rm 2}}{t_2} in this particular situation since t_0 = 0. Because the ball lands at the same height from which it was launched, by symmetry it spends half its flight time traveling up to the peak and the other half traveling back down. The flight time is determined by the initial vertical component of the velocity and by the acceleration. The flight time does not depend on whether the object is moving horizontally while it is in the air. Part D If a second ball were dropped from rest from height \texttip{y_{\rm max}}{y_max}, how long would it take to reach the ground? Ignore air resistance. Check all that apply. Hint 1. Kicking a ball of cliff; a related problem Consider two balls, one of which is dropped from rest off the edge of a cliff at the same moment that the other is kicked horizontally off the edge of the cliff. Which ball reaches the level ground at the base of the cliff first? Ignore air resistance. Hint 1. Comparing position, velocity, and acceleration of the two balls Both balls start at the same height and have the same initial y velocity (v_{0,y} = 0) as well as the same acceleration (\vec a = g downward). They differ only in their x velocity (one is 0, 0, 0, 0 0, 0, 0, -9.80 15.0, 0, 0, 0 15.0, 0, 0, -9.80 0, 26.0, 0, 0 0, 26.0, 0, -9.80 15.0, 26.0, 0, 0 15.0, 26.0, 0, -9.80 Typesetting math: 15% zero, the other nonzero). This difference will affect their x motion but not their y motion. ANSWER: ANSWER: Correct In projectile motion over level ground, it takes an object just as long to rise from the ground to the peak as it takes for it to fall from the peak back to the ground. The range \texttip{R}{R} of the ball refers to how far it moves horizontally, from just after it is launched until just before it lands. Range is defined as x_2 – x_0, or just \texttip{x_{\rm 2}}{x_2} in this particular situation since x_0 = 0. Range can be calculated as the product of the flight time \texttip{t_{\rm 2}}{t_2} and the x component of the velocity \texttip{v_{\mit x}}{v_x} (which is the same at all times, so v_x = v_{0,x}). The value of \texttip{v_{\mit x}}{v_x} can be found from the launch speed \texttip{v_{\rm 0}}{v_0} and the launch angle \texttip{\theta }{theta} using trigonometric functions, as was done in Part B. The flight time is related to the initial y component of the velocity, which may also be found from \texttip{v_{\rm 0}}{v_0} and \texttip{\theta }{theta} using trig functions. The following equations may be useful in solving projectile motion problems, but these equations apply only to a projectile launched over level ground from position (x_0 = y_0 = 0) at time t_0 = 0 with initial speed \texttip{v_{\rm 0}}{v_0} and launch angle \texttip{\theta }{theta} measured from the horizontal. As was the case above, \texttip{t_{\rm 2}}{t_2} refers to the flight time and \texttip{R}{R} refers to the range of the projectile. flight time: \large{t_2 = \frac{2 v_{0, y}}{g} = \frac{2 v_0 \sin(\theta)}{g}} range: \large{R = v_x t_2 = \frac{v_0^2 \sin(2\theta)}{g}} The ball that falls straight down strikes the ground first. The ball that was kicked so it moves horizontally as it falls strikes the ground first. Both balls strike the ground at the same time. \texttip{t_{\rm 0}}{t_0} t_1 – t_0 \texttip{t_{\rm 2}}{t_2} t_2 – t_1 \large{\frac{t_2 – t_0}{2}} Typesetting math: 15% In general, a high launch angle yields a long flight time but a small horizontal speed and hence little range. A low launch angle gives a larger horizontal speed, but less flight time in which to accumulate range. The launch angle that achieves the maximum range for projectile motion over level ground is 45 degrees. Part E Which of the following changes would increase the range of the ball shown in the original figure? Check all that apply. ANSWER: Correct A solid understanding of the concepts of projectile motion will take you far, including giving you additional insight into the solution of projectile motion problems numerically. Even when the object does not land at the same height from which is was launched, the rules given in the introduction will still be useful. Recall that air resistance is assumed to be negligible here, so this projectile motion analysis may not be the best choice for describing things like frisbees or feathers, whose motion is strongly influenced by air. The value of the gravitational free-fall acceleration \texttip{g}{g} is also assumed to be constant, which may not be appropriate for objects that move vertically through distances of hundreds of kilometers, like rockets or missiles. However, for problems that involve relatively dense projectiles moving close to the surface of the earth, these assumptions are reasonable. A World-Class Sprinter World-class sprinters can accelerate out of the starting blocks with an acceleration that is nearly horizontal and has magnitude 15 \;{\rm m}/{\rm s}^{2}. Part A How much horizontal force \texttip{F}{F} must a sprinter of mass 64{\rm kg} exert on the starting blocks to produce this acceleration? Express your answer in newtons using two significant figures. Increase \texttip{v_{\rm 0}}{v_0} above 30 \rm{m/s}. Reduce \texttip{v_{\rm 0}}{v_0} below 30 \rm{m/s}. Reduce \texttip{\theta }{theta} from 60 \rm{degrees} to 45 \rm{degrees}. Reduce \texttip{\theta }{theta} from 60 \rm{degrees} to less than 30 \rm{degrees}. Increase \texttip{\theta }{theta} from 60 \rm{degrees} up toward 90 \rm{degrees}. Typesetting math: 15% Hint 1. Newton’s 2nd law of motion According to Newton’s 2nd law of motion, if a net external force \texttip{F_{\rm net}}{F_net} acts on a body, the body accelerates, and the net force is equal to the mass \texttip{m}{m} of the body times the acceleration \texttip{a}{a} of the body: F_{\rm net} = ma. ANSWER: Co

