The spout heights in the container in the figure are 15cm, 30cm, 45cm, and 60cm.

Question

The spout heights in the container in the figure are 15cm, 30cm, 45cm, and 60cm. The water level is maintained at a 68cm height by an outside supply.

The spout heights in the container in the figure are 15cm, 30cm, 45cm, and 60cm. The water level is maintained at a 68cm height by an outside supply. Part A) What is the speed of the water out of the 15cm high hole? Part B) What is the speed of the water out of the 30cm high hole? Part C) What is the speed of the water out of the 45cm high hole? Part D) What is the speed of the water out of the 60cm high hole?

Explanation / Answer

The spout heights in the container in Fig. 9.35 are 9 cm, 20 cm, 30 cm, and 38 cm. The water level is maintained at a 45 cm height by an outside supply.

Figure 9.35:
http://www.webassign.net/wb/9-35.gif

(a) What is the speed of the water out of each hole? 9 cm hole 1 m/s
20 cm hole _____m/s
30 cm hole _____m/s
38 cm hole _____m/s
(b) Which water stream has the greatest range relative to the base of the container?

• ANS :

The spout heights in the container in Fig. 9.35 are 9 cm, 20 cm, 30 cm, and 38 cm. The water level is maintained at a 45 cm height by an outside supply.

Figure 9.35:
http://www.webassign.net/wb/9-35.gif

(a) What is the speed of the water out of each hole? 9 cm hole 1 m/s
20 cm hole _____m/s
30 cm hole _____m/s
38 cm hole _____m/s
(b) Which water stream has the greatest range relative to the base of the container?

• ANS :

The spout heights in the container in Fig. 9.35 are 9 cm, 20 cm, 30 cm, and 38 cm. The water level is maintained at a 45 cm height by an outside supply.

Figure 9.35:
http://www.webassign.net/wb/9-35.gif

(a) What is the speed of the water out of each hole? 9 cm hole 1 m/s
20 cm hole _____m/s
30 cm hole _____m/s
38 cm hole _____m/s
(b) Which water stream has the greatest range relative to the base of the container? Are the heights from the bottom or from the top?. I'm assuming they are from the bottom.

From Bernoulli's eqn we know that gh + 1/2v^2 = constant Note we can divide from both sides leaving gh + v^2/2 = constant

So gh + v^2/2)top = gh + v^2/2)at the various heights...but vtop= 0 and h at the various height = 0

So gh = v^2/2...So v = sqrt(2gh) [note h is the distance from the top to the listed height]

9cm hole v = sqrt(2*g*h) = sqrt(2*9.8*(0.45-0.09)) = 2.66m/s (I don't understand your ans of 1m/s)

20cm hole v = sqrt(2*g*h) = sqrt(2*9.8*(0.45-0.20)) = 2.21m/s

30 cm v = sqrt(2*g*h) = sqrt(2*9.8*(0.45-0.30)) = 1.71m/s

38cm v = sqrt(2*g*h) = sqrt(2*9.8*(0.45-0.38)) = 1.17m/s


b) to find the range of each we need the time of travel .We get this from y = 1/2*g*t^2 or
t= sqrt(2*y/g) and then x = v*t

For 9cm hole ..t = sqrt(2*.09/9.8) = 0.136s so x = 2.66*0.136 = 0.360m

For 20cm hole ..t = sqrt(2*.20/9.8) = 0.202s so x = 2.21*0.202 = 0.446m

For 30 cm hole ..t = sqrt(2*.30/9.8) = 0.247s so x = 1.71*0.247 = 0.423m

For 38 cm hole ..t = sqrt(2*.38/9.8) = 0.278s so x = 1.17*0.278 = 0.325m


So the 20 cm hole has the greatest range Are the heights from the bottom or from the top?. I'm assuming they are from the bottom.

From Bernoulli's eqn we know that gh + 1/2v^2 = constant Note we can divide from both sides leaving gh + v^2/2 = constant

So gh + v^2/2)top = gh + v^2/2)at the various heights...but vtop= 0 and h at the various height = 0

So gh = v^2/2...So v = sqrt(2gh) [note h is the distance from the top to the listed height]

9cm hole v = sqrt(2*g*h) = sqrt(2*9.8*(0.45-0.09)) = 2.66m/s (I don't understand your ans of 1m/s)

20cm hole v = sqrt(2*g*h) = sqrt(2*9.8*(0.45-0.20)) = 2.21m/s

30 cm v = sqrt(2*g*h) = sqrt(2*9.8*(0.45-0.30)) = 1.71m/s

38cm v = sqrt(2*g*h) = sqrt(2*9.8*(0.45-0.38)) = 1.17m/s


b) to find the range of each we need the time of travel .We get this from y = 1/2*g*t^2 or
t= sqrt(2*y/g) and then x = v*t

For 9cm hole ..t = sqrt(2*.09/9.8) = 0.136s so x = 2.66*0.136 = 0.360m

For 20cm hole ..t = sqrt(2*.20/9.8) = 0.202s so x = 2.21*0.202 = 0.446m

For 30 cm hole ..t = sqrt(2*.30/9.8) = 0.247s so x = 1.71*0.247 = 0.423m

For 38 cm hole ..t = sqrt(2*.38/9.8) = 0.278s so x = 1.17*0.278 = 0.325m


So the 20 cm hole has the greatest range Are the heights from the bottom or from the top?. I'm assuming they are from the bottom.

From Bernoulli's eqn we know that gh + 1/2v^2 = constant Note we can divide from both sides leaving gh + v^2/2 = constant

So gh + v^2/2)top = gh + v^2/2)at the various heights...but vtop= 0 and h at the various height = 0

So gh = v^2/2...So v = sqrt(2gh) [note h is the distance from the top to the listed height]

9cm hole v = sqrt(2*g*h) = sqrt(2*9.8*(0.45-0.09)) = 2.66m/s (I don't understand your ans of 1m/s)

20cm hole v = sqrt(2*g*h) = sqrt(2*9.8*(0.45-0.20)) = 2.21m/s

30 cm v = sqrt(2*g*h) = sqrt(2*9.8*(0.45-0.30)) = 1.71m/s

38cm v = sqrt(2*g*h) = sqrt(2*9.8*(0.45-0.38)) = 1.17m/s


b) to find the range of each we need the time of travel .We get this from y = 1/2*g*t^2 or
t= sqrt(2*y/g) and then x = v*t

For 9cm hole ..t = sqrt(2*.09/9.8) = 0.136s so x = 2.66*0.136 = 0.360m

For 20cm hole ..t = sqrt(2*.20/9.8) = 0.202s so x = 2.21*0.202 = 0.446m

For 30 cm hole ..t = sqrt(2*.30/9.8) = 0.247s so x = 1.71*0.247 = 0.423m

For 38 cm hole ..t = sqrt(2*.38/9.8) = 0.278s so x = 1.17*0.278 = 0.325m


So the 20 cm hole has the greatest range

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