the power output in watts and horsepower of a shot-putter who takes 1.20 s to accelerate the 7.27-kg shot from rest to 14.0 m/s, while raising it 0.800 m. (Do not include the power produced to accelerate his body.)
Step-by-step solution Step 1 of 5 The equation to find the power is, Here is the power, is the work done and is the time period. Step 2 of 5 The work done by the power shot putter is equals the sum of change in kinetic energy and potential energy of the shot. Therefore, Here mass of he shot, is the velocity of the shot, is the height attained by the shot and is the work done by the shot putter. Step 3 of 5 Use the above formula to find the power,
Textbook: Physics: Principles with Applications
Author: Douglas C. Giancoli
Physics: Principles with Applications was written by and is associated to the ISBN: 9780130606204. This full solution covers the following key subjects: accelerate, shot, power, produced, include. This expansive textbook survival guide covers 35 chapters, and 3914 solutions. Since the solution to 46PE from 7 chapter was answered, more than 333 students have viewed the full step-by-step answer. This textbook survival guide was created for the textbook: Physics: Principles with Applications, edition: 6. The full step-by-step solution to problem: 46PE from chapter: 7 was answered by , our top Physics solution expert on 03/03/17, 03:53PM. The answer to “the power output in watts and horsepower of a shot-putter who takes 1.20 s to accelerate the 7.27-kg shot from rest to 14.0 m/s, while raising it 0.800 m. (Do not include the power produced to accelerate his body.)” is broken down into a number of easy to follow steps, and 39 words.