In designing a backyard water fountain, a gardener wants a stream of water to ex

Question

In designing a backyard water fountain, a gardener wants a stream of water to exit from the bottom of one can and land in a second one as shown to the left. The top of the second can is 0.52 m below the hole in hte first can which has a depth of 0.18 m.

1) What is gauge pressure on the top of the water in the first can?

2) What is the guage pressure on stream of water exiting the bottom of the first can?

3) What is the velocity of the stream of water exiting the bottom of the first can?

4) How long does it take the water to fall to the second can?

5) How to the right of the first can should the center of the second can be placed?

In designing a backyard water fountain, a gardener wants a stream of water to exit from the bottom of one can and land in a second one as shown to the left. The top of the second can is 0.52 m below the hole in hte first can which has a depth of 0.18 m. 1) What is gauge pressure on the top of the water in the first can? 2) What is the guage pressure on stream of water exiting the bottom of the first can? 3) What is the velocity of the stream of water exiting the bottom of the first can? 4) How long does it take the water to fall to the second can? 5) How to the right of the first can should the center of the second can be placed?

Explanation / Answer

1) Pgauge = 0

2) Pgauge = 0

3) v = sqrt(2*g*h)

= sqrt(2*9.8*0.18)

= 1.88 m/s

4) let t is the time taken

H = 0.5*g*t^2

t = sqrt(2*H/g)

= sqrt(2*0.52/9.8)

= 0.326 s

5) D = v*t

= 1.88*0.326

= 0.61 m

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