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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Question
![**Problem Statement:**
Water flows out of a spigot at the bottom of a very wide storage tank at a velocity of 10.4 m/s. What is the depth of this storage tank?
**Analysis:**
In fluid dynamics, the velocity of fluid flowing out of an opening can be related to the depth of the fluid above the opening using Torricelli’s Law. This states that the speed \( v \) of efflux is given by:
\[ v = \sqrt{2gh} \]
where:
- \( v \) is the exit velocity (10.4 m/s in this case),
- \( g \) is the acceleration due to gravity (approximately 9.81 m/s² on Earth),
- \( h \) is the height (or depth) of the fluid above the opening.
**Solution:**
Rearrange the formula to solve for \( h \):
\[ h = \frac{v^2}{2g} \]
Substitute the known values:
\[ h = \frac{(10.4)^2}{2 \times 9.81} \]
\[ h = \frac{108.16}{19.62} \]
\[ h \approx 5.51 \, \text{meters} \]
Therefore, the depth of the storage tank is approximately 5.51 meters.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F2b632062-7ccc-4dd3-a093-b9e3993697da%2F80e60b49-1d36-4535-a37c-9c9e1dfc8173%2F1pof3de_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Problem Statement:**
Water flows out of a spigot at the bottom of a very wide storage tank at a velocity of 10.4 m/s. What is the depth of this storage tank?
**Analysis:**
In fluid dynamics, the velocity of fluid flowing out of an opening can be related to the depth of the fluid above the opening using Torricelli’s Law. This states that the speed \( v \) of efflux is given by:
\[ v = \sqrt{2gh} \]
where:
- \( v \) is the exit velocity (10.4 m/s in this case),
- \( g \) is the acceleration due to gravity (approximately 9.81 m/s² on Earth),
- \( h \) is the height (or depth) of the fluid above the opening.
**Solution:**
Rearrange the formula to solve for \( h \):
\[ h = \frac{v^2}{2g} \]
Substitute the known values:
\[ h = \frac{(10.4)^2}{2 \times 9.81} \]
\[ h = \frac{108.16}{19.62} \]
\[ h \approx 5.51 \, \text{meters} \]
Therefore, the depth of the storage tank is approximately 5.51 meters.
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