Extra Credit Due: 11:59pm on Thursday, May 15, 2014 You will receive no credit for items you complete after the assignment is due. Grading Policy A Man Running to Catch a Bus A man is running at speed (much less than the speed of light) to catch a bus already at a stop. At , when he is a distance from the door to the bus, the bus starts moving with the positive acceleration . Use a coordinate system with at the door of the stopped bus. Part A What is , the position of the man as a function of time? Answer symbolically in terms of the variables , , and . Hint 1. Which equation should you use for the man’s speed? Because the man’s speed is constant, you may use . ANSWER: c t = 0 b a x = 0 xman(t) b c t x(t) = x(0) + vt xman(t) = −b + ct Correct Part B What is , the position of the bus as a function of time? Answer symbolically in terms of and . Hint 1. Which equation should you use for the bus’s acceleration? Because the bus has constant acceleration, you may use . Recall that . ANSWER: Correct Part C What condition is necessary for the man to catch the bus? Assume he catches it at time . Hint 1. How to approach this problem If the man is to catch the bus, then at some moment in time , the man must arrive at the position of the door of the bus. How would you express this condition mathematically? ANSWER: xbus(t) a t x(t) = x(0) + v(0)t + (1/2)at2 vbus(0) = 0 xbus = 1 a 2 t2 tcatch tcatch Typesetting math: 15% Correct Part D Inserting the formulas you found for and into the condition , you obtain the following: , or . Intuitively, the man will not catch the bus unless he is running fast enough. In mathematical terms, there is a constraint on the man’s speed so that the equation above gives a solution for that is a real positive number. Find , the minimum value of for which the man will catch the bus. Express the minimum value for the man’s speed in terms of and . Hint 1. Consider the discriminant Use the quadratic equation to solve: . What is the discriminant (the part under the radical) of the solution for ? xman(tcatch) > xbus(tcatch) xman(tcatch) = xbus(tcatch) xman(tcatch) < xbus(tcatch) c = a  tcatch xman(t) xbus(t) xman(tcatch) = xbus(tcatch) −b+ct = a catch 1 2 t2 catch 1 a −c +b = 0 2 t2 catch tcatch c tcatch cmin c a b 1 a − c + b = 0 2 t2 catch tcatch tcatch Typesetting math: 15% Hint 1. The quadratic formula Recall: If then ANSWER: Hint 2. What is the constraint? To get a real value for , the discriminant must be greater then or equal to zero. This condition yields a constraint that exceed . ANSWER: Correct Part E Assume that the man misses getting aboard when he first meets up with the bus. Does he get a second chance if he continues to run at the constant speed ? Hint 1. What is the general quadratic equation? The general quadratic equation is , where , \texttip{B}{B}, and \texttip{C}{C} are constants. Depending on the value of the discriminant, \Delta = c^2-2ab, the equation may have Ax2 + Bx + C = 0 x = −B±B2−4AC 2A  = cc − 2ab tcatch c cmin cmin = (2ab) −−−−  c > cmin Ax2 + Bx + C = 0 A Typesetting math: 15% two real valued solutions 1. if \Delta > 0, 2. one real valued solution if \Delta = 0, or 3. two complex valued solutions if \Delta < 0. In this case, every real valued solution corresponds to a time at which the man is at the same position as the door of the bus. ANSWER: Correct Adding and Subtracting Vectors Conceptual Question Six vectors (A to F) have the magnitudes and directions indicated in the figure. Part A No; there is no chance he is going to get aboard. Yes; he will get a second chance Typesetting math: 15% Which two vectors, when added, will have the largest (positive) x component? Hint 1. Largest x component The two vectors with the largest x components will, when combined, give the resultant with the largest x component. Keep in mind that positive x components are larger than negative x components. ANSWER: Correct Part B Which two vectors, when added, will have the largest (positive) y component? Hint 1. Largest y component The two vectors with the largest y components will, when combined, give the resultant with the largest y component. Keep in mind that positive y components are larger than negative y components. ANSWER: C and E E and F A and F C and D B and D Typesetting math: 15% Correct Part C Which two vectors, when subtracted (i.e., when one vector is subtracted from the other), will have the largest magnitude? Hint 1. Subtracting vectors To subtract two vectors, add a vector with the same magnitude but opposite direction of one of the vectors to the other vector. ANSWER: Correct Tactics Box 3.1 Determining the Components of a Vector Learning Goal: C and D A and F E and F A and B E and D A and F A and E D and B C and D E and F Typesetting math: 15% To practice Tactics Box 3.1 Determining the Components of a Vector. When a vector \texttip{\vec{A}}{A_vec} is decomposed into component vectors \texttip{\vec{A}_{\mit x}}{A_vec_x} and \texttip{\vec{A}_{\mit y}}{A_vec_y} parallel to the coordinate axes, we can describe each component vector with a single number (a scalar) called the component. This tactics box describes how to determine the x component and y component of vector \texttip{\vec{A}}{A_vec}, denoted \texttip{A_{\mit x}}{A_x} and \texttip{A_{\mit y}}{A_y}. TACTICS BOX 3.1 Determining the components of a vector The absolute value |A_x| of the x component \texttip{A_{\mit x}}{A_x} is the magnitude of the component vector \texttip{\vec{A}_{\1. mit x}}{A_vec_x}. The sign of \texttip{A_{\mit x}}{A_x} is positive if \texttip{\vec{A}_{\mit x}}{A_vec_x} points in the positive x direction; it is negative if \texttip{\vec{A}_{\mit x}}{A_vec_x} points in the negative x direction. 2. 3. The y component \texttip{A_{\mit y}}{A_y} is determined similarly. Part A What is the magnitude of the component vector \texttip{\vec{A}_{\mit x}}{A_vec_x} shown in the figure? Express your answer in meters to one significant figure. ANSWER: Correct |A_x| = 5 \rm m Typesetting math: 15% Part B What is the sign of the y component \texttip{A_{\mit y}}{A_y} of vector \texttip{\vec{A}}{A_vec} shown in the figure? ANSWER: Correct Part C Now, combine the information given in the tactics box above to find the x and y components, \texttip{B_{\mit x}}{B_x} and \texttip{B_{\mit y}}{B_y}, of vector \texttip{\vec{B}}{B_vec} shown in the figure. Express your answers, separated by a comma, in meters to one significant figure. positive negative Typesetting math: 15% ANSWER: Correct Conceptual Problem about Projectile Motion Learning Goal: To understand projectile motion by considering horizontal constant velocity motion and vertical constant acceleration motion independently. Projectile motion refers to the motion of unpowered objects (called projectiles) such as balls or stones moving near the surface of the earth under the influence of the earth's gravity alone. In this analysis we assume that air resistance can be neglected. An object undergoing projectile motion near the surface of the earth obeys the following rules: An object undergoing projectile motion travels horizontally at a constant rate. That is, the x component of its velocity, \texttip{v_{\mit x}}{1. v_x}, is constant. An object undergoing projectile motion moves vertically with a constant downward acceleration whose magnitude, denoted by \texttip{g}{g}, is equal to 9.80 \rm{m/s^2} near the surface of the earth. Hence, the y component of its velocity, \texttip{v_{\mit y}}{v_y}, changes continuously. 