oldhw 01 - alexander (jra2623) – oldhomework 01 –...

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Unformatted text preview: alexander (jra2623) – oldhomework 01 – Turner – (92510) This print-out should have 4 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 1. Yes; it weighs about 4.5 lbs. ρ2 ρ1 R2 R1 M1 = M2 ρ1 ρ2 R2 R1 3 4. 001 10.0 points In one scene in the movie The Godfather II, a solid gold phone is passed around a large table for everyone to see. Suppose the volume of gold in the phone were equal to the volume of a 10-centimeter cube of gold. The density of gold is 19,300 kg/m3 . Could such a phone be casually passed around a table from hand to hand? What is the weight of the phone? 1 kg of mass is about 2.2 lb. M1 = M2 2 3. M1 5. = M2 ρ2 ρ1 R1 R2 2 M1 6. = M2 ρ1 ρ2 R1 R2 3 M1 = M2 M1 = 8. M2 M1 = 9. M2 ρ2 ρ1 ρ2 ρ1 ρ1 ρ2 R1 R2 R1 R2 R1 R2 3 M1 = M2 ρ1 ρ2 R2 R1 1 2 7. 2. Yes; it weighs about 9 lbs. 10. 3. No; it weighs about 90 lbs. correct Explanation: m ρ= , so V 4. No; it weighs about 45 lbs. correct Explanation: One cubic meter of gold has a mass of about 20,000 kg, which is roughly 45,000 pounds. A ten centimeter cube of gold is about 45 pounds; it would be difficult to pass around this phone. m = ρ V = (19300 kg/m3 )(10 cm)3 1m 100 cm × 3 2.2 lb 1 kg = 42.46 lbs . M1 = M2 M1 = M2 ρ2 ρ1 R2 R1 M1 = 2. M2 ρ1 ρ2 R1 R2 2 A1 2. = A2 3 R1 R2 R2 R1 3 2 A1 R2 =π A2 R1 R2 A1 = 4. A2 R1 3. 3 1. ρ2 = 3 ρ1 R1 3. ρ2 R2 003 (part 2 of 2) 10.0 points A1 of the circular areas What is the ratio A2 defined by the two equators? A1 1. =π A2 002 (part 1 of 2) 10.0 points Consider two planets with uniform mass distributions. The mass density and the radius of planet 1 are ρ1 and R1 , respectively, and those of planet 2 are ρ2 and R2 M1 What is the ratio of their masses? M2 4 3 π R1 3 4 3 π R2 3 ρ1 5. A1 = A2 A1 =π 6. A2 R1 R2 R1 R2 3 2 alexander (jra2623) – oldhomework 01 – Turner – (92510) 7. A1 = A2 8. A1 =π A2 A1 9. = A2 A1 = 10. A2 R2 R1 2 3 R2 R1 R1 R2 R1 R2 2 correct Explanation: A = π r 2 , so 2 2 A1 R1 πR1 = 2 = R2 . A2 πR2 2 004 10.0 points The distance between the sun and the Earth is about 1.5184 × 1011 m. Express this distance in kilometers. 1. None of these 2. 1.5184 × 108 km correct 3. 1.5184 × 109 km 4. 1.5184 × 106 km 5. 1.5184 × 1010 km 6. 1.5184 × 1011 km 7. 1.5184 × 107 km Explanation: 1 km = 1 × 103 m , so d = 1.5184 × 1011 m · = 1.5184 × 108 km . 1 km 1 × 103 m 2 ...
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