(a) d y d t = − 2 g B y π ( 2 R y − y 2 ) (b) For some constant C, 2 π 15 B 2 g ( 10 R y 3 / 2 − 3 y 5 / 2 ) = C − t (c) C = t e , the time at which the tank is empty. (d) That t e is proportional to R 5 / 2 and inversely proportional to B .

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Answer 1

Question

Chapter 10.1, Problem 69E

To determine

(a) $\frac{d y}{d t} = - \frac{\sqrt{2 g} B \sqrt{y}}{π (2 R y - y^{2})}$

(b) For some constant C, $\frac{2 π}{15 B \sqrt{2 g}} (10 R y^{3 / 2} - 3 y^{5 / 2}) = C - t$

(c) $C = t_{e}$ , the time at which the tank is empty.

(d) That $t_{e}$ is proportional to $R^{5 / 2}$ and inversely proportional to B.

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Answer 2

Question

Chapter 10.1, Problem 69E

To determine

(a) $\frac{d y}{d t} = - \frac{\sqrt{2 g} B \sqrt{y}}{π (2 R y - y^{2})}$

(b) For some constant C, $\frac{2 π}{15 B \sqrt{2 g}} (10 R y^{3 / 2} - 3 y^{5 / 2}) = C - t$

(c) $C = t_{e}$ , the time at which the tank is empty.

(d) That $t_{e}$ is proportional to $R^{5 / 2}$ and inversely proportional to B.

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See solution

Answer 3

Question

Chapter 10.1, Problem 69E

To determine

(a) $\frac{d y}{d t} = - \frac{\sqrt{2 g} B \sqrt{y}}{π (2 R y - y^{2})}$

(b) For some constant C, $\frac{2 π}{15 B \sqrt{2 g}} (10 R y^{3 / 2} - 3 y^{5 / 2}) = C - t$

(c) $C = t_{e}$ , the time at which the tank is empty.

(d) That $t_{e}$ is proportional to $R^{5 / 2}$ and inversely proportional to B.

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Answer 4

Question

Chapter 10.1, Problem 69E

To determine

(a) $\frac{d y}{d t} = - \frac{\sqrt{2 g} B \sqrt{y}}{π (2 R y - y^{2})}$

(b) For some constant C, $\frac{2 π}{15 B \sqrt{2 g}} (10 R y^{3 / 2} - 3 y^{5 / 2}) = C - t$

(c) $C = t_{e}$ , the time at which the tank is empty.

(d) That $t_{e}$ is proportional to $R^{5 / 2}$ and inversely proportional to B.

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Answer 5

Chapter 10.1, Problem 69E

To determine

(a) $\frac{d y}{d t} = - \frac{\sqrt{2 g} B \sqrt{y}}{π (2 R y - y^{2})}$

(b) For some constant C, $\frac{2 π}{15 B \sqrt{2 g}} (10 R y^{3 / 2} - 3 y^{5 / 2}) = C - t$

(c) $C = t_{e}$ , the time at which the tank is empty.

(d) That $t_{e}$ is proportional to $R^{5 / 2}$ and inversely proportional to B.

Answer 6

Chapter 10.1, Problem 69E

To determine

(a) $\frac{d y}{d t} = - \frac{\sqrt{2 g} B \sqrt{y}}{π (2 R y - y^{2})}$

(b) For some constant C, $\frac{2 π}{15 B \sqrt{2 g}} (10 R y^{3 / 2} - 3 y^{5 / 2}) = C - t$

(c) $C = t_{e}$ , the time at which the tank is empty.

(d) That $t_{e}$ is proportional to $R^{5 / 2}$ and inversely proportional to B.

Answer 7

Chapter 10.1, Problem 69E

To determine

(a) $\frac{d y}{d t} = - \frac{\sqrt{2 g} B \sqrt{y}}{π (2 R y - y^{2})}$

(b) For some constant C, $\frac{2 π}{15 B \sqrt{2 g}} (10 R y^{3 / 2} - 3 y^{5 / 2}) = C - t$

(c) $C = t_{e}$ , the time at which the tank is empty.

(d) That $t_{e}$ is proportional to $R^{5 / 2}$ and inversely proportional to B.

Answer 8

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Answer 9

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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. 5PQ 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

(a) d y d t = − 2 g B y π ( 2 R y − y 2 ) (b) For some constant C, 2 π 15 B 2 g ( 10 R y 3 / 2 − 3 y 5 / 2 ) = C − t (c) C = t e , the time at which the tank is empty. (d) That t e is proportional to R 5 / 2 and inversely proportional to B .

Solution Summary: The spherical tank of radius R is half filled with water. The water leaks through a hole in the bottom of area B.

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(a) d y d t = − 2 g B y π ( 2 R y − y 2 ) (b) For some constant C, 2 π 15 B 2 g ( 10 R y 3 / 2 − 3 y 5 / 2 ) = C − t (c) C = t e , the time at which the tank is empty. (d) That t e is proportional to R 5 / 2 and inversely proportional to B .

Solution Summary: The spherical tank of radius R is half filled with water. The water leaks through a hole in the bottom of area B.

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(a) dydt=−2gByπ(2Ry−y2) (b) For some constant C,2π15B2g(10Ry3/2−3y5/2)=C−t (c) C=te, the time at which the tank is empty. (d) That te is proportional to R5/2 and inversely proportional to B.

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Chapter 10 Solutions