We need to show that a. The water level at time t is y ( t ) = ( y 0 3 2 − 3 a B 2 g 2 π t ) 2 3 where y 0 is the water level at time t = 0 b. If the total volume of water in the tank has volume V at time t = 0 , then y 0 = 2 a V π . c. The tank is empty at time t e = ( 2 3 B g ) ( 2 π V 3 a ) 1 4 .

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Answer 1

Question

Chapter 10.1, Problem 49E

To determine

We need to show that

a. The water level at time t is $y (t) = {(y_{0}^{\frac{3}{2}} - \frac{3 a B \sqrt{2 g}}{2 π} t)}^{\frac{2}{3}}$ where $y_{0}$ is the water level at time $t = 0$

b. If the total volume of water in the tank has volume V at time $t = 0$ , then $y_{0} = \sqrt{\frac{2 a V}{π}}$ .

c. The tank is empty at time $t_{e} = (\frac{2}{3 B \sqrt{g}}) {(\frac{2 π V^{3}}{a})}^{\frac{1}{4}}$ .

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Answer 2

Question

Chapter 10.1, Problem 49E

To determine

We need to show that

a. The water level at time t is $y (t) = {(y_{0}^{\frac{3}{2}} - \frac{3 a B \sqrt{2 g}}{2 π} t)}^{\frac{2}{3}}$ where $y_{0}$ is the water level at time $t = 0$

b. If the total volume of water in the tank has volume V at time $t = 0$ , then $y_{0} = \sqrt{\frac{2 a V}{π}}$ .

c. The tank is empty at time $t_{e} = (\frac{2}{3 B \sqrt{g}}) {(\frac{2 π V^{3}}{a})}^{\frac{1}{4}}$ .

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Answer 3

Question

Chapter 10.1, Problem 49E

To determine

We need to show that

a. The water level at time t is $y (t) = {(y_{0}^{\frac{3}{2}} - \frac{3 a B \sqrt{2 g}}{2 π} t)}^{\frac{2}{3}}$ where $y_{0}$ is the water level at time $t = 0$

b. If the total volume of water in the tank has volume V at time $t = 0$ , then $y_{0} = \sqrt{\frac{2 a V}{π}}$ .

c. The tank is empty at time $t_{e} = (\frac{2}{3 B \sqrt{g}}) {(\frac{2 π V^{3}}{a})}^{\frac{1}{4}}$ .

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Answer 4

Question

Chapter 10.1, Problem 49E

To determine

We need to show that

a. The water level at time t is $y (t) = {(y_{0}^{\frac{3}{2}} - \frac{3 a B \sqrt{2 g}}{2 π} t)}^{\frac{2}{3}}$ where $y_{0}$ is the water level at time $t = 0$

b. If the total volume of water in the tank has volume V at time $t = 0$ , then $y_{0} = \sqrt{\frac{2 a V}{π}}$ .

c. The tank is empty at time $t_{e} = (\frac{2}{3 B \sqrt{g}}) {(\frac{2 π V^{3}}{a})}^{\frac{1}{4}}$ .

Expert Solution & Answer

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Answer 5

Chapter 10.1, Problem 49E

To determine

We need to show that

a. The water level at time t is $y (t) = {(y_{0}^{\frac{3}{2}} - \frac{3 a B \sqrt{2 g}}{2 π} t)}^{\frac{2}{3}}$ where $y_{0}$ is the water level at time $t = 0$

b. If the total volume of water in the tank has volume V at time $t = 0$ , then $y_{0} = \sqrt{\frac{2 a V}{π}}$ .

c. The tank is empty at time $t_{e} = (\frac{2}{3 B \sqrt{g}}) {(\frac{2 π V^{3}}{a})}^{\frac{1}{4}}$ .

Answer 6

Chapter 10.1, Problem 49E

To determine

We need to show that

a. The water level at time t is $y (t) = {(y_{0}^{\frac{3}{2}} - \frac{3 a B \sqrt{2 g}}{2 π} t)}^{\frac{2}{3}}$ where $y_{0}$ is the water level at time $t = 0$

b. If the total volume of water in the tank has volume V at time $t = 0$ , then $y_{0} = \sqrt{\frac{2 a V}{π}}$ .

c. The tank is empty at time $t_{e} = (\frac{2}{3 B \sqrt{g}}) {(\frac{2 π V^{3}}{a})}^{\frac{1}{4}}$ .

Answer 7

Chapter 10.1, Problem 49E

To determine

We need to show that

a. The water level at time t is $y (t) = {(y_{0}^{\frac{3}{2}} - \frac{3 a B \sqrt{2 g}}{2 π} t)}^{\frac{2}{3}}$ where $y_{0}$ is the water level at time $t = 0$

b. If the total volume of water in the tank has volume V at time $t = 0$ , then $y_{0} = \sqrt{\frac{2 a V}{π}}$ .

c. The tank is empty at time $t_{e} = (\frac{2}{3 B \sqrt{g}}) {(\frac{2 π V^{3}}{a})}^{\frac{1}{4}}$ .

Answer 8

Expert Solution & Answer

Answer 9

Expert Solution & Answer

Answer 10

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Answer 11

Ch. 10.1 - Prob. 1PQ Ch. 10.1 - Prob. 2PQ Ch. 10.1 - Prob. 3PQ Ch. 10.1 - Prob. 4PQ Ch. 10.1 - Prob. 1E Ch. 10.1 - Prob. 2E Ch. 10.1 - Prob. 3E Ch. 10.1 - Prob. 4E Ch. 10.1 - Prob. 5E Ch. 10.1 - Prob. 6E

Solution Summary: The author shows the water level at time t is y (t ) = (y 0 3 2 3 a B 2

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Solution Summary: The author shows the water level at time t is y (t ) = (y 0 3 2 3 a B 2

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We need to show that a. The water level at time t is y(t)=(y032−3aB2g2πt)23 where y0 is the water level at time t=0 b. If the total volume of water in the tank has volume V at time t=0, then y0=2aVπ. c. The tank is empty at time te=(23Bg)(2πV3a)14.

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