Practice Homework 1 Solutions

Practice Homework 1 Solutions - practice 01 – ALIBHAI...

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Unformatted text preview: practice 01 – ALIBHAI, ZAHID – Due: Jan 21 2007, 4:00 am 1 Question 1 part 1 of 1 10 points A newly discovered giant planet has an av- erage radius 19 times that of the Earth and a mass 201 times that of the Earth. Calculate the ratio of the new planet’s den- sity to the Earth’s density. Correct answer: 0 . 0293046 (tolerance ± 1 %). Explanation: Let : R n = 19 R E and m n = 201 m E . Density is the ratio of mass to volume, ρ = m V . A spherical planet of average radius R has volume 4 3 π R 3 and hence density ρ = m 4 3 π R 3 . For two planets of respective radii R 1 and R 2 and masses m 1 and m 2 we have ρ 1 ρ 2 = m 1 4 3 π R 3 1 m 2 4 3 π R 3 2 = parenleftbigg m 1 m 2 parenrightbigg parenleftbigg R 1 R 2 parenrightbigg 3 = 201 (19) 3 = . 0293046 . Question 2 part 1 of 1 10 points A sphere of metal has a radius of 6 . 3 cm and a density of 6 . 88 g / cm 3 . What is the mass of the sphere? Correct answer: 7206 . 07 g (tolerance ± 1 %). Explanation: Let : r = 6 . 3 cm and ρ = 6 . 88 g / cm 3 . Density is mass per unit volume, so ρ = m V m = ρ V = ρ 4 3 π r 3 = ( 6 . 88 g / cm 3 ) 4 3 π (6 . 3 cm) 3 = 7206 . 07 g . Question 3 part 1 of 1 10 points A cylinder, 16 cm long and 8 cm in radius, is made of two different metals bonded end- to-end to make a single bar. The densities are 4 . 6 g / cm 3 and 6 . 1 g / cm 3 . 1 6 c m 8 cm What length of the lighter metal is needed if the total mass is 17741 g? Correct answer: 6 . 24234 cm (tolerance ± 1 %). Explanation: Let : ℓ = 16 cm , r = 8 cm , ρ 1 = 4 . 6 g / cm 3 , ρ 2 = 6 . 1 g / cm 3 , and m = 17741 g . practice 01 – ALIBHAI, ZAHID – Due: Jan 21 2007, 4:00 am 2 Volume of a bar of radius r and length ℓ is V = π r 2 ℓ and its density is ρ = m V = m π r 2 ℓ so that m = ρ π r 2 ℓ ℓ x ℓ − x r Let x be the length of the lighter metal; then ℓ − x is the length of the heavier metal. Thus, m = m 1 + m 2 = ρ 1 π r 2 x + ρ 2 π r 2 ( ℓ − x ) = ρ 1 π r 2 x + ρ 2 π r 2 ℓ − ρ 2 π r 2 x . Therefore m − ρ 2 π r 2 ℓ = ρ 1 π r 2 x − ρ 2 π r 2 x and x π r 2 ( ρ 1 − ρ 2 ) = m − ρ 2 π r 2 ℓ . Consequently, x = m − ρ 2 π r 2 ℓ π r 2 ( ρ 1 − ρ 2 ) = (17741 g) − (6 . 1 g / cm 3 ) π (8 cm) 2 (16 cm) π (8 cm) 2 (4 . 6 g / cm 3 − 6 . 1 g / cm 3 ) = 6 . 24234 cm . Question 4 part 1 of 4 10 points Consider the following accepted constants: Radius of the moon=1 . 74 × 10 6 m. Radius of the sun=6 . 96 × 10 8 m. Average moon-earth distance=3 . 84 × 10 8 m. Average sun-earth distance=1 . 496 × 10 11 m. a) What is the apparent angle the diameter of the moon subtends, as seen from the earth, in radians? Correct answer: 0 . 0090625 rad (tolerance ± 1 %)....
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This note was uploaded on 10/13/2009 for the course PHY 303K taught by Professor Turner during the Spring '08 term at University of Texas.

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Practice Homework 1 Solutions - practice 01 – ALIBHAI...

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