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Unformatted text preview: whittiker (mcw968) – oldmidterm 01 – Turner – (58220) 1 This printout should have 20 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. 001 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 3 . 56 cm. Correct answer: 5 . 08319 cm. Explanation: Let : m Al = 2700 kg , m Fe = 7860 kg , and r Fe = 3 . 56 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 = (3 . 56 cm) 3 radicalBigg 7860 kg 2700 kg = 5 . 08319 cm . 002 10.0 points A piece of pipe has an outer radius of 4 . 6 cm, an inner radius of 2 . 8 cm, and length of 33 cm as shown in the figure. 3 3 c m 4 . 6 cm 2 . 8 cm b What is the mass of this pipe? Assume its density is 8 . 9 g / cm 3 . Correct answer: 12 . 2902 kg. Explanation: Let : r 1 = 4 . 6 cm , r 2 = 2 . 8 cm , ℓ = 33 cm , and ρ = 8 . 9 g / cm 3 . V = π r 2 1 ℓ − π r 2 2 ℓ = π ( r 2 1 − r 2 2 ) ℓ, so the density is ρ = m V m = ρV = ρπ ( r 2 1 − r 2 2 ) ℓ = (8 . 9 g / cm 3 ) π × [(4 . 6 cm) 2 − (2 . 8 cm) 2 ] × (33 cm) 1 kg 1000 g = 12 . 2902 kg . 003 10.0 points Suppose the volume V of some object happens to depend on time t according to the equation V ( t ) = At 3 + B t 2 , where A and B are some constants. Let L and T denote dimensions of length and time, respectively. whittiker (mcw968) – oldmidterm 01 – Turner – (58220) 2 What is the dimension of the constant A ? 1. L 3 · T 3 2. L 2 / T 3. L / T 4. L 3 / T 3 correct 5. L / T 3 Explanation: The term At 3 has dimensions of volume, so [ V ] = [ A ] [ t ] 3 [ A ] = [ V ] [ t ] 3 = L 3 T 3 . 004 10.0 points Consider a car which is traveling along a straight road with constant acceleration a . There are two checkpoints A and B which are a distance 85 . 5 m apart. The time it takes for the car to travel from A to B is 4 . 51 s. A B 2 . 45 m / s 2 85 . 5 m Find the velocity v B for the case where the acceleration is 2 . 45 m / s 2 . Correct answer: 24 . 4826 m / s. Explanation: Let : Δ x = 85 . 5 m , Δ t = 4 . 51 s , and a = 2 . 45 m / s 2 . ¯ v = v A + v B = Δ x Δ t v A = 2 Δ x Δ t − v B and v B = v A + a Δ t = parenleftbigg 2 Δ x Δ t − v B parenrightbigg + a Δ t 2 v B = 2 Δ x Δ t + a Δ t v B = Δ x Δ t + a Δ t 2 = 85 . 5 m 4 . 51 s + (2 . 45 m / s 2 ) (4 . 51 s) 2 = 24 . 4826 m / s ....
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This note was uploaded on 03/01/2010 for the course PHYS 52450 taught by Professor Turner during the Spring '10 term at University of Texas.
 Spring '10
 Turner
 Mass

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