2. An object undergoing projectile motion will undergo the horizontal and vertical motions described above from the instant it is launched until the instant it strikes the ground again. Even though the horizontal and vertical motions can be treated independently, they are related by the fact that they occur for exactly the same amount of time, namely the time \texttip{t}{t} the projectile is in the air. 3. The figure shows the trajectory (i.e., the path) of a ball undergoing projectile motion over level ground. The time t_0 = 0\;\rm{s} corresponds to the moment just after the ball is launched from position x_0 = 0\;\rm{m} and y_0 = 0\;\rm{m}. Its launch velocity, also called the initial velocity, is \texttip{\vec{v}_{\rm 0}}{v_vec_0}. Two other points along the trajectory are indicated in the figure. One is the moment the ball reaches the peak of its trajectory, at time \texttip{t_{\rm 1}}{t_1} with velocity \texttip{\vec{v}_{\rm 1}}{v_1_vec}. Its position at this moment is denoted by (x_1, y_1) or (x_1, y_{\max}) since it is at its maximum \texttip{B_{\mit x}}{B_x}, \texttip{B_{\mit y}}{B_y} = -2,-5 \rm m, \rm m Typesetting math: 15% The other point, at time \texttip{t_{\rm 2}}{t_2} with velocity \texttip{\vec{v}_{\rm 2}}{v_2_vec}, corresponds to the moment just before the ball strikes the ground on the way back down. At this time its position is (x_2, y_2), also known as (x_{\max}, y_2) since it is at its maximum horizontal range. Projectile motion is symmetric about the peak, provided the object lands at the same vertical height from which is was launched, as is the case here. Hence y_2 = y_0 = 0\;\rm{m}. Part A How do the speeds \texttip{v_{\rm 0}}{v_0}, \texttip{v_{\rm 1}}{v_1}, and \texttip{v_{\rm 2}}{v_2} (at times \texttip{t_{\rm 0}}{t_0}, \texttip{t_{\rm 1}}{t_1}, and \texttip{t_{\rm 2}}{t_2}) compare? ANSWER: Correct Here \texttip{v_{\rm 0}}{v_0} equals \texttip{v_{\rm 2}}{v_2} by symmetry and both exceed \texttip{v_{\rm 1}}{v_1}. This is because \texttip{v_{\rm 0}}{v_0} and \texttip{v_{\rm 2}}{v_2} include vertical speed as well as the constant horizontal speed. Consider a diagram of the ball at time \texttip{t_{\rm 0}}{t_0}. Recall that \texttip{t_{\rm 0}}{t_0} refers to the instant just after the ball has been launched, so it is still at ground level (x_0 = y_0= 0\;\rm{m}). However, it is already moving with initial velocity \texttip{\vec{v}_{\rm 0}}{v_0_vec}, whose magnitude is v_0 = 30.0\;{\rm m/s} and direction is \theta = 60.0\;{\rm degrees} counterclockwise from the positive x direction. \texttip{v_{\rm 0}}{v_0} = \texttip{v_{\rm 1}}{v_1} = \texttip{v_{\rm 2}}{v_2} > 0 \texttip{v_{\rm 0}}{v_0} = \texttip{v_{\rm 2}}{v_2} > \texttip{v_{\rm 1}}{v_1} = 0 \texttip{v_{\rm 0}}{v_0} = \texttip{v_{\rm 2}}{v_2} > \texttip{v_{\rm 1}}{v_1} > 0 \texttip{v_{\rm 0}}{v_0} > \texttip{v_{\rm 1}}{v_1} > \texttip{v_{\rm 2}}{v_2} > 0 \texttip{v_{\rm 0}}{v_0} > \texttip{v_{\rm 2}}{v_2} > \texttip{v_{\rm 1}}{v_1} = 0 Typesetting math: 15% Part B What are the values of the intial velocity vector components \texttip{v_{0,x}}{v_0, x} and \texttip{v_{0,y}}{v_0, y} (both in \rm{m/s}) as well as the acceleration vector components \texttip{a_{0,x}}{a_0, x} and \texttip{a_{0,y}}{a_0, y} (both in \rm{m/s^2})? Here the subscript 0 means “at time \texttip{t_{\rm 0}}{t_0}.” Hint 1. Determining components of a vector that is aligned with an axis If a vector points along a single axis direction, such as in the positive x direction, its x component will be its full magnitude, whereas its y component will be zero since the vector is perpendicular to the y direction. If the vector points in the negative x direction, its x component will be the negative of its full magnitude. Hint 2. Calculating the components of the initial velocity Notice that the vector \texttip{\vec{v}_{\rm 0}}{v_0_vec} points up and to the right. Since “up” is the positive y axis direction and “to the right” is the positive x axis direction, \texttip{v_{0,x}}{v_0, x} and \texttip{v_{0,y}}{v_0, y} will both be positive. As shown in the figure, \texttip{v_{0,x}}{v_0, x}, \texttip{v_{0,y}}{v_0, y}, and \texttip{v_{\rm 0}}{v_0} are three sides of a right triangle, one angle of which is \texttip{\theta }{theta}. Thus \texttip{v_{0,x}}{v_0, x} and \texttip{v_{0,y}}{v_0, y} can be found using the definition of the sine and cosine functions given below. Recall that v_0 = 30.0\;\rm{m/s} and \theta = 60.0\;\rm{degrees} and note that \large{\sin(\theta) = \frac{\rm{length\;of\;opposite\;side}}{\rm{length\;of\;hypotenuse}}} \large{= \frac{v_{0, y}}{v_0}}, \large{\cos(\theta) = \frac{\rm{length\;of\;adjacent\;side}}{\rm{length\;of\;hypotenuse}}} \large{= \frac{v_{0, x}}{v_0}.} What are the values of \texttip{v_{0,x}}{v_0, x} and \texttip{v_{0,y}}{v_0, y}? Enter your answers numerically in meters per second separated by a comma. ANSWER: ANSWER: 15.0,26.0 \rm{m/s} Typesetting math: 15% Correct Also notice that at time \texttip{t_{\rm 2}}{t_2}, just before the ball lands, its velocity components are v_{2, x} = 15\;\rm{m/s} (the same as always) and v_{2, y} = – 26.0\;\rm{m/s} (the same size but opposite sign from \texttip{v_{0,y}}{v_0, y} by symmetry). The acceleration at time \texttip{t_{\rm 2}}{t_2} will have components (0, -9.80 \rm{m/s^2}), exactly the same as at \texttip{t_{\rm 0}}{t_0}, as required by Rule 2. The peak of the trajectory occurs at time \texttip{t_{\rm 1}}{t_1}. This is the point where the ball reaches its maximum height \texttip{y_{\rm max}}{y_max}. At the peak the ball switches from moving up to moving down, even as it continues to travel horizontally at a constant rate. Part C What are the values of the velocity vector components \texttip{v_{1,x}}{v_1, x} and \texttip{v_{1,y}}{v_1, y} (both in \rm{m/s}) as well as the acceleration vector components \texttip{a_{1,x}}{a_1, x} and \texttip{a_{1,y}}{a_1, y} (both in \rm{m/s^2})? Here the subscript 1 means that these are all at time \texttip{t_{\rm 1}}{t_1}. ANSWER: 30.0, 0, 0, 0 0, 30.0, 0, 0 15.0, 26.0, 0, 0 30.0, 0, 0, -9.80 0, 30.0, 0, -9.80 15.0, 26.0, 0, -9.80 15.0, 26.0, 0, +9.80 Typesetting math: 15% Correct At the peak of its trajectory the ball continues traveling horizontally at a constant rate. However, at this moment it stops moving up and is about to move back down. This constitutes a downward-directed change in velocity, so the ball is accelerating downward even at the peak. The flight time refers to the total amount of time the ball is in the air, from just after it is launched (\texttip{t_{\rm 0}}{t_0}) until just before it lands (\texttip{t_{\rm 2}}{t_2}). Hence the flight time can be calculated as t_2 – t_0, or just \texttip{t_{\rm 2}}{t_2} in this particular situation since t_0 = 0. Because the ball lands at the same height from which it