172_practice_exam_1_2_key

172_practice_exam_1_2_key - Phys 172 Practice Exam 1 Summer...

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Phys 172 Practice Exam 1 Summer 2009 Exam 1 Equation Sheet () n R v p dt p d p F r r q q F s k F F F r r m Gm F v m p o by on elect spring i i net by on grav ˆ ˆ ˆ 4 1 ˆ 21 2 21 2 1 1 2 , 21 2 21 2 1 1 2 , G G G G G G G G G G = = = Δ = = = = πε γ d v f T m k sound s ω π = = = = 2 2 , T s i F k stress A Y L strain d L ≡= = Δ ------------------------------------------- CONSTANTS --------------------------------------------- 2 11 34 2 2 23 9 19 2 19 8 2 6.7 10 9.8 1.05 10 1 1.38 10 9 10 1.6 10 4 1 1 1.6 10 3 10 1 electron o Nm N Gg J s kg kg JN m kq C KC m eV J c s v c = = × = × = × = × = ⎛⎞ ⎜⎟ ⎝⎠ = G electron 93 1 k g me =− proton neutron hydrogen atom 1.7 27 kg mm m e ≈≈ M Earth = 24 61 0 × kg Radius of Earth = 6 6.4 10 × m M Moon = 22 71 0 × kg Distance from Earth to Moon = 8 41 0 × m M Sun = 30 21 0 × kg Distance from Earth to Sun = 11 1.5 10 × 1 u = 1.7e-27 kg Avagadro’s Number = 6 x 10 23 molecules/mole Typical atomic radius r 10 -10 m
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Problem 1. For the following speeds, determine if it is valid to approximate gamma as 1 with 0.5% precision. I. A spaceship flies by at 2.9E6 m/s. II. Superman soars back to Krypton at 5.0E7 m/s III. A toy car races down a track at 2.213 m/s 1. I and II only 2. I and III only 3. I, II and III 4. II and III only 5. I only 6. II only 7. III only As mentioned in class and in the book (pg 24), you can approximate gamma as being 1 up to 10% the speed of light. Problem 2. The position and velocity of an ant is observed at t = 12 [s] to be 1, 0, 2 rm = G and 1, 0, 2 cm v s =− G respectively. At t = 14 [s] the ant’s velocity is measured to be 2, 0, 3 cm v s G . Assuming a constant force during the observation period, what is the final position of the ant? 1. 1.03, 0, 2.05 m 2. 4.00, 0, 3.00 m 3. 1.14, 0, 1.95 m 4. 1.03, 0, 1.95 m 5. 1.14, 0, 2.05 m Here we use the position update formula ( ) fia v g rr vt = G GG . However, before doing so, we need to find the average velocity of the ant:
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2 1, 0, 2 2, 0, 3 2 35 ,0, 22 fi avg avg avg vv v v cm v s + = −+ = = GG G G G Note that the units on this velocity are centimeters (not meters); we’ll need to convert units. Moving on to the position update formula: () 13 5 , 0, 2 100 2 2 1.03,0,1.95 fia v g f f rr vt r rm =+ Δ =+ = G G G Problem 3 What is the net force net F G acting on the ant during the observation? The ant has a mass of 2E-5 kg. 1. * 17 , 0 , 57 kg m EE s −− 2. * , 0 , kg m s 3. * , 0 , kg m s 4. * , 0 , kg m s 5. * , 0 , 55 kg m s Here we use the momentum principle. Taking the ant as the system and the surroundings as the unspecified objects exerting the force.
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() 1 2 5 2,0, 3 1,0, 2 2 100 17 1 , 0 , 1 , 0 , net net net net net net pF t p F t mv F t E F FE E γ Δ= Δ Δ = Δ Δ =≈ Δ −− = =− G G G G G G G G G Problem 4 . Which of the following are vectors? I. 3 r + G II. /3 r G III. 3/ r G IV. 3 1. II only 2. II and III only 3. I , II and III only 4. III and IV only 5. II and IV only 6. I and IV only 7. II, III and IV only I. Can’t add a vector and a scalar II. This is a vector III. Division by a vector is not possible IV.
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172_practice_exam_1_2_key - Phys 172 Practice Exam 1 Summer...

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