Exam Practice key

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Dec 6, 2023

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Fall 2020 Physics 103 Answer Key Lab Final Exam Review Name_______________________ Section_____ In preparing for the lab final exam it is important to review all the lab answer keys that your lab instructor sent you. Due 17 for Sections 70-73 Due 18 for Sections 74-77 You will be sent an answer key for the Lab Final Exam Review on November 19. Equations: W=f x d Work done in stretching or compressing a spring = .5kx 2 k= spring constant N/m x= elongation or compression distance. Elastic Potential Energy stored in spring equals the work done. Gravitational Potential Energy= mgh m=mass in kg g=9.8 m/s 2 h=height in meters Kinetic Energy =.5mv 2 m=mass in kg v= velocity in m/s Ft=mΔt p=mv Equations in red were not on original practice exam. 1. a. Determine the standard deviation, average and range of the time data in the table below . Trial Times(s) 1 4.1 2 6.3 3 5.2 4 7 5 4.8 6 9

4.1 6.8 std.dev. p 1.57990 3 average 5.9125 range 4.9 b. Identify all time data, if any, that are more than two standard deviations from the mean. Show your work. There are none. Trial 6 (9 seconds), is the closest to two standard deviations about the mean. 2. a. What is the slope of the line on the graph above? Include units. You must show work. 2.5 g/cm 3 Δ mass 200g/Δ volume 80cm 3 b. Based on the graph, would you describe the relationship of mass and volume as a strong positive linear relationship. Justify your answer. Yes. The trendline slops upward and data points are all on the trendline. R value close to 1.

3. List and describe the types of scientific models. (go to https://www.texasgateway.org/resource/scientific-models for a quick review) Answer is clearly given in the link. 4. Describe the uses and limitations of scientific models. (go tohttps://www.texasgateway.org/resource/scientific-models for a quick review) Answer is clearly given in the link. a. Make a sketch of a velocity-time graph that represents the motion of a single coffee filter falling through a vacuum. Note the acceleration of 9.8 m/s 2 . Filter is in free fall. Graph A 0 2 4 6 8 10 12 0 20 40 60 80 100 120 f(x) = 9.8 x − 0 R² = 1 Velocity-Time time (s) velocity (m/s) b. Make a sketch of a velocity-time graph that represents a single coffee filter falling through air. Filter shows an increase in speed. As the speed increases, the air resistance increases. At some point the air resistance equals the force of gravity(mg) on the filter and it stops accelerating. The net force at this point is 0. a=net force/mass Graph B

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c. Examine the first velocity-time graph above(a) and determine the acceleration of the object. You must explain how you determined the acceleration. Slope of the line is 9.8 m/s 2 . Y= mx + b d. Calculate the area under the “curve” (line) from 0-6 s. velocity-time graph (A) Extend the trendline back through the origin. Area = !/2 bh. Use the equation to determine the value of the velocity a t=6s. base= 6s height = 58.80 m/s. Area=(6s) (58.8m/s) meters/2 = 176 m e. What quantity does the area represent? Choose one.  Acceleration  Distance  Velocity ,

4 . Students performed an experiment to establish the relationship between the force applied to a spring and the spring elongation. The data from their experiment is plotted below. 0 0.2 0.4 0.6 0.8 1 1.2 1.4 0 10 20 30 40 50 60 70 f(x) = 50.06 x − 0.32 R² = 1 Hooke's Law Elongation (m) Force(N) a. What is the value of the spring constant k? 50 N/m. b. How much energy is stored in the spring when it is elongated 1 m? You must show work. Using the area under the curve from 0-1 m. Area = ½ bh b = 1m h= 50 n Area = 25 Nm=25 J Keep in mind we have ignored frictional effects. Work done on spring equals energy stored in spring. Work done on string = 1/2kx 2 Work done (energy stored) = ( ½) (50 N/m) (1 2)