was launched, by symmetry it spends half its flight time traveling up to the peak and the other half traveling back down. The flight time is determined by the initial vertical component of the velocity and by the acceleration. The flight time does not depend on whether the object is moving horizontally while it is in the air. Part D If a second ball were dropped from rest from height \texttip{y_{\rm max}}{y_max}, how long would it take to reach the ground? Ignore air resistance. Check all that apply. Hint 1. Kicking a ball of cliff; a related problem Consider two balls, one of which is dropped from rest off the edge of a cliff at the same moment that the other is kicked horizontally off the edge of the cliff. Which ball reaches the level ground at the base of the cliff first? Ignore air resistance. Hint 1. Comparing position, velocity, and acceleration of the two balls Both balls start at the same height and have the same initial y velocity (v_{0,y} = 0) as well as the same acceleration (\vec a = g downward). They differ only in their x velocity (one is 0, 0, 0, 0 0, 0, 0, -9.80 15.0, 0, 0, 0 15.0, 0, 0, -9.80 0, 26.0, 0, 0 0, 26.0, 0, -9.80 15.0, 26.0, 0, 0 15.0, 26.0, 0, -9.80 Typesetting math: 15% zero, the other nonzero). This difference will affect their x motion but not their y motion. ANSWER: ANSWER: Correct In projectile motion over level ground, it takes an object just as long to rise from the ground to the peak as it takes for it to fall from the peak back to the ground. The range \texttip{R}{R} of the ball refers to how far it moves horizontally, from just after it is launched until just before it lands. Range is defined as x_2 – x_0, or just \texttip{x_{\rm 2}}{x_2} in this particular situation since x_0 = 0. Range can be calculated as the product of the flight time \texttip{t_{\rm 2}}{t_2} and the x component of the velocity \texttip{v_{\mit x}}{v_x} (which is the same at all times, so v_x = v_{0,x}). The value of \texttip{v_{\mit x}}{v_x} can be found from the launch speed \texttip{v_{\rm 0}}{v_0} and the launch angle \texttip{\theta }{theta} using trigonometric functions, as was done in Part B. The flight time is related to the initial y component of the velocity, which may also be found from \texttip{v_{\rm 0}}{v_0} and \texttip{\theta }{theta} using trig functions. The following equations may be useful in solving projectile motion problems, but these equations apply only to a projectile launched over level ground from position (x_0 = y_0 = 0) at time t_0 = 0 with initial speed \texttip{v_{\rm 0}}{v_0} and launch angle \texttip{\theta }{theta} measured from the horizontal. As was the case above, \texttip{t_{\rm 2}}{t_2} refers to the flight time and \texttip{R}{R} refers to the range of the projectile. flight time: \large{t_2 = \frac{2 v_{0, y}}{g} = \frac{2 v_0 \sin(\theta)}{g}} range: \large{R = v_x t_2 = \frac{v_0^2 \sin(2\theta)}{g}} The ball that falls straight down strikes the ground first. The ball that was kicked so it moves horizontally as it falls strikes the ground first. Both balls strike the ground at the same time. \texttip{t_{\rm 0}}{t_0} t_1 – t_0 \texttip{t_{\rm 2}}{t_2} t_2 – t_1 \large{\frac{t_2 – t_0}{2}} Typesetting math: 15% In general, a high launch angle yields a long flight time but a small horizontal speed and hence little range. A low launch angle gives a larger horizontal speed, but less flight time in which to accumulate range. The launch angle that achieves the maximum range for projectile motion over level ground is 45 degrees. Part E Which of the following changes would increase the range of the ball shown in the original figure? Check all that apply. ANSWER: Correct A solid understanding of the concepts of projectile motion will take you far, including giving you additional insight into the solution of projectile motion problems numerically. Even when the object does not land at the same height from which is was launched, the rules given in the introduction will still be useful. Recall that air resistance is assumed to be negligible here, so this projectile motion analysis may not be the best choice for describing things like frisbees or feathers, whose motion is strongly influenced by air. The value of the gravitational free-fall acceleration \texttip{g}{g} is also assumed to be constant, which may not be appropriate for objects that move vertically through distances of hundreds of kilometers, like rockets or missiles. However, for problems that involve relatively dense projectiles moving close to the surface of the earth, these assumptions are reasonable. A World-Class Sprinter World-class sprinters can accelerate out of the starting blocks with an acceleration that is nearly horizontal and has magnitude 15 \;{\rm m}/{\rm s}^{2}. Part A How much horizontal force \texttip{F}{F} must a sprinter of mass 64{\rm kg} exert on the starting blocks to produce this acceleration? Express your answer in newtons using two significant figures. Increase \texttip{v_{\rm 0}}{v_0} above 30 \rm{m/s}. Reduce \texttip{v_{\rm 0}}{v_0} below 30 \rm{m/s}. Reduce \texttip{\theta }{theta} from 60 \rm{degrees} to 45 \rm{degrees}. Reduce \texttip{\theta }{theta} from 60 \rm{degrees} to less than 30 \rm{degrees}. Increase \texttip{\theta }{theta} from 60 \rm{degrees} up toward 90 \rm{degrees}. Typesetting math: 15% Hint 1. Newton’s 2nd law of motion According to Newton’s 2nd law of motion, if a net external force \texttip{F_{\rm net}}{F_net} acts on a body, the body accelerates, and the net force is equal to the mass \texttip{m}{m} of the body times the acceleration \texttip{a}{a} of the body: F_{\rm net} = ma. ANSWER: Co

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The human body converts internal chemical energy into work and heat at rates of 60 to 125W (called the basal metabolic rate). This energy comes from food and is usually measured in kilocalories [1 kcal = 4.186 kJ]. (Note that the nutritional ‘Calorie’ listed on packaged food actually equals 1 kilocalorie). How many kilocalories of food energy does a person with a metoblic rate of 103.0 W require per day? Enter the numerical answer without units.

The human body converts internal chemical energy into work and heat at rates of 60 to 125W (called the basal metabolic rate). This energy comes from food and is usually measured in kilocalories [1 kcal = 4.186 kJ]. (Note that the nutritional ‘Calorie’ listed on packaged food actually equals 1 kilocalorie). How many kilocalories of food energy does a person with a metoblic rate of 103.0 W require per day? Enter the numerical answer without units.

The human body converts internal chemical energy into work and … Read More...
