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 0.55 m below the hole in hte first can which has a depth of 0.16 m.

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

pguage =_____________ N/m2

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

pguage =______________ N/m2

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

v =____________________ m/sec

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

t =____________________ sec

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

x =_______________ m

Explanation / Answer

1) Pgauge = 0

2) Pgauge = rgo*g*h

= 1000*9.8*0.16

= 1568 pa

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

= sqrt(2*9.8*0.16)

= 1.77 m./s

4) H = 0.5*g*t^2

==> t = sqrt(2*H/g)

= sqrt(2*0.55/9.8)

= 0.335 s

5) x = v*t = 1.77*0.335 = 0.593 m

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