b. Use the information given to solve for the numerical value of h. Show your work, When the spring is compressed, energy is stored in the spring. Amount of work or energy stored= ½ kx 2 (1/2) (100N/m) (.10 2 ) = .5Nm =.5 Joules .5 Joules =mgh .5J= 5 kg * 9.8 m/s 2 *h h=.01 m h=.01m= 1cm Note; Work is done to compress the spring and store energy in it. The spring transfers elastic potential energy to the block giving it kinetic energy. As block slides up the ramp some of the kinetic energy is being converted to gravitational potential energy. All the kinetic energy has been converted to gravitational potential energy when the block reaches the highest point and stops. 5. a. If you transfer equal amounts of heat to equal masses of sand and water which substance would have the greatest change in temperature? The specific heat of water is 4.19 J/g 0 C and sand has a specific heat of .800 J/g 0 C. Justify your answer. Sand because the specific heat of water is over 5 times that of sand. This means that for the same heat input, the sand would have a temperature change 5 times greater than water.

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b. If you had 200g of water and 200g of sand in separate well insulated containers and heated each of them using a 50-watt (J/s) heating element for 20 minutes, what would the temperature change of each substance be? Show your work. Sand Q=mcΔt Δt=Q/mc + Q=(50J/s) (20 minutes) (60 seconds/minute) = 60,000J 60,000J= mcΔt solve for Δt. 60,000J/mc=Δt mc= (200g) (.800 J/g 0 C) =160 J/ 0 C 60000J/160 J/ 0 C = 365 0 C Water Q=mcΔt Δt=Q/mc + Q=(50J/s) (20 minutes) (60 seconds/minute) = 60,000 j mc= (200g) (4.19 J/g 0 C) =833 J/ 0 C. Δt=60000J/833 J/ 0 C = 72 0 C

6. a. Is there a correlation between CO2 and the temperature anomaly. Positive or negative? Strong or weak. Justify your answer. Positive and strong. Trendline slops upward. Data points are close to the trendline b. Does the above graph prove that the temperature rise is due to increasing C02 levels. Justify your answer. Just because two variables display a relationship, does not mean that one variable causes the other.

7. Students investigated Newton’s Second Law. The force was kept constant and the mass being accelerated was changed. Students measured the resulting acceleration and generated the graph shown below. 0.5 1 1.5 2 2.5 3 3.5 0 2 4 6 8 10 12 14 f(x) = 10.01 x^-1.01 R² = 1 Acceleration-Mass Mass(kg) Acceleration (m/s/s) a. Newton’s Second Law States that a=f/m. Is above graph in agreement with Newton’s second law. Justify your answer. Note: x -1.007 equals 1/x 1.007 and a=f/m is equal to a=fm -1 y is acceleration y=10x -1 x -1 =1/x y=10/X a=f/m force is 10 N x is mass b. Assume that the exponent of x has been “rounded” and the equation now states that y=10x -1 . Determine the acceleration of a 4 kg mass. Show your work. y=10x -1 a=10/4 2.5 m/s 2 a=kgm/s 2 divided by a kg kgm/s 2 X 1/kg = m/s 2

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c. As stated, the force in the experiment was kept constant. What was the numerical value of the force? Explain how you determined the answer from the equation. Slope is 10 N y=mx + b m=slope 8. In terms of impulse, explain how crush zones, seatbelts, air bags, and safety helmets prevent injuries. Assume a car and its occupants are moving at 60mph. When they come to a stop the change in velocity will be 60 mph. The key is to consider the equation below. Ft=mΔv which can be rewritten F= (mΔv)/t. All the devices above make the time of losing the velocity greater (less acceleration). The greater the time, the less the force. Example: An 80 kg walking with a speed of 2.2 m/s walks into a wall. It doesn’t matter if the wall is padded or not, the person will have the same change in velocity. The padded wall increases the time to bring the person to a complete stop. Increasing the time decreases the force. 9. A 1500 kg automobile moving a 22 m/s is brought to a stop in .01. What was the change in velocity of the automobile? 22 m/s What was the acceleration(deceleration)? Δv/Δt 22m/s/.01= 2200m/s 2 What is the change in momentum? 1500kg X 22 m/s= 33,000 kgm/s

