The earth's orbital radius is about 1.5 x 10 m. At this distance from the sun, the energy flux of sunlight is about 1.4kW/m². How much mass does the sun radiate as light in one second? ) A student pedaling a bicycle full throttle produces about one-half horsepower of useful power. (1 hp is equivalent to 746 W.) The human body is about 25 percent efficient: 75 percent of the food burned is converted to heat and 25 percent is converted to useful work. How long will the student have to ride to lose one pound by conversion of mass to energy? How come you do not have to exercise this long to lose a pound?
The earth's orbital radius is about 1.5 x 10 m. At this distance from the sun, the energy flux of sunlight is about 1.4kW/m². How much mass does the sun radiate as light in one second? ) A student pedaling a bicycle full throttle produces about one-half horsepower of useful power. (1 hp is equivalent to 746 W.) The human body is about 25 percent efficient: 75 percent of the food burned is converted to heat and 25 percent is converted to useful work. How long will the student have to ride to lose one pound by conversion of mass to energy? How come you do not have to exercise this long to lose a pound?
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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How do I use E = MC^2 to solve these questions?
![### Educational Content: Exercise and Energy
#### Problem Statement:
**(b)**
The Earth's orbital radius is about \(1.5 \times 10^{11} \, \text{m}\). At this distance from the sun, the energy flux of sunlight is about \(1.4 \, \text{kW/m}^2\). How much mass does the sun radiate as light in one second?
---
**(c)**
A student pedaling a bicycle full throttle produces about one-half horsepower of *useful* power. (1 hp is equivalent to 746 W.) The human body is about 25 percent efficient: 75 percent of the food burned is converted to heat and 25 percent is converted to useful work.
- How long will the student have to ride to lose one pound by conversion of mass to energy?
- How come you do not have to exercise this long to lose a pound?
#### Detailed Explanation:
**(b)**
For part (b), you need to determine the mass energy conversion for the Sun’s radiated energy.
1. **Calculate the total power output of the Sun:**
- The energy flux (\(1.4 \, \text{kW/m}^2\)) at the Earth's distance can be used with the surface area of a sphere (with radius being Earth's distance from the Sun) to find the total power.
- The surface area of the sphere: \[ A = 4 \pi R^2 \]
- \( R = 1.5 \times 10^{11} \, \text{m} \)
- Total power \( P = 1.4 \times 10^3 \, \text{W/m}^2 \times 4 \pi (1.5 \times 10^{11} \, \text{m})^2 \)
2. **Relate power to energy:**
- Energy radiated per second (\(E\)) is equal to the power output.
3. **Convert energy to mass using Einstein’s equation \(E = mc^2\):**
- Solve for the mass \(m\): \( m = \frac{E}{c^2} \)
- \( c \) is the speed of light \( (3 \times 10^8 \, \text{m/s}) \)
**(c)**
For part (c](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F2c1389c8-c38b-4f11-a551-6266360614a0%2F1ebd186b-1d6b-4c21-8f2b-2c6b3970703d%2Fnok7gx_processed.png&w=3840&q=75)
Transcribed Image Text:### Educational Content: Exercise and Energy
#### Problem Statement:
**(b)**
The Earth's orbital radius is about \(1.5 \times 10^{11} \, \text{m}\). At this distance from the sun, the energy flux of sunlight is about \(1.4 \, \text{kW/m}^2\). How much mass does the sun radiate as light in one second?
---
**(c)**
A student pedaling a bicycle full throttle produces about one-half horsepower of *useful* power. (1 hp is equivalent to 746 W.) The human body is about 25 percent efficient: 75 percent of the food burned is converted to heat and 25 percent is converted to useful work.
- How long will the student have to ride to lose one pound by conversion of mass to energy?
- How come you do not have to exercise this long to lose a pound?
#### Detailed Explanation:
**(b)**
For part (b), you need to determine the mass energy conversion for the Sun’s radiated energy.
1. **Calculate the total power output of the Sun:**
- The energy flux (\(1.4 \, \text{kW/m}^2\)) at the Earth's distance can be used with the surface area of a sphere (with radius being Earth's distance from the Sun) to find the total power.
- The surface area of the sphere: \[ A = 4 \pi R^2 \]
- \( R = 1.5 \times 10^{11} \, \text{m} \)
- Total power \( P = 1.4 \times 10^3 \, \text{W/m}^2 \times 4 \pi (1.5 \times 10^{11} \, \text{m})^2 \)
2. **Relate power to energy:**
- Energy radiated per second (\(E\)) is equal to the power output.
3. **Convert energy to mass using Einstein’s equation \(E = mc^2\):**
- Solve for the mass \(m\): \( m = \frac{E}{c^2} \)
- \( c \) is the speed of light \( (3 \times 10^8 \, \text{m/s}) \)
**(c)**
For part (c
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