## Morgan Extra Pages Graphing with Excel to be carried out in a computer lab, 3rd floor Calloway Hall or elsewhere The Excel spreadsheet consists of vertical columns and horizontal rows; a column and row intersect at a cell. A cell can contain data for use in calculations of all sorts. The Name Box shows the currently selected cell (Fig. 1). In the Excel 2007 and 2010 versions the drop-down menus familiar in most software screens have been replaced by tabs with horizontally-arranged command buttons of various categories (Fig. 2) ___________________________________________________________________ Open Excel, click on the Microsoft circle, upper left, and Save As your surname. xlsx on the desktop. Before leaving the lab e-mail the file to yourself and/or save to a flash drive. Also e-mail it to your instructor. Figure 1. Parts of an Excel spreadsheet. Name Box Figure 2. Tabs. Click to buy NOW! PDF-XChange Viewer www.docu-track.com Click to buy NOW! PDF-XChange Viewer www.docu-track.com EXERCISE 1: BASIC OPERATIONS Click Save often as you work. 1. Type the heading “Edge Length” in Cell A1 and double click the crack between the A and B column heading for automatic widening of column A. Similarly, write headings for columns B and C and enter numbers in Cells A2 and A3 as in Fig. 3. Highlight Cells A2 and A3 by dragging the cursor (chunky plus-shape) over the two of them and letting go. 2. Note that there are three types of cursor crosses: chunky for selecting, barbed for moving entries or blocks of entries from cell to cell, and tiny (appearing only at the little square in the lower-right corner of a cell). Obtain a tiny arrow for Cell A3 and perform a plus-drag down Column A until the cells are filled up to 40 (in Cell A8). Note that the two highlighted cells set both the starting value of the fill and the intervals. 3. Click on Cell B2 and enter a formula for face area of a cube as follows: type =, click on Cell A2, type ^2, and press Enter (note the formula bar in Fig. 4). 4. Enter the formula for cube volume in Cell C2 (same procedure, but “=, click on A2, ^3, Enter”). 5. Highlight Cells B2 and C2; plus-drag down to Row 8 (Fig. 5). Do the numbers look correct? Click on some cells in the newly filled area and notice how Excel steps the row designations as it moves down the column (it can do it for horizontal plusdrags along rows also). This is the major programming development that has led to the popularity of spreadsheets. Figure 3. Entries. Figure 4. A formula. Figure 5. Plus-dragging formulas. Click to buy NOW! PDF-XChange Viewer www.docu-track.com Click to buy NOW! PDF-XChange Viewer www.docu-track.com 6. Now let’s graph the Face Area versus Edge Length: select Cells A1 through B8, choose the Insert tab, and click the Scatter drop-down menu and select “Scatter with only Markers” (Fig. 6). 7. Move the graph (Excel calls it a “chart”) that appears up alongside your number table and dress it up as follows: a. Note that some Chart Layouts have appeared above. Click Layout 1 and alter each title to read Face Area for the vertical axis, Edge Length for the horizontal and Face Area vs. Edge Length for the Graph Title. b. Activate the Excel Least squares routine, called “fitting a trendline” in the program: right click any of the data markers and click Add Trendline. Choose Power and also check “Display equation on chart” and “Display R-squared value on chart.” Fig. 7 shows what the graph will look like at this point. c. The titles are explicit, so the legend is unnecessary. Click on it and press the delete button to remove it. Figure 6. Creating a scatter graph. Figure 7. A graph with a fitted curve. Click to buy NOW! PDF-XChange Viewer www.docu-track.com Click to buy NOW! PDF-XChange Viewer www.docu-track.com 8. Now let’s overlay the Volume vs. Edge Length curve onto the same graph (optional for 203L/205L): Make a copy of your graph by clicking on the outer white area, clicking ctrl-c (or right click, copy), and pasting the copy somewhere else (ctrl-v). If you wish, delete the trendline as in Fig. 8. a. Right click on the outer white space, choose Select Data and click the Add button. b. You can type in the cell ranges by hand in the dialog box that comes up, but it is easier to click the red, white, and blue button on the right of each space and highlight what you want to go in. Click the red, white, and blue of the bar that has appeared, and you will bounce back to the Add dialog box. Use the Edge Length column for the x’s and Volume for the y’s. c. Right-click on any volume data point and choose Format Data Series. Clicking Secondary Axis will place its scale on the right of the graph as in Fig. 8. d. Dress up your graph with two axis titles (Layout-Labels-Axis Titles), etc. Figure 8. Adding a second curve and y-axis to the graph Click to buy NOW! PDF-XChange Viewer www.docu-track.com Click to buy NOW! PDF-XChange Viewer www.docu-track.com EXERCISE 2: INTERPRETING A LINEAR GRAPH Introduction: Many experiments are repeated a number of times with one of the parameters involved varied from run to run. Often the goal is to measure the rate of change of a dependent variable, rather than a particular value. If the dependent variable can be expressed as a linear function of the independent parameter, then the slope and yintercept of an appropriate graph will give the rate of change and a particular value, respectively. An example of such an experiment in PHYS.203L/205L is the first part of Lab 20, in which weights are added to the bottom of a suspended