Practice 01 Solutions

Practice 01 Solutions - practice 01 – GADHIA, TEJAS –...

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Unformatted text preview: practice 01 – GADHIA, TEJAS – Due: Jan 22 2007, 5:00 pm 1 Question 1 part 1 of 2 10 points The radius of a small ball is around 3 . 62538 cm. The radius of a basketball is about 3 . 31 times larger. What is the ratio of the surface area of the small ball and a basketball? Correct answer: 0 . 0912734 (tolerance ± 1 %). Explanation: The surface area of a sphere is 4 π R 2 , so the area ratio is A small ;all A basketball = 4 π R 2 small ball 4 π R 2 basketball = R 2 small ball R 2 basketball = 1 3 . 31 2 = . 0912734 . Question 2 part 2 of 2 10 points What is the ratio of their volume? Correct answer: 0 . 027575 (tolerance ± 1 %). Explanation: The volume of a sphere is 4 3 π R 3 , so the volume ratio is V small ball V basketball = 4 3 π R 3 small ball 4 3 π R 3 basketball = R 3 small ball R 3 basketball = 1 3 . 31 3 = . 027575 . Question 3 part 1 of 1 10 points A high fountain of water is located at the center of a circular pool as in the figure. A student walks around the pool and estimates its circumference to be 144 m. Next, the student stands at the edge of the pool and uses a protractor to gauge the angle of elevation of the top of the fountain to be 46 . 7 ◦ . How high is the fountain? Correct answer: 24 . 3203 m (tolerance ± 1 %). Explanation: R h θ The circumference of the pool is C = 2 πR so its radius is R = C 2 π = 144 m 2 π = 22 . 9183 m The radius and the height of the fountain form the adjacent and opposite sides of the angle of elevation, respectively. Thus the tangent of the angle can be used, and tan θ = h R so that h = R tan θ = 24 . 3203 m Question 4 part 1 of 1 10 points Two ants race across a table 75 cm long. One travels at 5 . 59 cm / s and the other at 3 . 99999 cm / s. When the first one crosses the finish line, how far behind is the second one? practice 01 – GADHIA, TEJAS – Due: Jan 22 2007, 5:00 pm 2 Correct answer: 21 . 3329 cm (tolerance ± 1 %). Explanation: Let : ℓ = 75 cm , v 1 = 5 . 59 cm / s , and v 2 = 3 . 99999 cm / s . You first have to compute the time it takes the first (faster) ant to cross the finish line: t = ℓ v 1 . Then you must compute the distance the slower ant covers in that time s 2 = v 2 t = v 2 ℓ v 1 . The slower ant is ℓ- s 2 = ℓ- v 2 ℓ v 1 = (75 cm)- (3 . 99999 cm / s) (75 cm) (5 . 59 cm / s) = 21 . 3329 cm from the finish line when the faster one crosses it....
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This note was uploaded on 03/09/2011 for the course PHY 303K taught by Professor Turner during the Spring '08 term at University of Texas.

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Practice 01 Solutions - practice 01 – GADHIA, TEJAS –...

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