om1 - hernandez (ejh742) – oldmidterm 01 – Turner –...

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Unformatted text preview: hernandez (ejh742) – oldmidterm 01 – Turner – (58120) 1 This print-out should have 20 questions. Multiple-choice 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 4 . 66 cm. Correct answer: 6 . 65384 cm. Explanation: Let : m Al = 2700 kg , m Fe = 7860 kg , and r Fe = 4 . 66 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 = (4 . 66 cm) 3 radicalBigg 7860 kg 2700 kg = 6 . 65384 cm . 002 10.0 points A piece of pipe has an outer radius of 4 . 7 cm, an inner radius of 2 . 9 cm, and length of 32 cm as shown in the figure. 3 2 c m 4 . 7 cm 2 . 9 cm b What is the mass of this pipe? Assume its density is 8 . 9 g / cm 3 . Correct answer: 12 . 2398 kg. Explanation: Let : r 1 = 4 . 7 cm , r 2 = 2 . 9 cm , ℓ = 32 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 . 7 cm) 2 − (2 . 9 cm) 2 ] × (32 cm) 1 kg 1000 g = 12 . 2398 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. hernandez (ejh742) – oldmidterm 01 – Turner – (58120) 2 What is the dimension of the constant A ? 1. L / T 3 2. L 3 / T 3 correct 3. L 2 / T 4. L / T 5. L 3 · 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 96 . 6 m apart. The time it takes for the car to travel from A to B is 4 . 19 s. A B 5 . 26 m / s 2 96 . 6 m Find the velocity v B for the case where the acceleration is 5 . 26 m / s 2 . Correct answer: 34 . 0746 m / s. Explanation: Let : Δ x = 96 . 6 m , Δ t = 4 . 19 s , and a = 5 . 26 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 = 96 . 6 m 4 . 19 s + (5 . 26 m / s 2 ) (4 . 19 s) 2 = 34 . 0746 m / s ....
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This note was uploaded on 06/14/2011 for the course PHY 303K taught by Professor Turner during the Spring '08 term at University of Texas.

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om1 - hernandez (ejh742) – oldmidterm 01 – Turner –...

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