Chapter 13 Practice Problems (Practice – no credit) Due: 11:59pm on Friday, May 16, 2014 You will receive no credit for items you complete after the assignment is due. Grading Policy A Matter of Some Gravity Learning Goal: To understand Newton’s law of gravitation and the distinction between inertial and gravitational masses. In this problem, you will practice using Newton’s law of gravitation. According to that law, the magnitude of the gravitational force between two small particles of masses and , separated by a distance , is given by , where is the universal gravitational constant, whose numerical value (in SI units) is . This formula applies not only to small particles, but also to spherical objects. In fact, the gravitational force between two uniform spheres is the same as if we concentrated all the mass of each sphere at its center. Thus, by modeling the Earth and the Moon as uniform spheres, you can use the particle approximation when calculating the force of gravity between them. Be careful in using Newton’s law to choose the correct value for . To calculate the force of gravitational attraction between two uniform spheres, the distance in the equation for Newton’s law of gravitation is the distance between the centers of the spheres. For instance, if a small object such as an elephant is located on the surface of the Earth, the radius of the Earth would be used in the equation. Note that the force of gravity acting on an object located near the surface of a planet is often called weight. Also note that in situations involving satellites, you are often given the altitude of the satellite, that is, the distance from the satellite to the surface of the planet; this is not the distance to be used in the formula for the law of gravitation. There is a potentially confusing issue involving mass. Mass is defined as a measure of an object’s inertia, that is, its ability to resist acceleration. Newton’s second law demonstrates the relationship between mass, acceleration, and the net force acting on an object: . We can now refer to this measure of inertia more precisely as the inertial mass. On the other hand, the masses of the particles that appear in the expression for the law of gravity seem to have nothing to do with inertia: Rather, they serve as a measure of the strength of gravitational interactions. It would be reasonable to call such a property gravitational mass. Does this mean that every object has two different masses? Generally speaking, yes. However, the good news is that according to the latest, highly precise, measurements, the inertial and the gravitational mass of an object are, in fact, equal to each other; it is an established consensus among physicists that there is only one mass after all, which is a measure of both the object’s inertia and its ability to engage in gravitational interactions. Note that this consensus, like everything else in science, is open to possible amendments in the future. In this problem, you will answer several questions that require the use of Newton’s law of gravitation. Part A Two particles are separated by a certain distance. The force of gravitational interaction between them is . Now the separation between the particles is tripled. Find the new force of gravitational Fg m1 m2 r Fg = G m1m2 r2 G 6.67 × 10−11 N m2 kg2 r r rEarth F  = m net a F0 interaction . Express your answer in terms of . ANSWER: Part B A satellite revolves around a planet at an altitude equal to the radius of the planet. The force of gravitational interaction between the satellite and the planet is . Then the satellite moves to a different orbit, so that its altitude is tripled. Find the new force of gravitational interaction . Express your answer in terms of . You did not open hints for this part. ANSWER: Part C A satellite revolves around a planet at an altitude equal to the radius of the planet. The force of gravitational interaction between the satellite and the planet is . Then the satellite is brought back to the surface of the planet. Find the new force of gravitational interaction . Express your answer in terms of . ANSWER: F1 F0 F1 = F0 F2 F0 F2 = F0 F4 F0 Typesetting math: 81% Part D Two satellites revolve around the Earth. Satellite A has mass and has an orbit of radius . Satellite B has mass and an orbit of unknown radius . The forces of gravitational attraction between each satellite and the Earth is the same. Find . Express your answer in terms of . ANSWER: Part E An adult elephant has a mass of about 5.0 tons. An adult elephant shrew has a mass of about 50 grams. How far from the center of the Earth should an elephant be placed so that its weight equals that of the elephant shrew on the surface of the Earth? The radius of the Earth is 6400 . ( .) Express your answer in kilometers. ANSWER: The table below gives the masses of the Earth, the Moon, and the Sun. Name Mass (kg) Earth Moon Sun F4 = m r 6m rb rb r rb = r km 1 ton = 103 kg r = km 5.97 × 1024 7.35 × 1022 1.99 × 1030 Typesetting math: 81% The average distance between the Earth and the Moon is . The average distance between the Earth and the Sun is . Use this information to answer the following questions. Part F Find the net gravitational force acting on the Earth in the Sun-Earth-Moon system during the new moon (when the moon is located directly between the Earth and the Sun). Express your answer in newtons to three significant figures. You did not open hints for this part. ANSWER: Part G Find the net gravitational force acting on the Earth in the Sun-Earth-Moon system during the full moon (when the Earth is located directly between the moon and the sun). Express your answer in newtons to three significant figures. ANSWER: ± Understanding Newton’s Law of Universal Gravitation Learning Goal: To understand Newton’s law of universal gravitation and be able to apply it in two-object situations and (collinear) three-object situations; to distinguish between the use of and . 3.84 × 108 m 1.50 × 1011 m Fnet Fnet = N Fnet Fnet = N Typesetting math: 81% G g In the late 1600s, Isaac Newton proposed a rule to quantify the attractive force known as gravity between objects that have mass, such as those shown in the figure. Newton’s law of universal gravitation describes the magnitude of the attractive gravitational force between two objects with masses and as , where is the distance between the centers of the two objects and is the gravitational constant. The gravitational force is attractive, so in the figure it pulls to the right on (toward ) and toward the left on (toward ). The gravitational force acting on is equal in size to, but exactly opposite in direction from, the gravitational force acting on , as required by Newton’s third law. The magnitude of both forces is calculated with the equation given above. The gravitational constant has the value and should not be confused with the magnitude of the gravitational free-fall acceleration constant, denoted by , which equals 9.80 near the surface of the earth. The size of in SI units is tiny. This means that gravitational forces are sizeable only in the vicinity of very massive objects, such as the earth. You are in fact gravitationally attracted toward all the objects around you, such as the computer you are using, but the size of that force is too small to be noticed without extremely sensitive equipment. Consider the earth following its nearly circular orbit (dashed curve) about the sun. The earth has mass and the sun has mass . They are separated, center to center, by . Part A What is the size of the gravitational force acting on the earth due to the sun? Express your answer in newtons. F  g m1 m2 Fg = G( ) m1m2 r2 r G m1 m2 m2 m1 m1 m2 G G = 6.67 × 10−11 N m2/kg2 g m/s2 G mearth = 5.98 × 1024 kg msun = 1.99 × 1030 kg r = 93 million miles = 150 million km Typesetting math: 81% You did not open hints for this part. ANSWER: Part B This question will be shown after you complete previous question(s). Part C This question will be shown after you complete previous question(s). Part D This question will be shown after you complete previous question(s). Part E This question will be shown after you complete previous question(s). Part F N Typesetting math: 81% This question will be shown after you complete previous question(s). Understanding Mass and Weight Learning Goal: To understand the distinction between mass and weight and to be able to calculate the weight of an object from its mass and Newton’s law of gravitation. The concepts of mass and weight are often confused. In fact, in everyday conversations, the word “weight” often replaces “mass,” as in “My weight is seventy-five kilograms” or “I need to lose some weight.” Of course, mass and weight are related; however, they are also very different. Mass, as you recall, is a measure of an object’s inertia (ability to resist acceleration). Newton’s 2nd law demonstrates the relationship among an object’s mass, its acceleration, and the net force acting on it: . Mass is an intrinsic property of an object and is independent of the object’s location. Weight, in contrast, is defined as the force due to gravity acting on the object. That force depends on the strength of the gravitational field of the planet: , where is the weight of an object, is the mass of that object, and is the local acceleration due to gravity (in other words, the strength of the