spring (Figure 9). This experiment shows that a spring exerts a force Fs proportional to the distance stretched y = (y-yo), a relationship known as Hooke’s Law: Fs = – k(y – yo) (Eq. 1) where k is called the Hooke’s Law constant. The minus sign shows that the spring opposes any push or pull on it. In Lab 20 Fs is equal to (- Mg) and y is given by the reading on a meter stick. Masses were added to the bottom of the spring in 50-g increments giving weights in newtons of 0.49, 0.98, etc. The weight pan was used as the pointer for reading y and had a mass of 50 g, so yo could not be directly measured. For convenient graphing Equation 1 can be rewritten: -(Mg) = – ky + kyo Or (Mg) = ky – kyo (Eq. 1′) Procedure 1. On your spreadsheet note the tabs at the bottom left and double-click Sheet1. Type in “Basics,” and then click the Sheet2 tab to bring up a fresh worksheet. Change the sheet name to “Linear Fit” and fill in data as in this table. Hooke’s Law Experiment y (m) -Fs = Mg (N) 0.337 0.49 0.388 0.98 0.446 1.47 0.498 1.96 0.550 2.45 2. Highlight the cells with the numbers, and graph (Mg) versus y as in Steps 6 and 7 of the Basics section. Your Trendline this time will be Linear of course. If you are having trouble remembering what’s versus what, “y” looks like “v”, so what comes before the “v” of “versus” goes on the y (vertical) axis. Yes, this graph is confusing: the horizontal (“x”) axis is distance y, and the “y” axis is something else. 3. Click on the Equation/R2 box on the graph and highlight just the slope, that is, only the number that comes before the “x.” Copy it (control-c is a fast way to Figure 9. A spring with a weight stretching it Click to buy NOW! PDF-XChange Viewer www.docu-track.com Click to buy NOW! PDF-XChange Viewer www.docu-track.com do it) and paste it (control-v) into an empty cell. Do likewise for the intercept (including the minus sign). SAVE YOUR FILE! 5. The next steps use the standard procedure for obtaining information from linear data. Write the general equation for a straight line immediately below a hand-written copy of Equation 1′ then circle matching items: (Mg) = k y + (- k yo) (Eq. 1′) y = m x + b Note the parentheses around the intercept term of Equation 1′ to emphasize that the minus sign is part of it. Equating above and below, you can create two useful new equations: slope m = k (Eq. 2) y-intercept b = -kyo (Eq. 3) 6. Solve Equation 2 for k, that is, rewrite left to right. Then substitute the value for slope m from your graph, and you have an experimental value for the Hooke’s Law constant k. Next solve Equation 3 for yo, substitute the value for intercept b from your graph and the value of k that you just found, and calculate yo. 7. Examine your linear graph for clues to finding the units of the slope and the yintercept. Use these units to find the units of k and yo. 8. Present your values of k and yo with their units neatly at the bottom of your spreadsheet. 9. R2 in Excel, like r in our lab manual and Corr. in the LoggerPro software, is a measure of how well the calculated line matches the data points. 1.00 would indicate a perfect match. State how good a match you think was made in this case? 10. Do the Homework, Further Exercises on Interpreting Linear Graphs, on the following pages. Click to buy NOW! PDF-XChange Viewer www.docu-track.com Click to buy NOW! PDF-XChange Viewer www.docu-track.com Eq.1 M m f M a g , (Eq.2) M slope m g (Eq.3) M b f Morgan Extra Pages Homework: Graph Interpretation Exercises EXAMPLE WITH COMPLETE SOLUTION In PHYS.203L and 205L we do Lab 9 Newton’s Second Law on Atwood’s Machine using a photogate sensor (Fig. 1). The Atwood’s apparatus can slow the rate of fall enough to be measured even with primitive timing devices. In our experiment LoggerPro software automatically collects and analyzes the data giving reliable measurements of g, the acceleration of gravity. The equation governing motion for Atwood’s Machine can be written: where a is the acceleration of the masses and string, g is the acceleration of gravity, M is the total mass at both ends of the string, m is the difference between the masses, and f is the frictional force at the hub of the pulley wheel. In this exercise you are given a graph of a vs. m obtained in this experiment with the values of M and the slope and intercept (Fig. 2). The goal is to extract values for acceleration of gravity g and frictional force f from this information. To analyze the graph we write y = mx + b, the general equation for a straight line, directly under Equation 1 and match up the various parameters: Equating above and below, you can create two new equations: and y m x b M m f M a g Figure 1. The Atwood’s Machine setup (from the LoggerPro handout). Figure 2. Graph of acceleration versus mass difference; data from a Physics I experiment. Atwood’s Machine M = 0.400 kg a = 24.4 m – 0.018 R2 = 0.998 0.00 0.20 0.40 0.60 0.80 1.00 1.20 1.40 0.000 0.010 0.020 0.030 0.040 0.050 0.060 m (kg) a (m/s2) Click to buy NOW! PDF-XChange Viewer www.docu-track.com Click to buy NOW! PDF-XChange Viewer www.docu-track.com 2 2 9.76 / 0.400 24.4 /( ) m s kg m kg s g Mm To handle Equation 2 it pays to consider what the units of the slope are. A slope is “the rise over the run,“ so its units must be the units of the vertical axis divided by those of the horizontal axis. In this case: Now let’s solve Equation 2 for g and substitute the values of total mass M and of the slope m from the graph: Using 9.80 m/s2 as the Baltimore accepted value for g, we can