# 9 - homework 09 – JACOBS AMBER – Due Apr 8 2008 1:00 am...

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Unformatted text preview: homework 09 – JACOBS, AMBER – Due: Apr 8 2008, 1:00 am 1 Question 1, chap 14, sect 7. part 1 of 2 10 points A rock weighs 110 N in air and has a volume of 0 . 0029 m 3 . The acceleration of gravity is 9 . 8 m / s 2 . What is its apparent weight when sub- merged in water? Correct answer: 81 . 58 N (tolerance ± 1 %). Explanation: The buoyant force for any liquid is defined to be F = ρV g but it is equal to the weight lost by the object, so W − W a = ρV g W a = W − ρV g Dimensional analysis for F : kg m 3 · m 3 · m s 2 = kg · m s 2 = N apparent weight is W a = W − ρ w V g Question 2, chap 14, sect 7. part 2 of 2 10 points If it is submerged in a liquid with a density exactly 1 . 9 times that of water, what will be its new apparent weight? Correct answer: 56 . 002 N (tolerance ± 1 %). Explanation: liquid, the apparent weight is W a 2 = W − ρ n V g Question 3, chap 14, sect 10. part 1 of 4 10 points Assume: The fluid is incompressible and non-viscous. Shown below is a cross-section of a ver- tical view of a pipe discharging a fluid into the atmosphere at its highest elevation. The pipe diameter increases and then remains con- stant. P i is the pressure and bardbl vectorv i bardbl is the speed of the fluid, at locations i = w , u , z , and y . w u z y The relationship between the magnitude of the velocity v ≡ bardbl vectorv bardbl at position w and u is 1. bardbl vectorv u bardbl = bardbl vectorv w bardbl . 2. bardbl vectorv u bardbl > bardbl vectorv w bardbl . 3. bardbl vectorv u bardbl < bardbl vectorv w bardbl . correct 4. indeterminable, not enough information available. Explanation: The fluid is incompressible, so bardbl vectorv w bardbl > bardbl vectorv u bardbl . Question 4, chap 14, sect 10. part 2 of 4 10 points The relationship between the pressure P at position u and z is 1. P z = P u . 2. P z > P u . correct 3. P z < P u . 4. indeterminable, not enough information available. Explanation: Bernoulli’s equation P + 1 2 ρv 2 + ρg y = constant . At a lower elevation the pressure Δ P = ρg y is greater. Thus, P z > P u . homework 09 – JACOBS, AMBER – Due: Apr 8 2008, 1:00 am 2 Question 5, chap 14, sect 10. part 3 of 4 10 points The relationship between the pressure P at position w and y is 1. P y < P w . 2. P y > P w . 3. indeterminable, not enough information available. correct 4. P y = P w . Explanation: The pressure at w and y may be compa- rable. Not enough information is available to determine which is larger, if they are not equal. Both P w < P u and P y < P u , see Parts 4 & 2. Question 6, chap 14, sect 10. part 4 of 4 10 points The relationship between the pressure P at position w and u is 1. P u < P w . 2. P u > P w . correct 3. P u = P w ....
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## This note was uploaded on 04/28/2008 for the course PHY 317k taught by Professor Kopp during the Spring '07 term at University of Texas.

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9 - homework 09 – JACOBS AMBER – Due Apr 8 2008 1:00 am...

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