gravitational field at the location of the object). Weight, unlike mass, is not an intrinsic property of the object; it is determined by both the object and its location. Part A Which of the following quantities represent mass? Check all that apply. ANSWER: Fnet = ma w = mg w m g 12.0 lbs 0.34 g 120 kg 1600 kN 0.34 m 411 cm 899 MN Typesetting math: 81% Part B This question will be shown after you complete previous question(s). Using the universal law of gravity, we can find the weight of an object feeling the gravitational pull of a nearby planet. We can write an expression , where is the weight of the object, is the gravitational constant, is the mass of that object, is mass of the planet, and is the distance from the center of the planet to the object. If the object is on the surface of the planet, is simply the radius of the planet. Part C The gravitational field on the surface of the earth is stronger than that on the surface of the moon. If a rock is transported from the moon to the earth, which properties of the rock change? ANSWER: Part D This question will be shown after you complete previous question(s). Part E If acceleration due to gravity on the earth is , which formula gives the acceleration due to gravity on Loput? You did not open hints for this part. ANSWER: w = GmM/r2 w G m M r r mass only weight only both mass and weight neither mass nor weight g Typesetting math: 81% Part F This question will be shown after you complete previous question(s). Part G This question will be shown after you complete previous question(s). Part H This question will be shown after you complete previous question(s). ± Weight on a Neutron Star Neutron stars, such as the one at the center of the Crab Nebula, have about the same mass as our sun but a much smaller diameter. g 1.7 5.6 g 1.72 5.6 g 1.72 5.62 g 5.6 1.7 g 5.62 1.72 g 5.6 1.72 Typesetting math: 81% Part A If you weigh 655 on the earth, what would be your weight on the surface of a neutron star that has the same mass as our sun and a diameter of 19.0 ? Take the mass of the sun to be = 1.99×1030 , the gravitational constant to be = 6.67×10−11 , and the acceleration due to gravity at the earth’s surface to be = 9.810 . Express your weight in newtons. You did not open hints for this part. ANSWER: ± Escape Velocity Learning Goal: To introduce you to the concept of escape velocity for a rocket. The escape velocity is defined to be the minimum speed with which an object of mass must move to escape from the gravitational attraction of a much larger body, such as a planet of total mass . The escape velocity is a function of the distance of the object from the center of the planet , but unless otherwise specified this distance is taken to be the radius of the planet because it addresses the question “How fast does my rocket have to go to escape from the surface of the planet?” Part A The key to making a concise mathematical definition of escape velocity is to consider the energy. If an object is launched at its escape velocity, what is the total mechanical energy of the object at a very large (i.e., infinite) distance from the planet? Follow the usual convention and take the gravitational potential energy to be zero at very large distances. You did not open hints for this part. ANSWER: N km ms kg G N m2/kg2 g m/s2 wstar wstar = N m M R Etotal Typesetting math: 81% Consider the motion of an object between a point close to the planet and a point very very far from the planet. Indicate whether the following statements are true or false. Part B Angular momentum about the center of the planet is conserved. ANSWER: Part C Total mechanical energy is conserved. ANSWER: Part D Kinetic energy is conserved. ANSWER: Etotal = true false true false Typesetting math: 81% Part E This question will be shown after you complete previous question(s). Part F This question will be shown after you complete previous question(s). A Satellite in a Circular Orbit Consider a satellite of mass that orbits a planet of mass in a circle a distance from the center of the planet. The satellite’s mass is negligible compared with that of the planet. Indicate whether each of the statements in this problem is true or false. Part A The information given is sufficient to uniquely specify the speed, potential energy, and angular momentum of the satellite. You did not open hints for this part. ANSWER: true false m1 m2 r true false Typesetting math: 81% Part B The total mechanical energy of the satellite is conserved. You did not open hints for this part. ANSWER: Part C The linear momentum vector of the satellite is conserved. You did not open hints for this part. ANSWER: Part D The angular momentum of the satellite about the center of the planet is conserved. You did not open hints for this part. ANSWER: true false true false Typesetting math: 81% Part E The equations that express the conservation laws of total mechanical energy and linear momentum are sufficient to solve for the speed necessary to maintain a circular orbit at without using . You did not open hints for this part. ANSWER: At the Galaxy’s Core Astronomers have observed a small, massive object at the center of our Milky Way galaxy. A ring of material orbits this massive object; the ring has a diameter of about 15 light years and an orbital speed of about 200 . Part A Determine the mass of the massive object at the center of the Milky Way galaxy. Take the distance of one light year to be . Express your answer in kilograms. You did not open hints for this part. true false R F = ma true false km/s M 9.461 × 1015 m Typesetting math: 81% ANSWER: Part B This question will be shown after you complete previous question(s). Part C This question will be shown after you complete previous question(s). Part D This question will be shown after you complete previous question(s). Part E This question will be shown after you complete previous question(s). Properties of Circular Orbits Learning Goal: To teach you how to find the parameters characterizing an object in a circular orbit around a much heavier body like the earth. M = kg Typesetting math: 81% The motivation for Isaac Newton to discover his laws of motion was to explain the properties of planetary orbits that were observed by Tycho Brahe and analyzed by Johannes Kepler. A good starting point for understanding this (as well as the speed of the space shuttle and the height of geostationary satellites) is the simplest orbit–a circular one. This problem concerns the properties of circular orbits for a satellite orbiting a planet of mass . For all parts of this problem, where appropriate, use for the universal gravitational constant. Part A Find the orbital speed for a satellite in a circular orbit of radius . Express the orbital speed in terms of , , and . You did not open hints for this part. ANSWER: Part B Find the kinetic energy of a satellite with mass in a circular orbit with radius . Express your answer in terms of \texttip{m}{m}, \texttip{M}{M}, \texttip{G}{G}, and \texttip{R}{R}. ANSWER: Part C M G v R G M R v = K m R \texttip{K}{K} = Typesetting math: 81% This question will be shown after you complete previous question(s). Part D Find the orbital period \texttip{T}{T}. Express your answer in terms of \texttip{G}{G}, \texttip{M}{M}, \texttip{R}{R}, and \texttip{\pi }{pi}. You did not open hints for this part. ANSWER: Part E This question will be shown after you complete previous question(s). Part F Find \texttip{L}{L}, the magnitude of the angular momentum of the satellite with respect to the center of the planet. Express your answer in terms of \texttip{m}{m}, \texttip{M}{M}, \texttip{G}{G}, and \texttip{R}{R}. You did not open hints for this part. ANSWER: \texttip{T}{T} = Typesetting math: 81% Part G The quantities \texttip{v}{v}, \texttip{K}{K}, \texttip{U}{U}, and \texttip{L}{L} all represent physical quantities characterizing the orbit that depend on radius \texttip{R}{R}. Indicate the exponent (power) of the radial dependence of the absolute value of each. Express your answer as a comma-separated list of exponents corresponding to \texttip{v}{v}, \texttip{K}{K}, \texttip{U}{U}, and \texttip{L}{L}, in that order. For example, -1,-1/2,-0.5,-3/2 would mean v \propto R^{-1}, K \propto R^{-1/2}, and so forth. You did not open hints for this part. ANSWER: Score Summary: Your score on this assignment is 0%. You received 0 out of a possible total of 0 points. \texttip{L}{L} = Typesetting math: 81%