calculate the percent error: A similar process with Equation 3 leads to a value for f, the frictional force at the hub of the pulley wheel. Note that the units of intercept b are simply whatever the vertical axis units are, m/s2 in this case. Solving Equation 3 for f: EXERCISE 1 The Picket Fence experiment makes use of LoggerPro software to calculate velocities at regular time intervals as the striped plate passes through the photogate (Fig. 3). The theoretical equation is v = vi + at (Eq. 4) where vi = 0 (the fence is dropped from rest) and a = g. a. Write Equation 4 with y = mx + b under it and circle matching factors as in the Example. b. What is the experimental value of the acceleration of gravity? What is its percent error from the accepted value for Baltimore, 9.80 m/s2? c. Does the value of the y-intercept make sense? d. How well did the straight Trendline match the data? 2 / 2 kg s m kg m s 0.4% 100 9.80 9.76 9.80 100 . . . % Acc Exp Acc Error kg m s mN kg m s f Mb 7.2 10 / 7.2 0.400 ( 0.018 / ) 3 2 2 Figure 3. Graph of speed versus time as calculated by LoggerPro as a picket fence falls freely through a photogate. Picket Fence Drop y = 9.8224x + 0.0007 R2 = 0.9997 0 2 4 6 8 10 12 0 0.2 0.4 0.6 0.8 1 1.2 t (s) v (m/s) Click to buy NOW! PDF-XChange Viewer www.docu-track.com Click to buy NOW! PDF-XChange Viewer www.docu-track.com EXERCISE 2 This is an electrical example from PHYS.204L/206L, potential difference, V, versus current, I (Fig. 4). The theoretical equation is V = IR (Eq. 5) and is known as “Ohm’s Law.” The unit symbols stand for volts, V, and Amperes, A. The factor R stands for resistance and is measured in units of ohms, symbol (capital omega). The definition of the ohm is: V (Eq. 6) By coincidence the letter symbols for potential (a quantity ) and volts (its unit) are identical. Thus “voltage” has become the laboratory slang name for potential. a. Rearrange the Ohm’s Law equation to match y = mx + b.. b. What is the experimental resistance? c. Comment on the experimental intercept: is its value reasonable? EXERCISE 3 This graph (Fig. 5) also follows Ohm’s Law, but solved for current I. For this graph the experimenter held potential difference V constant at 15.0V and measured the current for resistances of 100, 50, 40, and 30 Solve Ohm’s Law for I and you will see that 1/R is the logical variable to use on the x axis. For units, someone once jokingly referred to a “reciprocal ohm” as a “mho,” and the name stuck. a. Rearrange Equation 5 solved for I to match y = mx + b. b. What is the experimental potential difference? c. Calculate the percent difference from the 15.0 V that the experimenter set on the power supply (the instrument used for such experiments). d. Comment on the experimental intercept: is its value reasonable? Figure 4. Graph of potential difference versus current; data from a Physics II experiment. The theoretical equation, V = IR, is known as “Ohm’s Law.” Ohm’s Law y = 0.628x – 0.0275 R2 = 0.9933 0 0.1 0.2 0.3 0.4 0 0.1 0.2 0.3 0.4 0.5 0.6 Current, I (A) Potential difference, V (V) Figure 5. Another application of Ohm’s Law: a graph of current versus the inverse of resistance, from a different electric circuit experiment. Current versus (1/Resistance) y = 14.727x – 0.2214 R2 = 0.9938 0 100 200 300 400 500 600 5 10 15 20 25 30 35 R-1 (millimhos) I (milliamperes) Click to buy NOW! PDF-XChange Viewer www.docu-track.com Click to buy NOW! PDF-XChange Viewer www.docu-track.com EXERCISE 4 The Atwood’s Machine experiment (see the solved example above) can be done in another way: keep mass difference m the same and vary the total mass M (Fig. 6). a. Rewrite Equation 1 and factor out (1/M). b. Equate the coefficient of (1/M) with the experimental slope and solve for acceleration of gravity g. c. Substitute the values for slope, mass difference, and frictional force and calculate the experimental of g. d. Derive the units of the slope and show that the units of g come out as they should. e. Is the value of the experimental intercept reasonable? EXERCISE 5 In the previous two exercises the reciprocal of a variable was used to make the graph come out linear. In this one the trick will be to use the square root of a variable (Fig. 7). In PHYS.203L and 205L Lab 19 The Pendulum the theoretical equation is where the period T is the time per cycle, L is the length of the string, and g is the acceleration of gravity. a. Rewrite Equation 7 with the square root of L factored out and placed at the end. b. Equate the coefficient of √L with the experimental slope and solve for acceleration of gravity g. c. Substitute the value for slope and calculate the experimental of g. d. Derive the units of the slope and show that the units of g come out as they should. e. Is the value of the experimental intercept reasonable? 2 (Eq . 