What is the impulse? 33,000 Ns How much force was needed to stop the automobile? Impulse=ft f=33000Ns/.01 10. A 15 kg “medicine ball” is thrown with a velocity of 5 m/s is caught by a 60 kg girl wearing roller skates. After she catches the ball, she holds on the ball. What is the velocity of the ball and the girl after she catches the ball . Use the numbers given in the question above, not those in the diagram below, to solve the problem. You must show work, M1V1 + M2V2 = M1V1’ + M2V2’ Before = After (15 kg) (5m/s) +( 60kg) (0m/s) = (15Kg) (V1’) + (60kg) V2’ V1’ =V2” because object are “stuck” together after the collision (15 kg) (5m/s) +( 60kg) (0m/s) =(75kg) (V) 75kgm/s + 0 =(75kg) V V=1 m/s

11. The height of the ramp is 40 cm. What is the velocity of the red block when it reaches the bottom of the ramp? You do not need to know the mass of the block to do this problem. mgh=1/2mv 2 dividing each side of the equation will cancel m. You can do this because they are the same m. gh=1/2v 2 V=(2gh) 1/2 V = 2.8 m/s 12, The spring shown in the diagram below was compressed. The energy stored in the spring was used to launch the ball upwards to a height of 1 m. The ball has a mass of .2 kg. The spring in the diagram has completely uncompressed and has transferred all its energy to the ball. The 1m is the maximum height. a. Calculate the gravitational potential energy the ball has when it reached the 1m height? mgh=.2kg X 9.8m/s 2 X 1m= 1.96 J .

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b. How much kinetic energy did the ball have when it left the spring. mgh=1/2 mv 2 = 1.96 J The ball’s maximum height is 1m. At that point the kinetic=0 because the velocity is 0. Ball has to stop before it comes back down. At the moment the ball leaves the spring all the energy is kinetic energy. As the ball rises some of the kinetic energy is being transformed to gravitational potential energy. At the highest point all the kinetic energy has been transformed to gravitation potential energy. c. What was the velocity of the ball when it left the spring. 1.96J= 1/2mv 2 1.96J= (1/2) (.2kg) (v 2 ) v=4.42 m/s 13. A student decided to study the behavior of a pendulum. He hypothesizes that the mass of the pendulum bob has no effect on the time for the pendulum to move back and forth. Your task is to collect and analyze pendulum data and determine if the data supports the hypothesis. Tasks: Go to https://phet.colorado.edu/sims/html/pendulum-lab/latest/pendulum-lab_en.html Select Lab Set the friction control half way to maximum. Note that the time to make one back and forth cycle is called the period.

Generate a mass-period data table and a graph with an equation that relates the variables, including an r^2 value. Do the results support the hypothesis? Justify you answer. For this type of question, it is important that your conclusion is based on your data and graph. You may not have the same data as shown below. R 2 =.1318 R=.36 Weak relationship at best. Based on the standard deviation, there is very little spread in the period values. If the friction had been turned off, the period of different masses would be the same. mass(kg ) period(s) 0.10 1.6810 0.40 1.6807 0.60 1.6791 0.70 1.6820 0.90 1.6808 1.10 1.6835 1.50 1.6814 1.30 1.6821 1.40 1.6809 Average 1.6813 St.Dev. S 0.001139 0.00 0.20 0.40 0.60 0.80 1.00 1.20 1.40 1.60 1.6760 1.6770 1.6780 1.6790 1.6800 1.6810 1.6820 1.6830 1.6840 f(x) = 0 x + 1.68 R² = 0.13 Mass-Period Mass (kg) Period (s)

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