You are in charge of the evacuation and repair of a hemisphere storage tank with a radius measuring 10 ft. The tank is full of benzene, with a weight density of 56- lb ft³ a. How much work would it take to empty the tank? Answer: Round your answer for work to 2 decimals. b. A firm you contacted says it can empty the tank for $0.005 per ft-lb of work. How much will the bill. be for the job? Answer: S Units: Select an answer
You are in charge of the evacuation and repair of a hemisphere storage tank with a radius measuring 10 ft. The tank is full of benzene, with a weight density of 56- lb ft³ a. How much work would it take to empty the tank? Answer: Round your answer for work to 2 decimals. b. A firm you contacted says it can empty the tank for $0.005 per ft-lb of work. How much will the bill. be for the job? Answer: S Units: Select an answer
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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![### Problem: Work Required to Empty a Hemispherical Storage Tank
#### Scenario
You are in charge of the evacuation and repair of a hemispherical storage tank with a radius measuring 10 ft. The tank is full of benzene, with a weight density of 56 lb/ft³.
#### Questions
a. How much work would it take to empty the tank?
**Answer:**
\[ \boxed{\hspace{4cm}} \]
**Units:** \[ \boxed{\text{Select an answer}} \]
**Instruction:** Round your answer for work to 2 decimals.
b. A firm you contacted says it can empty the tank for $0.005 per ft·lb of work. How much will the bill be for the job?
**Answer:** \[ \$\boxed{\hspace{4cm}} \]
### Explanation
**Work Calculation Details:**
1. **Volume of the Hemispherical Tank:**
- Formula for the volume of a hemisphere:
\[
V = \frac{2}{3} \pi r^3
\]
- Given radius \( r = 10 \, \text{ft} \):
\[
V = \frac{2}{3} \pi (10)^3 = \frac{2}{3} \pi (1000) = \frac{2000}{3} \pi \, \text{ft}^3
\]
2. **Weight of Benzene:**
- Weight density of benzene \( = 56 \, \text{lb/ft}^3 \)
- Total weight of benzene = Volume \( \times \) weight density
\[
W = \frac{2000}{3} \pi \times 56 \, \text{lb}
\]
3. **Work to Empty the Tank:**
- Work required to empty the tank involves calculating the integral of the force exerted over the distance the liquid is lifted.
\[
W = \int_0^{10} \pi (10^2 - y^2) \cdot 56y \, dy
\]
**Cost Calculation Details:**
- Cost per unit of work \( = \$ 0.005 \, \text{per ft} \cdot \text{lb} \)
- Total cost \( = \text{Work}](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F2386874c-e724-4957-85a1-36ae7eec2eac%2Fedeced70-2d17-4e18-881f-fb4269750299%2F5rt1qli_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Problem: Work Required to Empty a Hemispherical Storage Tank
#### Scenario
You are in charge of the evacuation and repair of a hemispherical storage tank with a radius measuring 10 ft. The tank is full of benzene, with a weight density of 56 lb/ft³.
#### Questions
a. How much work would it take to empty the tank?
**Answer:**
\[ \boxed{\hspace{4cm}} \]
**Units:** \[ \boxed{\text{Select an answer}} \]
**Instruction:** Round your answer for work to 2 decimals.
b. A firm you contacted says it can empty the tank for $0.005 per ft·lb of work. How much will the bill be for the job?
**Answer:** \[ \$\boxed{\hspace{4cm}} \]
### Explanation
**Work Calculation Details:**
1. **Volume of the Hemispherical Tank:**
- Formula for the volume of a hemisphere:
\[
V = \frac{2}{3} \pi r^3
\]
- Given radius \( r = 10 \, \text{ft} \):
\[
V = \frac{2}{3} \pi (10)^3 = \frac{2}{3} \pi (1000) = \frac{2000}{3} \pi \, \text{ft}^3
\]
2. **Weight of Benzene:**
- Weight density of benzene \( = 56 \, \text{lb/ft}^3 \)
- Total weight of benzene = Volume \( \times \) weight density
\[
W = \frac{2000}{3} \pi \times 56 \, \text{lb}
\]
3. **Work to Empty the Tank:**
- Work required to empty the tank involves calculating the integral of the force exerted over the distance the liquid is lifted.
\[
W = \int_0^{10} \pi (10^2 - y^2) \cdot 56y \, dy
\]
**Cost Calculation Details:**
- Cost per unit of work \( = \$ 0.005 \, \text{per ft} \cdot \text{lb} \)
- Total cost \( = \text{Work}
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