7) g T L Figure 6. Graph of acceleration versus the reciprocal of total mass; data from a another Physics I experiment. Atwood’s Machine m = 0.020 kg f = 7.2 mN y = 0.1964x – 0.0735 R2 = 0.995 0.400 0.600 0.800 1.000 2.000 2.500 3.000 3.500 4.000 4.500 5.000 1/M (1/kg) a (m/s2) Effect of Pendulum Length on Period y = 2.0523x – 0.0331 R2 = 0.999 0.400 0.800 1.200 1.600 2.000 2.400 0.00 0.10 0.20 0.30 0.40 0.50 0.60 0.70 0.80 0.90 1.00 1.10 L1/2 (m1/2) T (s) Figure 7. Graph of period T versus the square root of pendulum length; data from a Physics I experiment. Click to buy NOW! PDF-XChange Viewer www.docu-track.com Click to buy NOW! PDF-XChange Viewer www.docu-track.com EXERCISE 6 In Exercise 5 another approach would have been to square both sides of Equation 7 and plot T2 versus L. Lab 20 directs us to use that alternative. It involves another case of periodic or harmonic motion with a similar, but more complicated, equation for the period: where T is the period of the bobbing (Fig. 8), M is the suspended mass, ms is the mass of the spring, k is a measure of stiffness called the spring constant, and C is a dimensionless factor showing how much of the spring mass is effectively bobbing. a. Square both sides of Equation 8 and rearrange it to match y = mx + b. b. Write y = mx + b under your rearranged equation and circle matching factors as in the Example. c. Write two new equations analogous to Equations 2 and 3 in the Example. Use the first of the two for calculating k and the second for finding C from the data of Fig. 9. d. A theoretical analysis has shown that for most springs C = 1/3. Find the percent error from that value. e. Derive the units of the slope and intercept; show that the units of k come out as N/m and that C is dimensionless. 2 (Eq . 8) k T M Cm s Figure 8. In Lab 20 mass M is suspended from a spring which is set to bobbing up and down, a good approximation to simple harmonic motion (SHM), described by Equation 8. Lab 20: SHM of a Spring Mass of the spring, ms = 25.1 g y = 3.0185x + 0.0197 R2 = 0.9965 0.0000 0.2000 0.4000 0.6000 0.8000 1.0000 0 0.05 0.1 0.15 0.2 0.25 0.3 M (kg) T 2 2 Figure 9. Graph of the square of the period T2 versus suspended mass M data from a Physics I experiment. Click to buy NOW! PDF-XChange Viewer www.docu-track.com Click to buy NOW! PDF-XChange Viewer www.docu-track.com EXERCISE 7 This last exercise deals with an exponential equation, and the trick is to take the logarithm of both sides. In PHYS.204L/206L we do Lab 33 The RC Time Constant with theoretical equation: where V is the potential difference at time t across a circuit element called a capacitor (the is dropped for simplicity), Vo is V at t = 0 (try it), and (tau) is a characteristic of the circuit called the time constant. a. Take the natural log of both sides and apply the addition rule for logarithms of a product on the right-hand side. b. Noting that the graph (Fig. 10) plots lnV versus t, arrange your equation in y = mx + b order, write y = mx + b under it, and circle the parts as in the Example. c. Write two new equations analogous to Equations 2 and 3 in the Example. Use the first of the two for calculating and the second for finding lnVo and then Vo. d. Note that the units of lnV are the natural log of volts, lnV. As usual derive the units of the slope and interecept and use them to obtain the units of your experimental V and t. V V e (Eq. 9) t o Figure 10. Graph of a logarithm versus time; data from Lab 33, a Physics II experiment. Discharge of a Capacitor y = -9.17E-03x + 2.00E+00 R2 = 9.98E-01 0.00 0.50 1.00 1.50 2.00 2.50

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## 1 MECE2320U-THERMODYNAMICS HOMEWORK # 5 Instructor: Dr. Ibrahim Dincer Assignment Date: Thursday, 22 October 2015 Assignment Type: Individual Due Date: Thursday, 29 October 2015 (3.00 pm latest, leave in dropbox 8) 1) As shown in figure, the inlet and outlet conditions of a steam turbine are given. The heat loss from turbine is 35 kJ per kg of steam. a) Show all the state points on T-v diagram b) Write mass and energy balance equations c) Calculate the turbine work 2) As shown in figure, refrigerant R134a enters to a compressor. Write both mass and energy balance equations. Calculate the compressor work and the mass flow rate of refrigerant. 3) As shown in figure, the heat exchanger uses the heat of hot exhaust gases to produce steam. Where, 15% of heat is lost to the surroundings. Exhaust gases enters the heat exchanger at 500°C. Water enters at 15°C as saturated liquid and exit at saturated vapor at 2 MPa. Mass flow rate of water is 0.025 kg/s, and for exhaust gases, it is 0.42 kg/s. The specific heat for exhaust gases is 1.045 kJ/kg K, which can be treated as ideal gas. 1 Turbine 2 ? 1 = 1 ??/? ?1 = 1 ??? ?1 = 300 ℃ ?1 = 40 ?/? ? ??? =? ????? = 35 ??/?? ?2 = 150 ??? ?2 = 0.9 ?2 = 180 ?/? 1 Compressor 2 ???? ???? = 1.3 ?3/??? ?1 = 100 ??? ?1 = −20 ℃ ? ?? =? ? ???? = 3 ?? ?2 = 800 ??? ?2 = 60 ℃ 2 a) Write mass and energy balance equations. b) Calculate the rate of heat transfer to the water. c) Calculate the exhaust gases exit temperature. 4) As shown in figure, two refrigerant R134a streams mix in a mixing chamber. If the mass flow rate of cold stream is twice that of the hot stream. a) Write mass and energy balance equations. b) Calculate the temperature of the mixture at the exit of the mixing chamber c) Calculate the quality at the exit of the mixing chamber 5) As shown in figure, an air conditioning system requires airflow at the main supply duct at a rate of 140 m3/min. The velocity inside circular duct is not to exceed 9 m/s. Assume that the fan converts 85% of electrical energy it consumes into kinetic energy of air. a) Write mass and energy balance equations. b) Calculate the size of electric motor require to drive the fan c) Calculate the diameter of the main duct ?2 = 1 ??? ?2 = 90 ℃ ?1 = 1 ??? ?1 = 30 ℃ ?3 =? ?3 =? 140 ?3/??? 9 ?/? Air Fan

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## Determine the maximum theoritical speed that may be achieved over a distance of 110 m by a car starting from rest assuming there is no slipping. The cofficient of static friction between the tires and pavement is 0.75 .assume (a) front-wheel drive, (b) near-wheel drive.

