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Get Full Access to University Physics - 13 Edition - Chapter 2 - Problem 47e
Get Full Access to University Physics - 13 Edition - Chapter 2 - Problem 47e

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# A 15-kg rock is dropped from rest on the earth and reaches

ISBN: 9780321675460 31

## Solution for problem 47E Chapter 2

University Physics | 13th Edition

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University Physics | 13th Edition

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Problem 47E

A 15-kg rock is dropped from rest on the earth and reaches the ground in 1.75 s. When it is dropped from the same height on Saturn’s satellite Enceladus, the rock reaches the ground in 18.6 s. What is the acceleration due to gravity on Enceladus?

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Step 1 of 3

Solution 47E Let the height of drop be h. For a body, dropped from rest, this height is given by 1 2 h = 2t g = 2h/t …..(1) Therefore, from the equation 1, acceleration due to gravity is inversely proportional to the square of time. Time for earth t = e.75 s Time for Enceladus t enc = 18.6 s 2 For earth, g = 2h/t e …..(2) For Enceladus, g enc = 2h/t enc2 …..(3) Dividing equation (2) by equation (3), 2 g/g enc = (tenc/te) g enc= g × (t et enc2 Substituting g , t and t values in this equation, e enc 2 2 g enc= 9.8 × (1.75/18.6) m/s g enc= 8.67 × 10 2 m/s 2 2 2 Therefore, the acceleration due to gravity on Enceladus is 8.67 × 10 m/s .

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