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sol01 - Question 1 chap 1 sect 1 part 1 of 1 10 points A...

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Question 1, chap 1, sect 1. part 1 of 1 10 points A newly discovered Jupiter-like planet has an average radius 11 . 2 times that of the Earth and a mass 313 times that of the Earth. Calculate the ratio of new planet’s mass density to the mass density of the Earth. Correct answer: 0 . 222787 (tolerance ± 1 %). Explanation: Let : R np = 11 . 2 R E and M np = 313 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 = 313 (11 . 2) 3 = 0 . 222787 . Question 2, chap 1, sect 1. part 1 of 1 10 points A spherical balloon with radius r inches has volume 4 3 πr 3 . Find a function that represents the amount of air required to inflate the balloon from a radius of r inches to a radius of r + 5 inches. 1. 4 3 π ( r + 5) 3 2. 4 3 π ( 75 r + 5 r 2 + 125 ) 3. 4 3 πr 3 4. 4 3 π ( 25 r + 15 r 2 + 125 ) 5. 4 3 π ( 75 r + 15 r 2 + 125 ) correct Explanation: Question 3, chap 1, sect 2. part 1 of 1 10 points What is the mass of water required to fill a circular hot tub 4 m in diameter and 1 . 9 m deep? Correct answer: 23864 kg (tolerance ± 1 %). Explanation: Let : d = 4 m , l = 1 . 9 m , and ρ = 1000 kg / m 3 . The volume of the tub is V =( πR 2 ) l = π d 2 4 l = π (4 m) 2 4 1 . 9 m = 23 . 864 m 3 . The mass is m = ρ V = (1000 kg / m 3 ) (23 . 864 m 3 ) = 23864 kg .
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Question 4, chap 1, sect 3. 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 0 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. F = m g bracketleftbigg 1 + v r g bracketrightbigg correct 2. t = - v 0 + radicalBig v 2 0 + 2 a x a + v 2 a 2 t 3. g = F m + ρ m g V 4. F = m ω 2 r 5. v = radicalbig 2 g y + radicalbigg r F m 6. y = A cos( k x - ω t ) 7. v = ± ω radicalbig A 2 - x 2 8. v 0 = radicalbig v 2 - 2 a x 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 = m ω 2 r : The dimension of force is [ F ] = ML T 2 . The dimension of right hand side is bracketleftbig m ω 2 r bracketrightbig = M parenleftbigg 1 T parenrightbigg 2 L = M L T 2 . So it is also dimensionally correct. Check v = radicalbig 2 g y + radicalbigg r F m : Since bracketleftbigg F m bracketrightbigg = L T 2 = [ g ], and [ y ] = [ r ] = L , bracketleftBig radicalbig 2 g y bracketrightBig = bracketleftBigg radicalbigg r F m bracketrightBigg = radicalbigg L 2 T 2 = L T .
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