## 1A. You administer an IV with 3 liters of 50 mM NaCl to a person whose osmolarity is 300 mOsM and whose total body water is 30 L. Fill in the table below: 3 L of 50 mM NaCl Total body ECF ICF Solute (osmoles) Volume (L) Concentration (OsM) 1B. The same person from the previous problem instead is given 1 liter of an IV contained 250 mOsM NaCl and 50 mOsM urea. Com Total body ECF ICF Solute Volume Concentration 2. You isolate intact mitochondria as described in class and equilibrate them in a buffered solution at pH 9, containing 0.1 M KCl and ADP plus Pi but without succinate. You then collect them by centrifugation, and quickly resuspend them in a new buffer at pH 7, without KCl , but with valinomycin (a K+ ionophore). Note: the K+ rushing out will create a huge positive charge differential. a. Describe what happens to proton concentrations in the intermembrane space and the matrix at each step of the study. b. What do you predict will be the result on oxygen consumption and the production of ATP? 3. A negatively charged nutrient (equivalent charge of one electron) is actively transported from the outside to the inside of a cell membrane; i.e. a cell captures energy from the hydrolysis of ATP in order to bring a molecule from the outside of the cell, where it is present at a low concentration, to the inside of the cell, where it is present at higher concentration. If the molecular species to be transported is present at a concentration of 34.5 nM on the outside of the cell, the potential on the outside of the cell is +75 mV, the potential on the inside of the cell is -35 mV, and the efficiency at which energy from the hydrolysis of ATP is captured for this active transport process is 59%, what is the maximum concentration of the transported species that may be achieved inside the cell? 4. . ATP + H2O -> ADP + Pi G0 = -7.3 kcal/mol In a chemical system that has two different solute concentrations, the Gibbs free energy that is available to do work is: ΔG = RT ln [C1/C2], where R and T are the gas constant (2 cal/mol K) and temperature (Kelvin). C1 and C2 refer to the concentrations (e.g. molarities, M) of a solute on different sides of a membrane. (a) For a one unit difference in pH across a cellular membrane, what is the energy (in kcal/mol) that is available to do chemical work? (b) This gradient is to be used to drive the reaction synthesis of ATP from ADP and Pi. A concentration gradient of any solute has potential energy. When the solute is charged, a voltage is also established across the membrane, which also adds to the total potential energy. What fraction of the energy needed to drive the reaction is provided by the voltage across the membrane?

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## Which of the following is not true about hysterectomy? Question 41 options: There is no risk of having to repeat the procedure in months or years. There is growing concern that it can be overused as a treatment for DUB. In 50% of hysterectomies performed to treat DUB, there is no cancer present. There is no change in sexual function or sex drive after hysterectomy.

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## You and a friend each drive 42 km. You travel at 89. km/h, your friend at 94 km/h. How long will your friend wait for you at the end of the trip

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## Question 1 When using NTFS as a file system, what can be used to control the amount of hard disk space each user on the machine can have as a maximum? Answer Logical drives Extended partitions Disk quotas Security Center Question 2 Pin 1 of the floppy cable connects to pin 34 of the controller. Answer True False Question 3 What is the primary cause of hard drive failures? Answer Heat Dust Dirty laser lens Moving parts Question 4 The DBR contains the system files. Answer True False Question 5 A spanned volume requires a minimum of three hard drives. Answer True False Question 6 Which situation would not be appropriate for the use of SSDs? Answer A military operation where fast access to data is critical A medical imaging office that needs high-capacity storage A manufacturing plant with heat-sensitive equipment A research facility where noise must be kept to a minimum Question 7 Why are SSDs more susceptible than mechanical hard drives to electrostatic discharge? Answer The internal battery of the SSD provides additional current. SSDs are memory. The voltage level of the SSD is lower than a mechanical hard drive. The SSD is a more fragile component. Question 8 A motherboard has two PATA IDE connectors, A and B. A is nearer the edge than B. The IDE cable from A connects to a 500GB hard drive and then to a 200GB hard drive. The IDE cable from B connects to an R/W optical drive and then to a Blu-ray optical drive. Assuming the setup is optimal, which of the following describes the 500GB hard drive? Answer Primary slave Secondary slave Primary master Secondary master Question 9 The primary IDE motherboard connection normally uses I/O address 1F0 -1F7h and IRQ 15. Answer True False Question 10 A cable with a twist is used when installing two floppy drives. Answer True False Question 11 What does partitioning the hard drive mean? Answer Dividing the hard drive up into three different sections: one for each type of file system Preparing the drive to be mounted Giving the hard drive a drive letter and/or allowing the hard drive to be seen as more than one drive Preparing the drive for an operating system Question 12 The Network Engineering Technology departmental secretary is getting a new computer funded by a grant. The old computer is being moved by the PC technicians to give to the new program facilitator in another department. Which one of the following is most likely to be used before the program facilitator uses the computer? Answer Check Now Tool Backup Tool Disk Management Tool BitLocker Question 13 What is CHKDSK? Answer A command used to scan the disk for viruses during off hours A program used to defragment the hard drive A program used to locate and identify lost clusters A command used to verify the validity of the drive surface before installing a file system or an operating system Question 14 When a disk has been prepared to store data, it has been Answer Cleaned Tracked Enabled Formatted Question 15 Where would you go to enable a SATA port? Answer CMOS BIOS Disk Management Tool Task Manager Question 16 The Windows boot