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Unformatted text preview: oldmidterm 01 – FIERRO, JEFFREY – Due: Feb 11 2008, 11:00 pm 1 Question 1, chap 1, sect 5. part 1 of 1 10 points A newly discovered Jupiterlike planet has an average radius 11 . 1times that of the Earth and a mass 314 times that of the Earth. Calculate the ratio of new planet’s mass density to the mass density of the Earth. Correct answer: 0 . 229594 (tolerance ± 1 %). Explanation: Let : R np = 11 . 1 R E and M np = 314 M E . The volume of a sphere of radius R is V = 4 3 π R 3 . The (average) density ρ of a body is the ratio of the body’s mass to its volume, ρ = M V . Planets are spherical, so the (average) density of a planet of a given mass M and a given radius R is ρ = M 4 3 π R 3 . Comparing the newly discovered planet to the Earth, we have ρ np ρ E = M np 4 3 π R 3 np M E 4 3 π R 3 E = M np M E parenleftbigg R np R E parenrightbigg 3 = 314 (11 . 1) 3 = . 229594 . Question 2, chap 1, sect 5. part 1 of 1 10 points A piece of wire has a density of 6 . 1 g / cm. What is the mass of 11 . 9 cm of the wire? Correct answer: 72 . 59 g (tolerance ± 1 %). Explanation: Let : ρ = 6 . 1 g / cm and ℓ = 11 . 9 cm . Since g cm · cm = g, we must multiply the den sity by the length of the wire to obtain the mass. This can also be determined by the definition of linear mass density: ρ = m ℓ , so that m = ρ ℓ = (6 . 1 g / cm) (11 . 9 cm) = 72 . 59 g . Question 3, chap 1, sect 6. part 1 of 1 10 points Identify the equation below which is dimen sionally incorrect . A, x, y and r have units of length. k here has units of inverse length. v and v have units of velocity. a and g have units of acceleration. ω has units of inverse time. t is time, m is mass, V is volume, ρ is density, and F is force. 1. v = radicalbig v 2 − 2 a x 2. g = F m + ρ m g V 3. y = A cos( k x − ω t ) 4. F = mg bracketleftbigg 1 + v r g bracketrightbigg correct 5. t = − v + radicalBig v 2 + 2 a x a + v 2 a 2 t 6. v = ± ω radicalbig A 2 − x 2 7. v = radicalbig 2 g y + radicalbigg r F m 8. F = mω 2 r oldmidterm 01 – FIERRO, JEFFREY – Due: Feb 11 2008, 11:00 pm 2 Explanation: • Check y = A cos( k x − ω t ): Since k is an inverse length, k x is dimension less, and similarly, ωt is also dimensionless, so k x − ω t is dimensionless, as an argument of a trigonometry function must always be. The value of a trigonometry function ( e.g. cos( k x − ω t )) is also dimensionless. So the right hand side has the dimension of length. [ y ] = L [ A cos( k x )] = L. This equation is dimensionally correct. • Check F = mω 2 r : The dimension of force is [ F ] = ML T 2 . The dimension of right hand side is bracketleftbig mω 2 r bracketrightbig = M parenleftbigg 1 T parenrightbigg 2 L = M L T 2 ....
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This homework help was uploaded on 04/03/2008 for the course PHY 303K taught by Professor Turner during the Spring '08 term at University of Texas.
 Spring '08
 Turner
 Mass

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