Physics Test 1 solutions

Physics Test 1 solutions - Version 032/AACAA – Test #1...

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Unformatted text preview: Version 032/AACAA – Test #1 – Antoniewicz – (57420) 1 This print-out should have 14 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points The velocity of a particle moving along the x axis is given by v x = a t − b t 3 for t > , where a = 33 m / s 2 , b = 3 . 3 m / s 4 , and t is in s. What is the acceleration of the particle when it achieves its maximum displacement in the positive x direction? 1. -54.0 2. -66.0 3. -68.0 4. -64.0 5. -62.0 6. -60.0 7. -50.0 8. -58.0 9. -52.0 10. -56.0 Correct answer: − 66 m / s 2 . Explanation: The time when it achieves maximum dis- placement is when its velocity is 0: v = a t − b t 3 = 0 t ( a − b t 2 ) = 0 . t = radicalbigg a b = radicalBigg 33 m / s 2 3 . 3 m / s 4 = 3 . 16228 s . The acceleration at this particular time is a x = d v x dt = d ( a t − b t 3 ) dt = a − 3 b t 2 = a − 3 a = − 2 a = − 2 (33 m / s 2 ) = − 66 m / s 2 . 002 10.0 points One cubic meter (1.0 m 3 ) of aluminum has a mass of 2700 kg, and a cubic meter of iron has a mass of 7860 kg. Find the radius of a solid aluminum sphere that has the same mass as a solid iron sphere of radius 2 . 87 cm. 1. 2.91284 2. 6.45394 3. 6.06841 4. 6.33971 5. 4.09796 6. 5.49727 7. 6.82518 8. 4.78334 9. 5.64006 10. 3.78383 Correct answer: 4 . 09796 cm. Explanation: Let : m Al = 2700 kg , m Fe = 7860 kg , and r Fe = 2 . 87 cm . Density is ρ = m V . Since the masses are the same, ρ Al V Al = ρ Fe V Fe ρ Al parenleftbigg 4 3 π r 3 Al parenrightbigg = ρ Fe parenleftbigg 4 3 π r 3 Fe parenrightbigg parenleftbigg r Al r Fe parenrightbigg 3 = ρ Fe ρ Al r Al = r Fe 3 radicalbigg ρ Fe ρ Al = (2 . 87 cm) 3 radicalBigg 7860 kg 2700 kg = 4 . 09796 cm . 003 (part 1 of 3) 4.0 points A motorboat heads due east at 14 . 8 m/s across a river that flows toward the south at a speed of 3 . 4 m/s. a) What is the magnitude of the resultant velocity relative to an observer on the shore? 1. 14.2201 Version 032/AACAA – Test #1 – Antoniewicz – (57420) 2 2. 14.5561 3. 13.3664 4. 12.2972 5. 11.7004 6. 13.8495 7. 13.132 8. 14.8946 9. 15.7166 10. 15.1855 Correct answer: 15 . 1855 m / s. Explanation: 14 . 8 m / s 3 . 4m / s v be θ Note: Figure is not drawn to scale. Basic Concepts: vectorv be = vectorv br + vectorv re Since the velocities are perpendicular, v 2 be = v 2 br + v 2 re Given: Let north and east be positive: v br = 14 . 8 m / s v re = − 3 . 4 m / s Solution: v be = radicalBig (14 . 8 m / s) 2 + ( − 3 . 4 m / s) 2 = 15 . 1855 m / s 004 (part 2 of 3) 3.0 points004 (part 2 of 3) 3....
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This note was uploaded on 04/07/2011 for the course PHY 303K taught by Professor Turner during the Spring '08 term at University of Texas.

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Physics Test 1 solutions - Version 032/AACAA – Test #1...

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