Chapter 13 Practice Problems (Practice – no credit) Due: 11:59pm on Friday, May 16, 2014 You will receive no credit for items you complete after the assignment is due. Grading Policy A Matter of Some Gravity Learning Goal: To understand Newton’s law of gravitation and the distinction between inertial and gravitational masses. In this problem, you will practice using Newton’s law of gravitation. According to that law, the magnitude of the gravitational force between two small particles of masses and , separated by a distance , is given by , where is the universal gravitational constant, whose numerical value (in SI units) is . This formula applies not only to small particles, but also to spherical objects. In fact, the gravitational force between two uniform spheres is the same as if we concentrated all the mass of each sphere at its center. Thus, by modeling the Earth and the Moon as uniform spheres, you can use the particle approximation when calculating the force of gravity between them. Be careful in using Newton’s law to choose the correct value for . To calculate the force of gravitational attraction between two uniform spheres, the distance in the equation for Newton’s law of gravitation is the distance between the centers of the spheres. For instance, if a small object such as an elephant is located on the surface of the Earth, the radius of the Earth would be used in the equation. Note that the force of gravity acting on an object located near the surface of a planet is often called weight. Also note that in situations involving satellites, you are often given the altitude of the satellite, that is, the distance from the satellite to the surface of the planet; this is not the distance to be used in the formula for the law of gravitation. There is a potentially confusing issue involving mass. Mass is defined as a measure of an object’s inertia, that is, its ability to resist acceleration. Newton’s second law demonstrates the relationship between mass, acceleration, and the net force acting on an object: . We can now refer to this measure of inertia more precisely as the inertial mass. On the other hand, the masses of the particles that appear in the expression for the law of gravity seem to have nothing to do with inertia: Rather, they serve as a measure of the strength of gravitational interactions. It would be reasonable to call such a property gravitational mass. Does this mean that every object has two different masses? Generally speaking, yes. However, the good news is that according to the latest, highly precise, measurements, the inertial and the gravitational mass of an object are, in fact, equal to each other; it is an established consensus among physicists that there is only one mass after all, which is a measure of both the object’s inertia and its ability to engage in gravitational interactions. Note that this consensus, like everything else in science, is open to possible amendments in the future. In this problem, you will answer several questions that require the use of Newton’s law of gravitation. Part A Two particles are separated by a certain distance. The force of gravitational interaction between them is . Now the separation between the particles is tripled. Find the new force of gravitational Fg m1 m2 r Fg = G m1m2 r2 G 6.67 × 10−11 N m2 kg2 r r rEarth F  = m net a F0 interaction . Express your answer in terms of . ANSWER: Part B A satellite revolves around a planet at an altitude equal to the radius of the planet. The force of gravitational interaction between the satellite and the planet is . Then the satellite moves to a different orbit, so that its altitude is tripled. Find the new force of gravitational interaction . Express your answer in terms of . You did not open hints for this part. ANSWER: Part C A satellite revolves around a planet at an altitude equal to the radius of the planet. The force of gravitational interaction between the satellite and the planet is . Then the satellite is brought back to the surface of the planet. Find the new force of gravitational interaction . Express your answer in terms of . ANSWER: F1 F0 F1 = F0 F2 F0 F2 = F0 F4 F0 Typesetting math: 81% Part D Two satellites revolve around the Earth. Satellite A has mass and has an orbit of radius . Satellite B has mass and an orbit of unknown radius . The forces of gravitational attraction between each satellite and the Earth is the same. Find . Express your answer in terms of . ANSWER: Part E An adult elephant has a mass of about 5.0 tons. An adult elephant shrew has a mass of about 50 grams. How far from the center of the Earth should an elephant be placed so that its weight equals that of the elephant shrew on the surface of the Earth? The radius of the Earth is 6400 . ( .) Express your answer in kilometers. ANSWER: The table below gives the masses of the Earth, the Moon, and the Sun. Name Mass (kg) Earth Moon Sun F4 = m r 6m rb rb r rb = r km 1 ton = 103 kg r = km 5.97 × 1024 7.35 × 1022 1.99 × 1030 Typesetting math: 81% The average distance between the Earth and the Moon is . The average distance between the Earth and the Sun is . Use this information to answer the following questions. Part F Find the net gravitational force acting on the Earth in the Sun-Earth-Moon system during the new moon (when the moon is located directly between the Earth and the Sun). Express your answer in newtons to three significant figures. You did not open hints for this part. ANSWER: Part G Find the net gravitational force acting on the Earth in the Sun-Earth-Moon system during the full moon (when the Earth is located directly between the moon and the sun). Express your answer in newtons to three significant figures. ANSWER: ± Understanding Newton’s Law of Universal Gravitation Learning Goal: To understand Newton’s law of universal gravitation and be able to apply it in two-object situations and (collinear) three-object situations; to distinguish between the use of and . 3.84 × 108 m 1.50 × 1011 m Fnet Fnet = N Fnet Fnet = N Typesetting math: 81% G g In the late 1600s, Isaac Newton proposed a rule to quantify the attractive force known as gravity between objects that have mass, such as those shown in the figure. Newton’s law of universal gravitation describes the magnitude of the attractive gravitational force between two objects with masses and as , where is the distance between the centers of the two objects and is the gravitational constant. The gravitational force is attractive, so in the figure it pulls to the right on (toward ) and toward the left on (toward ). The gravitational force acting on is equal in size to, but exactly opposite in direction from, the gravitational force acting on , as required by Newton’s third law. The magnitude of both forces is calculated with the equation given above. The gravitational constant has the value and should not be confused with the magnitude of the gravitational free-fall acceleration constant, denoted by , which equals 9.80 near the surface of the earth. The size of in SI units is tiny. This means that gravitational forces are sizeable only in the vicinity of very massive objects, such as the earth. You are in fact gravitationally attracted toward all the objects around you, such as the computer you are using, but the size of that force is too small to be noticed without extremely sensitive equipment. Consider the earth following its nearly circular orbit (dashed curve) about the sun. The earth has mass and the sun has mass . They are separated, center to center, by . Part A What is the size of the gravitational force acting on the earth due to the sun? Express your answer in newtons. F  g m1 m2 Fg = G( ) m1m2 r2 r G m1 m2 m2 m1 m1 m2 G G = 6.67 × 10−11 N m2/kg2 g m/s2 G mearth = 5.98 × 1024 kg msun = 1.99 × 1030 kg r = 93 million miles = 150 million km Typesetting math: 81% You did not open hints for this part. ANSWER: Part B This question will be shown after you complete previous question(s). Part C This question will be shown after you complete previous question(s). Part D This question will be shown after you complete previous question(s). Part E This question will be shown after you complete previous question(s). Part F N Typesetting math: 81% This question will be shown after you complete previous question(s). Understanding Mass and Weight Learning Goal: To understand the distinction between mass and weight and to be able to calculate the weight of an object from its mass and Newton’s law of gravitation. The concepts of mass and weight are often confused. In fact, in everyday conversations, the word “weight” often replaces “mass,” as in “My weight is seventy-five kilograms” or “I need to lose some weight.” Of course, mass and weight are related; however, they are also very different. Mass, as you recall, is a measure of an object’s inertia (ability to resist acceleration). Newton’s 2nd law demonstrates the relationship among an object’s mass, its acceleration, and the net force acting on it: . Mass is an intrinsic property of an object and is independent of the object’s location. Weight, in contrast, is defined as the force due to gravity acting on the object. That force depends on the strength of the gravitational field of the planet: , where is the weight of an object, is the mass of that object, and is the local acceleration due to gravity (in other words, the strength of the gravitational field at the location of the object). Weight, unlike mass, is not an intrinsic property of the object; it is determined by both the object and its location. Part A Which of the following quantities represent mass? Check all that apply. ANSWER: Fnet = ma w = mg w m g 12.0 lbs 0.34 g 120 kg 1600 kN 0.34 m 411 cm 899 MN Typesetting math: 81% Part B This question will be shown after you complete previous question(s). Using the universal law of gravity, we can find the weight of an object feeling the gravitational pull of a nearby planet. We can write an expression , where is the weight of the object, is the gravitational constant, is the mass of that object, is mass of the planet, and is the distance from the center of the planet to the object. If the object is on the surface of the planet, is simply the radius of the planet. Part C The gravitational field on the surface of the earth is stronger than that