/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 94 (a) How much work does a major-l... [FREE SOLUTION] | 91Ó°ÊÓ

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(a) How much work does a major-league pitcher do on the baseball when he throws a \(90.0 \mathrm{mi} / \mathrm{h}(40.2 \mathrm{m} / \mathrm{s})\) fastball? The mass of a baseball is 153 g. (b) How many fastballs would a pitcher have to throw to "burn off' a 1520 -Calorie meal? (1 Calorie \(=1000\) cal \(=1\) kcal.)Assume that \(80.0 \%\) of the chemical energy in the food is converted to thermal energy and only \(20.0 \%\) becomes the kinetic energy of the fastballs.

Short Answer

Expert verified
Answer: The pitcher would have to throw 10272 fastballs to burn off the 1520-Calorie meal.

Step by step solution

01

Convert the mass of the baseball to kg

Given the mass of the baseball as 153 g, we need to convert the mass to kg. 1 kg = 1000 g. So, \(m = \frac{153}{1000} = 0.153 \mathrm{kg}\).
02

Calculate the final kinetic energy of the fastball

The final kinetic energy of the baseball is given by the formula \(K = \frac{1}{2}mv^2\), where m is the mass of the baseball, and v is its final speed. Here, \(m = 0.153\mathrm{kg}\), and \(v = 40.2 \mathrm{m/s}\). So, \(K = \frac{1}{2}(0.153 \mathrm{kg})(40.2 \mathrm{m/s})^2 = 123.9141 \mathrm{J}\).
03

Calculate the work done on the baseball by the pitcher

Since the work done on the baseball is equal to the change in kinetic energy, and the initial kinetic energy is zero (the baseball is initially at rest), the work done by the pitcher is equal to the final kinetic energy. Thus, \(W = 123.9141 \mathrm{J}\).
04

Convert the energy from the meal to energy for fastballs

The energy from a 1520-Calorie meal needs to be converted to the energy for fastballs. First, we need to convert Calories to joules. We know that 1 Calorie = 1000 cal = 1 kcal, and 1 cal = 4.184 J. So, \(E_\text{meal} = 1520 \cdot 1000 \cdot 4.184 \mathrm{J} = 6364800 \mathrm{J}\). Since only 20% of this energy goes into the kinetic energy of fastballs, we have: \(E_\text{fastballs} = 0.20 \times 6364800 \mathrm{J} = 1272960 \mathrm{J}\).
05

Calculate the number of fastballs required to burn off the meal

To find out how many fastballs a pitcher needs to throw to burn off this meal, we need to divide the total energy spent on the fastballs (\(E_\text{fastballs}\)) by the work done on one fastball (\(W\)). Thus, \(N = \frac{1272960 \mathrm{J}}{123.9141 \mathrm{J/fastball}} = 10271.73 \mathrm{fastballs}\). Since we cannot have a fraction of a fastball, we round this up to get the final answer: \(N = 10272\) fastballs.

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