OLDHW37 - oldhomewk 37 PAPAGEORGE MATT Due 4:00 am Question...

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oldhomewk 37 – PAPAGEORGE, MATT – Due: Apr 28 2008, 4:00 am 1 Question 1, chap 18, sect 3. part 1 of 1 10 points In testing a new material for shielding spacecraft, 149 small ball bearings, each mov- ing at a supersonic speed of 415.7 m/s, collide head-on and elastically with the material dur- ing a 0.590 min interval. If the bearings each have a mass of 8.6 g and the area of the tested material is 0.74 m 2 , what pressure is exerted on the material? Correct answer: 40 . 6687 Pa (tolerance ± 1 %). Explanation: Basic Concepts: P = F A Conservation of momentum: Δ p = F Δ t Elastic collision: v f = - v i Given: Δ t = 0 . 590 min N = 149 m = 8 . 6 g v = 415 . 7 m / s A = 0 . 74 m 2 Solution: Δ p = m v f - m v i = m ( - v i ) - m v i = - 2 m v i and F = Δ p Δ t For one bearing, P = vextendsingle vextendsingle vextendsingle Δ p Δ t vextendsingle vextendsingle vextendsingle A = 2 m v A Δ t For all of the bearings, P = N parenleftbigg 2 m v A Δ t parenrightbigg = 2(149)(8 . 6 g)(415 . 7 m / s) (0 . 74 m 2 )(0 . 59 min) · parenleftbigg 1 min 60 s parenrightbigg parenleftbigg 1 kg 1000 g parenrightbigg = 40 . 6687 Pa Question 2, chap 18, sect 4. part 1 of 2 10 points A simple U-tube that is open at both ends is partially filled with a liquid of density 614 kg / m 3 . Water is then poured into one arm of the tube, forming a column 15 cm in height, as shown in the following diagram. The density of heavy liquid is 1000 kg / m 3 . h 1 15 cm light liquid 614 kg / m 3 heavy liquid 1000 kg / m 3 What is the difference, h 1 , in the heights of the two liquid surfaces? Correct answer: 9 . 42997 cm (tolerance ± 1 %). Explanation: Basic Concepts: gauge pressure, varia- tion of pressure with depth Because the liquid in the U-tube is static, the pressure exerted by the heavy liquid column of height in the right branch of the tube must balance the pressure exerted by the liquid of height h 1 + in the left branch. Therefore, P 0 + ( + h 1 ) ρ g = P 0 + ℓ ρ h g .
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oldhomewk 37 – PAPAGEORGE, MATT – Due: Apr 28 2008, 4:00 am 2 Solving for h 1 , h 1 = bracketleftbigg ρ h ρ - 1 bracketrightbigg
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