partition is the partition that must contain the majority of the operating system. Answer True False Question 17 Two considerations when adding or installing a floppy drive are an available drive bay and an available power connector. Answer True False Question 18 What is the difference between a SATA 2 and a SATA 3 hard drive? Answer The SATA 3 has a different power connector. The SATA 3 device transmits more simultaneous bits than SATA 2. The SATA 3 device transmits data faster. SATA 3 will always be a larger capacity drive. The SATA 3 device will be physically smaller. Question 19 What command would be used in Windows 7 to repair a partition table? Answer FDISK FORMAT FIXBOOT bootrec /FixMbr FIXMBR Question 20 What file system is optimized for optical media? Answer exFAT FAT32 CDFS NTFS Question 21 One of the most effective ways of increasing computer performance is to increase the size of virtual memory. Answer True False Question 22 Older PATA IDE cables and the Ultra ATA/66 cable differ by Answer Where the twist occurs The number of conductors The number of pins The number of devices they can connect to Question 23 Which of the following is NOT important in assigning SCSI IDs? Answer The hard drive that the system boots to may have a preset ID. ID priority must match the order of appearance on the SCSI chain. All devices must have unique IDs. Slower devices should have higher priority IDs. Question 24 The ATA standard is associated with the SCSI interface. Answer True False Question 25 A striped volume requires a minimum of two hard drives. Answer True False

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## Chapter 8 Practice Problems (Practice – no credit) Due: 12:59pm on Friday, April 18, 2014 You will receive no credit for items you complete after the assignment is due. Grading Policy Circular Launch A ball is launched up a semicircular chute in such a way that at the top of the chute, just before it goes into free fall, the ball has a centripetal acceleration of magnitude 2 . Part A How far from the bottom of the chute does the ball land? Your answer for the distance the ball travels from the end of the chute should contain . You did not open hints for this part. ANSWER: g R Normal Force and Centripetal Force Ranking Task A roller-coaster track has six semicircular “dips” with different radii of curvature. The same roller-coaster cart rides through each dip at a different speed. Part A For the different values given for the radius of curvature and speed , rank the magnitude of the force of the roller-coaster track on the cart at the bottom of each dip. Rank from largest to smallest. To rank items as equivalent, overlap them. You did not open hints for this part. ANSWER: D = R v Two Cars on a Curving Road Part A A small car of mass and a large car of mass drive along a highway. They approach a curve of radius . Both cars maintain the same acceleration as they travel around the curve. How does the speed of the small car compare to the speed of the large car as they round the curve? You did not open hints for this part. m 4m R a vS vL ANSWER: Part B Now assume that two identical cars of mass drive along a highway. One car approaches a curve of radius at speed . The second car approaches a curve of radius at a speed of . How does the magnitude of the net force exerted on the first car compare to the magnitude of the net force exerted on the second car? You did not open hints for this part. ANSWER: ± A Ride on the Ferris Wheel A woman rides on a Ferris wheel of radius 16 that maintains the same speed throughout its motion. To better understand physics, she takes along a digital bathroom scale (with memory) and sits on it. When she gets off the ride, she uploads the scale readings to a computer and creates a graph of scale reading versus time. Note that the graph has a minimum value of 510 and a maximum value of 666 . vS = 1 4 vL vS = 1 2 vL vS = vL vS = 2vL vS = 4vL m 2R v 6R 3v F1 F2 F1 = 1 3 F2 F1 = 3 4 F2 F1 = F2 F1 = 3F2 F1 = 27F2 m N N Part A What is the woman’s mass? Express your answer in kilograms. You did not open hints for this part. ANSWER: ± Mass on Turntable A small metal cylinder rests on a circular turntable that is rotating at a constant speed as illustrated in the diagram . The small metal cylinder has a mass of 0.20 , the coefficient of static friction between the cylinder and the turntable is 0.080, and the cylinder is located 0.15 from the center of the turntable. Take the magnitude of the acceleration due to gravity to be 9.81 . m = kg kg m m/s2 Part A What is the maximum speed that the cylinder can move along its circular path without slipping off the turntable? Express your answer numerically in meters per second to two significant figures. 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. vmax vmax = m/s

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## Assignment 6 Due: 11:59pm on Friday, March 7, 2014 You will receive no credit for items you complete after the assignment is due. Grading Policy Conceptual Question 7.7 A small car is pushing a large truck. They are speeding up. Part A Is the force of the truck on the car larger than, smaller than, or equal to the force of the car on the truck? ANSWER: Correct Conceptual Question 7.12 The figure shows two masses at rest. The string is massless and the pulley is frictionless. The spring scale reads in . Assume that = 4 . The force of the truck on the car is larger than the force of the car on the truck. The force of the truck on the car is equal to the force of the car on the truck. The force of the truck on the car is smaller than the force of the car on the truck. kg m kg Part A What is the reading of the scale? Express your answer to one significant figure and include the appropriate units. ANSWER: Correct Problem 7.1 A weightlifter stands up at constant speed from a squatting position while holding a heavy barbell across his shoulders. Part A Draw a free-body diagram for the barbells. Draw the force vectors with their tails at the dot. The orientation of your vectors will be graded. The exact length of your vectors will not be graded but the relative length of one to the other will be graded. ANSWER: m = 4 kg Correct Part B Draw a free-body diagram for the weight lifter. Draw the force vectors with their tails at the dot. The orientation of your vectors will be graded. The exact length of your vectors will not be graded but the relative length of one