Quiz 1 Solutions

# Quiz 1 Solutions - schroeder(cs39963 Quiz 1 Units...

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schroeder (cs39963) – Quiz 1: Units, Dimensions, Conversions – Balasubramanya – (1401310) 1 This print-out should have 22 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. This is your first quiz for credit. 001 10.0 points One tenth (0.1) of a meter is called a 1. decimeter. correct 2. dekameter. 3. centimeter. 4. millimeter. Explanation: 002 10.0 points A flat circular plate of copper has a radius of 0 . 268 m and a mass of 40 . 3 kg. What is the thickness of the plate? Correct answer: 0 . 0200002 m. Explanation: Let : m = 40 . 3 kg and r = 0 . 268 m . The density of copper is ρ = 8 . 93 × 10 3 kg / m 3 . ρ = m V , for a flat circular plate, so V = π r 2 h = m ρ h = m ρ ( π r 2 ) = 40 . 3 kg (8930 kg / m 3 ) π (0 . 268 m) 2 = 0 . 0200002 m . 003 10.0 points Consider a cube of soft, spongy material. Which piece below has the larger density? 1. Densities are the same. 2. Unable to determine 3. cutting out a piece of the cube that has one-eighth the volume 4. compressing the cube until it has one- eighth the volume correct Explanation: ρ 1 = m V Compressing the cube results in a denser ma- terial. Compared to the piece cut out, the compressed piece has density ρ 2 = m 1 8 V = 8 m V = 8 ρ 1 . 004 10.0 points A piece of wire has a density of 5 . 5 g / cm. What is the mass of 10 . 6 cm of the wire? Correct answer: 58 . 3 g. Explanation: Let : μ = 5 . 5 g / cm and = 10 . 6 cm . This is a linear mass density: μ = m m = μ ℓ = (5 . 5 g / cm) (10 . 6 cm) = 58 . 3 g .

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schroeder (cs39963) – Quiz 1: Units, Dimensions, Conversions – Balasubramanya – (1401310) 2 005 (part 1 of 2) 10.0 points Consider two planets with uniform mass dis- tributions. The mass density and the radius of planet 1 are ρ 1 and R 1 , respectively, and those of planet 2 are ρ 2 and R 2 What is the ratio M 1 M 2 of their masses? 1. M 1 M 2 = parenleftbigg ρ 2 ρ 1 parenrightbiggparenleftbigg R 1 R 2 parenrightbigg 2 2. M 1 M 2 = parenleftbigg ρ 1 ρ 2 parenrightbiggparenleftbigg R 2 R 1 parenrightbigg 3 3. M 1 M 2 = parenleftbigg ρ 2 ρ 1 parenrightbiggparenleftbigg R 1 R 2 parenrightbigg 4. M 1 M 2 = parenleftbigg ρ 1 ρ 2 parenrightbiggparenleftbigg R 2 R 1 parenrightbigg 2 5. M 1 M 2 = parenleftbigg ρ 1 ρ 2 parenrightbiggparenleftbigg R 1 R 2 parenrightbigg 3 correct 6.
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