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# Practice Exam 1 for General Physics 1 PHYS113

KSU

GPA 3.3

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This 8 page Study Guide was uploaded by Ericah Notetaker on Tuesday March 29, 2016. The Study Guide belongs to PHYS113 at Kansas State University taught by Horton Smith, Glenn A. in Fall 2015. Since its upload, it has received 68 views. For similar materials see General Physics 1 in Art at Kansas State University.

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Date Created: 03/29/16

Quiz instructions You should have both an exam and a machine-gradeable answer card (Scantron). On the other side of this page is an official “physics fact sheet”. Detach it if you wish. Not all of the information on the “fact sheet” will be useful. Before you start: ● Fill in the bubbles for your Wildcat ID number. Scantron cards without ID number bubbles filled out won't be counted. ● Fill in the the test version number (written in big letters before the first question), and the answer sheet number (1). ● Also print your name, the course, and the date in the box at the top of the card. ● Also write your name on the first and last pages of the exam. (If you detach this page, write your name on the next page, and also the last page.) ● Take a deep breath. Relax and pay attention. It's really not that bad. Please note: ● Your score on the multiple choice questions will be determined by your answers on the answer card. However, it is a good idea to mark your answers on the exam sheet too, for your own reference. ● Do not fold or roll the card. Unreadable cards won't be counted. ● Mark your responses on the answer card by filling in the bubbles completely with a #2 pencil. ● Do not fill in more than one response for each question. ● If you erase an answer, make sure you erase it completely. ● If you need a new card, please ask. When you are done: ● Recheck your filled-in circles against your exam sheet. ● Be sure that you have answered all questions, that you have all of the pages of the exam, and that both the question number and your response match agree with your exam sheet. You should see the words “The End” after the last question. ● Turn in both the Scantron card and your exam at the front of the room. ● Your exam will be returned to you after all exams are scanned and checked and the written part is graded. ● Please write your name on your exam. Constants SI prefixes Gravity at Earth's surface g = 9.80 m/s , down -12 12 -11 3 -1 -2 pico 10-9 p tera 10 9 T Gravitational constant G = 6.674 × 10 5 kg 2 nano 10 -6 n giga 10 6 G Atmospheric pressure 1 atm = 1.013×10 N/m micro 10 μ mega 10 M Boltzmann's constant k = 1.38×102J/K milli 10-3 m kilo 10 3 k Ideal gas constant R = 8.31 J/(mole·K) centi 10 -2 c hecto 10 2 h Avagadro's number NA= 6.02×1023 Speed of light c = 2.998 × 10 m/s Trigonometry sin θ = “opposite” / “hypotenuse” Kinematics, dynamics, energy, momentum cos θ = “adjacent” / “hypotenuse” Δ ⃗ ω= Δθ hypotenuse = opposite + adjacent ⃗= Δt Δt Δω t hy a= Δ⃗v α= Δt s po e n Δt o use ⃗ θ(t)=θ +ω t+ α1 t2 p θ ∑ F=ma ⃗ 0 0 2 const o adjacent 1 2 2 2 ⃗(t)=⃗0+v⃗0+ 2 ⃗const ω =ω +0αΔθ 2 2 ∑ τ=I α Impulses and collisions v =v0+2aΔ x Impulse: F Δ t=Δ ⃗ 1 2 KE rotationω 2 KE linear mv 2 Collision: ∑ ⃗pinitial ⃗final 2 W=τΔθ W=F dcosθ Fd Simple harmonic motion and waves p=mv L=I ω ⃗ ⃗ Period: T=2π inertia √ restoring Some kinds of potential energy Uniform gravity: PE Gmgy Spring: F=k x Pendulum: F=mg x/L Spring: PE = kx 2 Frequency: f =1/T S 2 Oscillator: v0/ A=2π f Waves: v= f λ Geometry of circles and spheres Standing: f =n(v/2L),any n 360 degrees = 2π radians = 1 full circle n Circumference of a circle: 2πr f nn(v/4L),odd n Rolling distance = (radius)*(angle in radians) Small angles: (linear size of object) ≈ (arc length) Area of a sphere: 4πr Other definitions and relations Density: ρ=m/V Pressure: P=F/ A Circular or rotational motion v2 2 In a fluid: Δ P=ρ gΔ y v=rω, acentrip =rω Buoyancy: F Bρ fluidg r 2 τ=r Fsinθ, I= ∑ mr Gravity (sphere): F =G m 1 2 G r2 Static equilibrium Friction: F Fμ F N ax=a y0 → ∑ F x ∑ F y0 Ideal gas: PV=nRT=NkT 3 α=0 → ∑ τ=0 Avg transl. KE: KE= kT 2 Name: Date: General Physics Exam 1 Practice, version 1 1. A snowball is thrown straight up, then it comes down and is caught at the same height it started. (Draw a picture if it helps visualize the problem.) Which of the following statements about the motion of the snowball must be FALSE? A. The snowball's position just before it is caught could be the same as just after it is thrown. B. The snowball's velocity just before it is caught could be the same as just after it is thrown. C. The snowball's speed just before it is caught could be the same as just after it is thrown. D. The snowball's acceleration just before it is caught could be the same as just after it is thrown. 