to the other will be graded. ANSWER: Correct Problem 7.6 Block A in the figure is sliding down the incline. The rope is massless, and the massless pulley turns on frictionless bearings, but the surface is not frictionless. The rope and the pulley are among the interacting objects, but you’ll have to decide if they’re part of the system. Part A Draw a free-body diagram for the block A. The orientation of your vectors will be graded. The exact length of your vectors will not be graded. ANSWER: Correct Part B Draw a free-body diagram for the block B. The orientation of your vectors will be graded. The exact length of your vectors will not be graded. ANSWER: Correct A Space Walk Part A An astronaut is taking a space walk near the shuttle when her safety tether breaks. What should the astronaut do to get back to the shuttle? Hint 1. How to approach the problem Newton’s 3rd law tells us that forces occur in pairs. Within each pair, the forces, often called action and reaction, have equal magnitude and opposite direction. Which of the actions suggested in the problem will result in the force pushing the astronaut back to the shuttle? ANSWER: Correct As the astronaut throws the tool away from the shuttle, she exerts a force in the direction away from the shuttle. Then, by Newton’s 3rd law, the tool will exert an opposite force on her. Thus, as she throws the tool, a force directed toward the shuttle will act on the astronaut. Newton’s 2nd law stipulates that the astronaut would acquire an acceleration toward the shuttle. Part B Assuming that the astronaut can throw any tool with the same force, what tool should be thrown to get back to the shuttle as quickly as possible? You should consider how much mass is left behind as the object is thrown as well as the mass of the object itself. Hint 1. How to approach the problem Recall that the force acting on the astronaut is equal in magnitude and opposite in direction to the force that she exerts on the tool. Hint 2. Newton’s 2nd law Newton’s 2nd law states that . If force is held constant, then acceleration is inversely proportional to mass. For example, when the same force is applied to objects of different mass, the object with the largest mass will experience the smallest acceleration. ANSWER: Attempt to “swim” toward the shuttle. Take slow steps toward the shuttle. Take a tool from her tool belt and throw it away from the shuttle. Take the portion of the safety tether still attached to her belt and throw it toward the shuttle. F = ma Correct The force that acts on the astronaut must equal in magnitude the force that she exerts on the tool. Therefore, if she exerts the same force on any tool, the force acting on her will be independent of the mass of the tool. However, the acceleration that the astronaut would acquire is inversely proportional to her mass since she is acted upon by a constant force. If she throws the tool with the largest mass, the remaining mass (the astronaut plus her remaining tools) would be smallest—and the acceleration the greatest! Part C If the astronaut throws the tool with a force of 16.0 , what is the magnitude of the acceleration of the astronaut during the throw? Assume that the total mass of the astronaut after she throws the tool is 80.0 . Express your answer in meters per second squared. Hint 1. Find the force acting on the astronaut What is the magnitude of the force acting on the astronaut as she throws the tool? Express your answer in newtons. ANSWER: Hint 2. Newton’s 2nd law An object of mass acted upon by a net force has an acceleration given by . ANSWER: The tool with the smallest mass. The tool with the largest mass. Any tool, since the mass of the tool would make no difference. N a kg F F = 16.0 N m F a F = ma a = 0.200 m/s2 Correct Problem 7.10 Blocks with masses of 2 , 4 , and 6 are lined up in a row on a frictionless table. All three are pushed forward by a 60 force applied to the 2 block. Part A How much force does the 4 block exert on the 6 block? Express your answer to one significant figure and include the appropriate units. ANSWER: Correct Part B How much force does the 4 block exert on the 2 block? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Problem 7.9 A 1000 car pushes a 2100 truck that has a dead battery. When the driver steps on the accelerator, the drive wheels of the car push against the ground with a force of 4500 . Rolling friction can kg kg kg N kg kg kg F = 30 N kg kg F = 50 N kg kg N be neglected. Part A What is the magnitude of the force of the car on the truck? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Part B What is the magnitude of the force of the truck on the car? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Atwood Machine Special Cases An Atwood machine consists of two blocks (of masses and ) tied together with a massless rope that passes over a fixed, perfect (massless and frictionless) pulley. In this problem you’ll investigate some special cases where physical variables describing the Atwood machine take on limiting values. Often, examining special cases will simplify a problem, so that the solution may be found from inspection or from the results of a problem you’ve already seen. For all parts of this problem, take upward to be the positive direction and take the gravitational constant, , to be positive. F = 3000 N F = 3000 N m1 m2 g Part A Consider the case where and are both nonzero, and . Let be the magnitude of the tension in the rope connected to the block of mass , and let be the magnitude of the tension in the rope connected to the block of mass . Which of the following statements is true? ANSWER: Correct Part B Now, consider the special case where the block of mass is not present. Find the magnitude, , of the tension in the rope. Try to do this without equations; instead, think about the physical consequences. Hint 1. How to approach the problem If the block of mass is not present, and the rope connecting the two blocks is massless, will the motion of the block of mass be any different from free fall? Hint 2. Which physical law to use Use Newton’s 2nd law on the block of mass . m1 m2 m2 > m1 T1 m1 T2 m2 is always equal to . is greater than by an amount independent of velocity. is greater than but the difference decreases as the blocks increase in velocity. There is not enough information to determine the relationship between and . T1 T2 T2 T1 T2 T1 T1 T2 m1 T m1 m2 m2 ANSWER: Correct Part C For the same