Kaplunovsky - hw 09

Kaplunovsky - hw 09 - homework 09 Due 4:00 am Question 1...

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homework 09 – – Due: Mar 29 2007, 4:00 am 1 Question 1, chap 10, sect 2. part 1 of 1 10 points Consider a 31 m high pyramide with a square base 48 m by 48 m. The pyramid is made of tightly-Ftting stones of uniform density 2730 kg / m 3 . Calculate the net potential energy of this pyramide. Use g = 9 . 8 m / s 2 and count height from the pyramid’s base. Correct answer: 4 . 93643 × 10 9 J (tolerance ± 1 %). Explanation: Gravitational potential energy of a contin- uous body is U = MgH cm (1) where M is the net mass of the body and H cm is the height of its center of mass. ±or a pyramid of uniform density, H cm = 1 4 × h, (2) while the mass is M = ρ × Volume = ρ × L 2 h 3 , (4) therefore U = ρ × L 2 h 3 × g × h 4 = ρ g L 2 h 2 12 = 4 . 93643 × 10 9 J . (5) Alternative solution: Another way to calculate the net potential energy is to integrate: U = i gz dm = iii gzρ dxdy dz (6) Note that the ranges of the integrals over x and y depend on z (and generally on each other), which makes for a very complicated triple integral. However, for constant density ρ and constant g , the integral simpliFes to U = i z A ( z ) dz (7) where A ( z ) is the area of the horizontal cross- section of the body at height z . ±or the pyramid, the cross-section is a square with a side ( z ) = L × p 1 z h P , (8) hence A ( z ) = L 2 × p 1 z h P 2 . (9) Substituting this formula into the integral (7) gives U = × h i 0 z × L 2 p 1 z h P 2 dz = gρL 2 × h i 0 ± z 2 h × z 2 + 1 h 2 × z 3 ² = gρL 2 × h i 0 ± h 2 2 2 h × h 3 3 + 1 h 2 × h 4 4 ² = gρL 2 × h 2 12 = 4 . 93643 × 10 9 J . (10) Question 2, chap 10, sect 1. part 1 of 1 8 points A(n) 53 . 3 kg astronaut becomes separated from the shuttle, while on a space walk. She Fnds herself 55 m away from the shuttle and moving with zero speed relative to the shuttle. She has a(n) 0 . 733 kg camera in her hand and decides to get back to the shuttle by throwing the camera at a speed of 12 m / s in the direction away from the shuttle. How long will it take for her to reach the shuttle? Answer in minutes. Correct answer: 5 . 55462 min (tolerance ± 1 %). Explanation:

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homework 09 – – Due: Mar 29 2007, 4:00 am 2 Let : M = 53 . 3 kg , m = 0 . 733 kg , d = 55 m , and v = 12 m / s . Because of conservation of linear momentum, we have 0 = M V + m ( v ) mv = M V where V is the velocity of the astronaut and it has a direction toward the shuttle. V = mv M = (0 . 733 kg) (12 m / s) 53 . 3 kg = 0 . 165028 m / s . And the time it takes for her to reach the shuttle t = d V = 55 m 0 . 165028 m / s · 1 min 60 s = 5 . 55462 min . Question 3, chap 11, sect 2. part 1 of 1 8 points A large 2 . 22 kg bird, searching for food, spots a large, 0 . 32 kg bug hovering in its vicinity. If the bird is traveling at 8 . 5 m / s, what will be his velocity immediately after nabbing the bug?
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Kaplunovsky - hw 09 - homework 09 Due 4:00 am Question 1...

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