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 .

Question

Want to see the full answer?

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 $t_{e} = (\frac{2}{3 B \sqrt{g}}) {(\frac{2 π V^{3}}{a})}^{\frac{1}{4}}$ .

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Students have asked these similar questions

Solve the next ED: (see image)

Write an equation for the polynomial graphed below. It will probably be easiest to leave your "a" value as a fraction. 8 7 + 9+ H 6 5 4 3 + 3 2 1 (-30) (-1,0) (1,0) (3,0) + -5 -4 -3 -2 2 3 4 7 2 -1 -2 3 (0,-3) f(x) = 456 -4 -5 -6+

Write an equation for the polynomial graphed below 5+ 4 - 3 2 1 + + -5 4-3 -2 -1 1 2 3 4 5 -1 -2 y(x) = -3 -4 5 -5+ Q

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 $t_{e} = (\frac{2}{3 B \sqrt{g}}) {(\frac{2 π V^{3}}{a})}^{\frac{1}{4}}$ .

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

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 $t_{e} = (\frac{2}{3 B \sqrt{g}}) {(\frac{2 π V^{3}}{a})}^{\frac{1}{4}}$ .

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

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

Want to see the full answer?

Check out a sample textbook solution

See solution

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

Want to see the full answer?

Check out a sample textbook solution

See solution

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

Concept explainers

Videos

Want to see the full answer?

Chapter 10 Solutions

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

Concept explainers

Videos

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.

Want to see the full answer?

Chapter 10 Solutions