special case (the block of mass not present), what is the acceleration of the block of mass ? Express your answer in terms of , and remember that an upward acceleration should be positive. ANSWER: Correct Part D Next, consider the special case where only the block of mass is present. Find the magnitude, , of the tension in the rope. ANSWER: Correct Part E For the same special case (the block of mass not present) what is the acceleration of the end of the rope where the block of mass would have been attached? Express your answer in terms of , and remember that an upward acceleration should be positive. T = 0 m1 m2 g a2 = -9.80 m1 T T = 0 m2 m2 g ANSWER: Correct Part F Next, consider the special case . What is the magnitude of the tension in the rope connecting the two blocks? Use the variable in your answer instead of or . ANSWER: Correct Part G For the same special case ( ), what is the acceleration of the block of mass ? ANSWER: Correct Part H Finally, suppose , while remains finite. What value does the the magnitude of the tension approach? Hint 1. Acceleration of block of mass a2 = 9.80 m1 = m2 = m m m1 m2 T = mg m1 = m2 = m m2 a2 = 0 m1 m2 m1 As becomes large, the finite tension will have a neglible effect on the acceleration, . If you ignore , you can pretend the rope is gone without changing your results for . As , what value does approach? ANSWER: Hint 2. Acceleration of block of mass As , what value will the acceleration of the block of mass approach? ANSWER: Hint 3. Net force on block of mass What is the magnitude of the net force on the block of mass . Express your answer in terms of , , , and any other given quantities. Take the upward direction to be positive. ANSWER: ANSWER: Correct Imagining what would happen if one or more of the variables approached infinity is often a good way to investigate the behavior of a system. m1 T a1 T a1 m1 a1 a1 = -9.80 m2 m1 m2 a2 = 9.80 m2 Fnet m2 T m2 g Fnet = T − m2g T = 2m2g Problem 7.17 A 5.9 rope hangs from the ceiling. Part A What is the tension at the midpoint of the rope? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Problem 7.23 The sled dog in figure drags sleds A and B across the snow. The coefficient of friction between the sleds and the snow is 0.10. Part A If the tension in rope 1 is 100 , what is the tension in rope 2? Express your answer to two significant figures and include the appropriate units. ANSWER: kg T = 29 N N T2 = 180 N Correct Enhanced EOC: Problem 7.31 Two packages at UPS start sliding down the ramp shown in the figure. Package A has a mass of 4.50 and a coefficient of kinetic friction of 0.200. Package B has a mass of 11.0 and a coefficient of kinetic friction of 0.150. You may want to review ( pages 177 – 181) . For help with math skills, you may want to review: Vector Components Part A How long does it take package A to reach the bottom? Express your answer with the appropriate units. Hint 1. How to approach the problem Start by drawing force identification diagrams for package A and package B separately. What are the four forces acting on each block? Which of the forces are related by Newton’s third law? Draw separate free-body diagrams for block A and for block B. What is a good coordinate system to use to describe the motion of the blocks down the ramp? Label your coordinate system on the free-body diagram. In your coordinate system, compute the x and y components of each force on block A. What are the x and y components of the net force on block A? What are the x and y components of the net force on block B? Given that the coefficient of friction of block A is greater than the coefficient of friction of block B, do you think the blocks will stay together as they slide down the ramp? Assuming that they do stay together, how is the acceleration of the two blocks related? (We can check this assumption later.) Using the components of the forces and Newton’s second law, what is the acceleration of the blocks? What is the initial velocity of the blocks? Given the initial velocity and the acceleration, 20 kg kg how long does it take block A to go the given distance? To check that the blocks do indeed stay together, solve for the force of block B on block A. If the force is directed toward the bottom of the ramp, then the blocks stay together. ANSWER: Correct Problem 7.33 The 1.0 kg block in the figure is tied to the wall with a rope. It sits on top of the 2.0 kg block. The lower block is pulled to the right with a tension force of 20 N. The coefficient of kinetic friction at both the lower and upper surfaces of the 2.0 kg block is = 0.420. 1.48 s μk Part A What is the tension in the rope holding the 1.0 kg block to the wall? Express your answer with the appropriate units. ANSWER: Correct Part B What is the acceleration of the 2.0 kg block? Express your answer with the appropriate units. ANSWER: Correct Problem 7.38 The 100 kg block in figure takes 5.60 to reach the floor after being released from rest. 4.12 N 1.77 m s2 s Part A What is the mass of the block on the left? Express your answer with the appropriate units. ANSWER: Correct Problem 7.41 Figure shows a block of mass m resting on a 20 slope. The block has coefficients of friction 0.82 and 0.51 with the surface. It is connected via a massless string over a massless, frictionless pulley to a hanging block of mass 2.0 . Part A What is the minimum mass that will stick and not slip? 98.7 kg kg m Express your answer to three significant figures and include the appropriate units. ANSWER: Correct If you need to use the rounded answer you submitted here in a subsequent part, instead use the full precision answer and only round as a final step before submitting an answer. Part B If this minimum mass is nudged ever so slightly, it will start being pulled up the incline. What acceleration will it have? Express your answer to three significant figures and include the appropriate units. ANSWER: Correct Problem 7.46 A house painter uses the chair and pulley arrangement of the figure to lift himself up the side of a house. The painter’s mass is 75 and the chair’s mass is 12 . m = 1.80 kg a = 1.35 m s2 kg kg Part A With what force must he pull down on the rope in order to accelerate upward at 0.22 ? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Score Summary: Your score on this assignment is 98.6%. You received 104.5 out of a possible total of 106 points. m/s2 F = 440 N

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