Pad 8:20 PM * 96% a flipitphysics.com Example Bobber For assistance with this pr

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Pad 8:20 PM * 96% a flipitphysics.com Example Bobber For assistance with this problem, see Example 15-9 Standard Exercse Multi Tube Standard Exercise Wet Penguins Standard Exercise Aspirator Standard Exercise Faucet Standard Exercse Dam and Plug Standard Exercae Water Stream from Can Apparent Weight in Wate 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 0.59 m below the hole in hte first can which has a depth of 0.16 m. Standard Exercse Hydraulic Lift 1) What is gauge pressure on the top of the water in the first can? N/m2 Submit 2) What is the guage pressure on stream of water exiting the bottom of the first can? Pguage N/m2 Submit 3) What is the velocity of the stream of water exiting the bottom of the first can? m/sec Submit 4) How long does it take the waer toal to the second can? sec Submit 5) How to the right of the first can should the center of the second can be placed? mSubmit Copyright © 2018 Freeman worth Publishers . a division of Macmillan Learning About I Tech Support | Find Your Local Sales Rep Terms Of Use Privacy Policy macmillan learning

Explanation / Answer

The answer to part (1) and (2) should be zero as the potential energy is being converted to kinetic energy

(3) vx = sqrt (2gh)

vx = sqrt (2*9.8*0.16)

vx = 1.7708 m/s

(4) Final velocity in y-direction

v = sqrt (2*9.8*0.59)

v = 3.4 m/s

Now, t = 3.4 / 9.8 = 034699 seconds

(5) x = vxt

x = 1.7708*0.34699

x = 0.61447 m

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