2. A sheep runs in a straight line, moving 177 m east in 12.3 s. It then turns and runs 69 m west in 5.83 s. What was the sheep's average speed over the trip? A. 5.96 m/s, calculated from ((177 m)-(69 m))/((12.3 s)+(5.83 s)) B. 38 m/s, calculated from ((177 m)+(69 m))/((12.3 s)-(5.83 s)) C. 16.7 m/s, calculated from ((177 m)-(69 m))/((12.3 s)-(5.83 s)) D. 13.6 m/s, calculated from ((177 m)+(69 m))/((12.3 s)+(5.83 s)) 3. What was the sheep's average velocity over the whole trip? A. 5.96 m/s, calculated from ((177 m)-(69 m))/((12.3 s)+(5.83 s)) B. 38 m/s, calculated from ((177 m)+(69 m))/((12.3 s)-(5.83 s)) C. 16.7 m/s, calculated from ((177 m)-(69 m))/((12.3 s)-(5.83 s)) D. 13.6 m/s, calculated from ((177 m)+(69 m))/((12.3 s)+(5.83 s)) page 1 of 6 4. A rock is dropped from a 200 m tall tower. What is the change in its velocity from the time it is dropped until 1.3 s later? For this problem, let the positive direction be up. 2 2 A. -118 m/s , calculated from -(200 m)/(1.3 s) B. -12.7 m/s, calculated from (-9.8 m/s )*(1.3 s) 2 2 C. 208 m, calculated from (200 m)-0.5*(-9.8 m/s )*(1.3 s) D. -8.28 m/s, calculated from 0.5*(-9.8 m/s )*(1.3 s) 2 E. -8.28 m, calculated from 0.5*(-9.8 m/s )*(1.3 s) 2 5. What is the change in the same rock's velocity from the 1.3 s after it is dropped until 0.79 s after that? A. -12.7 m/s, calculated from (-9.8 m/s )*(1.3 s) 2 B. -7.74 m/s, calculated from (-9.8 m/s )*(0.79 s) C. 253 m/s, calculated from (200 m)/(0.79 s) 2 D. -20.5 m/s, calculated from (-9.8 m/s )*((1.3 s)+(0.79 s)) E. -8.28 m/s, calculated from 0.5*(-9.8 m/s )*(1.3 s) 2 A stone is thrown almost straight upward from the top of a cliff. It leaves the thrower's hand at A, reaches its highest point at B, passes the thrower's hand at C, and lands at D. page 2 of 6 6. At point B, the acceleration of the ball is A. positive. B. upward. C. negative. D. zero. E. downward. 7. On the trip from B to D, the velocity and acceleration of the stone A. both point in the same direction. B. are both increasing. C. are both constant. D. point in opposite directions. 8. If the height of the cliff is 9.2 m and the distance from A to B is 2.6 m, what is the net displacement between points A and D? A. 14 m B. 6.6 m down C. 9.2 m down D. 0 m E. 12 m up 9. What is the net displacement between points A and C? A. 0 m B. 6.6 m down C. 12 m up D. 9.2 m down E. 14 m 10. What is the average velocity of the stone over the whole trip between points A and C? A. 8 m/s up 2 B. -0.27 s C. 5 m/s down D. 5 m/s up E. 0 m/s 11. If an object has constant velocity A. its direction must be constant, but its speed may change. B. it must have a constant nonzero acceleration. C. its speed must be constant, but its direction may change. D. its speed and direction must both be constant. page 3 of 6 12. If an object is traveling east and slowing down, its acceleration is A. west in any coordinate system. B. positive in any coordinate system. C. east in any coordinate system. D. negative in any coordinate system. 13. An "angry bird" is launched by a slingshot. The bird's initial velocity has a horizontal component 16 m/s and a vertical component 15 m/s (up). For some reason, the bird keeps its wings folded, so it travels like a cannonball. What is the horizontal component of the "angry bird"'s velocity at the highest point in its trajectory? A. 16 m/s B. 22 m/s C. 15 m/s D. 0 m/s 14. What is the vertical component of the "angry bird"'s velocity at the highest point in its trajectory? [Hint, for practice only: which way is it traveling at the top of its trajectory?] A. 16 m/s B. 0 m/s C. 15 m/s D. 22 m/s 15. How much time does it take to reach the top of its trajectory? [Hint, for practice only: this is the time it takes its velocity to change from 15 m/s to whatever it is at the top of its trajectory.] A. 98 m, calculated from 2*((16 m/s) +(15 m/s) )/(9.8 m/s ) 2 2 B. 1.5 s, calculated from (15 m/s)/(9.8 m/s ) C. 11 m, calculated from (15 m/s) /(2*(9.8 m/s )) 2 2 D. 22 m/s, calculated from sqrt((16 m/s) +(15 m/s) ) E. 13 m, calculated from (16 m/s) /(2*(9.8 m/s )) 16. A jet plane with a mass of 204000 kg travels east at 963 km/hr. Air resistance exerts a force of 941 kN to the west, while the jet engines provide a thrust on the airplane of 941 kN to the east. What is the change in the plane's horizontal velocity after 383 s? [Hint for practice: this is kind of a trick question, or a really easy question, depending on how you look at it.] A. 963 kph west B. -5.4×10 +03 kph east C. 963 kph east D. 482 kph east E. 0 kph page 4 of 6 17. Ann and Bill are sitting next to each other on a frictionless ice surface. Ann pushes Bill's shoulder with a force of 52 N. What force does Bill's shoulder exert on Ann? A. It depends on how Bill moves his shoulder. B. It depends on their masses. C. 52 N D. 0 N 18. An object with mass 73 kg experiences a net force of 475 N. What is its acceleration? A. It depends on the kind of force. B. 6.5 m/s , calculated from (475 N)/(73 kg) +04 2 C. 3.5×10 m/s , calculated from (475 N)*(73 kg) D. It depends on what the object is. 2 E. 0.15 m/s , calculated from (73 kg)/(475 N) page 5 of 6 Written answer question [Note: your solution will be graded for the correctness of the work shown as well as on the final answer. Show and explain all work! Be sure to include proper units in all answers!] -23 (i) The mass of a carbon atom is approximately 1.99g. What is the mass of a carbon atom in kg? (ii) How much force would it take to accelerate a single carbon atom at 9.8 m/s ? The End page 6 of 6

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