on the surface of the moon. If a rock is transported from the moon to the earth, which properties of the rock change? ANSWER: Part D This question will be shown after you complete previous question(s). Part E If acceleration due to gravity on the earth is , which formula gives the acceleration due to gravity on Loput? You did not open hints for this part. ANSWER: w = GmM/r2 w G m M r r mass only weight only both mass and weight neither mass nor weight g Typesetting math: 81% Part F This question will be shown after you complete previous question(s). Part G This question will be shown after you complete previous question(s). Part H This question will be shown after you complete previous question(s). ± Weight on a Neutron Star Neutron stars, such as the one at the center of the Crab Nebula, have about the same mass as our sun but a much smaller diameter. g 1.7 5.6 g 1.72 5.6 g 1.72 5.62 g 5.6 1.7 g 5.62 1.72 g 5.6 1.72 Typesetting math: 81% Part A If you weigh 655 on the earth, what would be your weight on the surface of a neutron star that has the same mass as our sun and a diameter of 19.0 ? Take the mass of the sun to be = 1.99×1030 , the gravitational constant to be = 6.67×10−11 , and the acceleration due to gravity at the earth’s surface to be = 9.810 . Express your weight in newtons. You did not open hints for this part. ANSWER: ± Escape Velocity Learning Goal: To introduce you to the concept of escape velocity for a rocket. The escape velocity is defined to be the minimum speed with which an object of mass must move to escape from the gravitational attraction of a much larger body, such as a planet of total mass . The escape velocity is a function of the distance of the object from the center of the planet , but unless otherwise specified this distance is taken to be the radius of the planet because it addresses the question “How fast does my rocket have to go to escape from the surface of the planet?” Part A The key to making a concise mathematical definition of escape velocity is to consider the energy. If an object is launched at its escape velocity, what is the total mechanical energy of the object at a very large (i.e., infinite) distance from the planet? Follow the usual convention and take the gravitational potential energy to be zero at very large distances. You did not open hints for this part. ANSWER: N km ms kg G N m2/kg2 g m/s2 wstar wstar = N m M R Etotal Typesetting math: 81% Consider the motion of an object between a point close to the planet and a point very very far from the planet. Indicate whether the following statements are true or false. Part B Angular momentum about the center of the planet is conserved. ANSWER: Part C Total mechanical energy is conserved. ANSWER: Part D Kinetic energy is conserved. ANSWER: Etotal = true false true false Typesetting math: 81% Part E This question will be shown after you complete previous question(s). Part F This question will be shown after you complete previous question(s). A Satellite in a Circular Orbit Consider a satellite of mass that orbits a planet of mass in a circle a distance from the center of the planet. The satellite’s mass is negligible compared with that of the planet. Indicate whether each of the statements in this problem is true or false. Part A The information given is sufficient to uniquely specify the speed, potential energy, and angular momentum of the satellite. You did not open hints for this part. ANSWER: true false m1 m2 r true false Typesetting math: 81% Part B The total mechanical energy of the satellite is conserved. You did not open hints for this part. ANSWER: Part C The linear momentum vector of the satellite is conserved. You did not open hints for this part. ANSWER: Part D The angular momentum of the satellite about the center of the planet is conserved. You did not open hints for this part. ANSWER: true false true false Typesetting math: 81% Part E The equations that express the conservation laws of total mechanical energy and linear momentum are sufficient to solve for the speed necessary to maintain a circular orbit at without using . You did not open hints for this part. ANSWER: At the Galaxy’s Core Astronomers have observed a small, massive object at the center of our Milky Way galaxy. A ring of material orbits this massive object; the ring has a diameter of about 15 light years and an orbital speed of about 200 . Part A Determine the mass of the massive object at the center of the Milky Way galaxy. Take the distance of one light year to be . Express your answer in kilograms. You did not open hints for this part. true false R F = ma true false km/s M 9.461 × 1015 m Typesetting math: 81% ANSWER: Part B This question will be shown after you complete previous question(s). Part C This question will be shown after you complete previous question(s). Part D This question will be shown after you complete previous question(s). Part E This question will be shown after you complete previous question(s). Properties of Circular Orbits Learning Goal: To teach you how to find the parameters characterizing an object in a circular orbit around a much heavier body like the earth. M = kg Typesetting math: 81% The motivation for Isaac Newton to discover his laws of motion was to explain the properties of planetary orbits that were observed by Tycho Brahe and analyzed by Johannes Kepler. A good starting point for understanding this (as well as the speed of the space shuttle and the height of geostationary satellites) is the simplest orbit–a circular one. This problem concerns the properties of circular orbits for a satellite orbiting a planet of mass . For all parts of this problem, where appropriate, use for the universal gravitational constant. Part A Find the orbital speed for a satellite in a circular orbit of radius . Express the orbital speed in terms of , , and . You did not open hints for this part. ANSWER: Part B Find the kinetic energy of a satellite with mass in a circular orbit with radius . Express your answer in terms of \texttip{m}{m}, \texttip{M}{M}, \texttip{G}{G}, and \texttip{R}{R}. ANSWER: Part C M G v R G M R v = K m R \texttip{K}{K} = Typesetting math: 81% This question will be shown after you complete previous question(s). Part D Find the orbital period \texttip{T}{T}. Express your answer in terms of \texttip{G}{G}, \texttip{M}{M}, \texttip{R}{R}, and \texttip{\pi }{pi}. You did not open hints for this part. ANSWER: Part E This question will be shown after you complete previous question(s). Part F Find \texttip{L}{L}, the magnitude of the angular momentum of the satellite with respect to the center of the planet. Express your answer in terms of \texttip{m}{m}, \texttip{M}{M}, \texttip{G}{G}, and \texttip{R}{R}. You did not open hints for this part. ANSWER: \texttip{T}{T} = Typesetting math: 81% Part G The quantities \texttip{v}{v}, \texttip{K}{K}, \texttip{U}{U}, and \texttip{L}{L} all represent physical quantities characterizing the orbit that depend on radius \texttip{R}{R}. Indicate the exponent (power) of the radial dependence of the absolute value of each. Express your answer as a comma-separated list of exponents corresponding to \texttip{v}{v}, \texttip{K}{K}, \texttip{U}{U}, and \texttip{L}{L}, in that order. For example, -1,-1/2,-0.5,-3/2 would mean v \propto R^{-1}, K \propto R^{-1/2}, and so forth. You did not open hints for this part. ANSWER: Score Summary: Your score on this assignment is 0%. You received 0 out of a possible total of 0 points. \texttip{L}{L} = Typesetting math: 81%

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Ball Bearings Inc. faces costs of production as follow: Q TFC TVC AFC AVC ATC MC 0 $100 $0 1 100 50 2 100 70 3 100 90 4 100 140 5 100 200 6 100 360 a) Complete the above Table by calculating the Company’s Average Fixed Costs (AFC), Average Variable Costs (AVC), Average Total Costs (ATC) and Marginal Cost (MC). Show all your calculations.b) The price of a case of ball bearings is $50. Seeing that she can’t make a profit, the chief executive officer (CEO) decides to shut down operations. What are the firm’s profits/losses? Was this a wise decision? Explain. c) Vaguely remembering his introductory Economics course, the chief financial officer tells the CEO, it is better to produce 1 case of ball bearings because marginal revenue equals marginal cost at that quantity. What are the firm’s profits/losses at that level of production? Was this the best decision? Explain.

Ball Bearings Inc. faces costs of production as follow: Q TFC TVC AFC AVC ATC MC 0 $100 $0 1 100 50 2 100 70 3 100 90 4 100 140 5 100 200 6 100 360 a) Complete the above Table by calculating the Company’s Average Fixed Costs (AFC), Average Variable Costs (AVC), Average Total Costs (ATC) and Marginal Cost (MC). Show all your calculations.b) The price of a case of ball bearings is $50. Seeing that she can’t make a profit, the chief executive officer (CEO) decides to shut down operations. What are the firm’s profits/losses? Was this a wise decision? Explain. c) Vaguely remembering his introductory Economics course, the chief financial officer tells the CEO, it is better to produce 1 case of ball bearings because marginal revenue equals marginal cost at that quantity. What are the firm’s profits/losses at that level of production? Was this the best decision? Explain.

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