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Through the tunnel at a speed of 3.0 m/s until it reaches a narrow section where...

Question

Through the tunnel at a speed of 3.0 m/s until it reaches a narrow section where...


through the tunnel at a speed of 3.0 m/s until it reaches a narrow section where the tunnels diameter is 2.70 m. The pressure in the narrow section is 82 kPa. Use the continuity equation to find the waters speed in the narrow section of the tunnel. Then find the pressure in the wide portion of the tunnel. An enormous open vat owned by a British cider company has a volume of nearly 7000 m3. If a small hole is drilled halfway down the side of this vat when it is full of cider, the cider will hit the ground 19.7 m away from the bottom of the vat. How tall is the vat? 3. N- The tallest cooling tower in the world is in Germany. Consider a water pipe coming down from the top of the tower. The pipe is punctured near the bottom, which causes the water to flow from the puncture hole with a speed of 59 m/s. How high is the tower? Assume the pressure inside the pipe is equal to the pressure outside the pipe 4.


through the tunnel at a speed of 3.0 m/s until it reaches a narrow section where the tunnel's diameter is 2.70 m. The pressure in the narrow section is 82 kPa. Use the continuity equation to find the water's speed in the narrow section of the tunnel. Then find the pressure in the wide portion of the tunnel. An enormous open vat owned by a British cider company has a volume of nearly 7000 m3. If a small hole is drilled halfway down the side of this vat when it is full of cider, the cider will hit the ground 19.7 m away from the bottom of the vat. How tall is the vat? 3. N- The tallest cooling tower in the world is in Germany. Consider a water pipe coming down from the top of the tower. The pipe is punctured near the bottom, which causes the water to flow from the puncture hole with a speed of 59 m/s. How high is the tower? Assume the pressure inside the pipe is equal to the pressure outside the pipe 4.

Answers

Here ,

let the velocity of flow from the vat is v

height of the vat is h

Now, time taken to fall , t = sqrt(2h/(2 *g))

t = sqrt(h/g)

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

v = sqrt(g*h)

Now, as distance = speed * time

19.7 = sqrt(h/9.8) * sqrt(9.8 * h)

h = 19.7 m

the height of the vat is 19.7 m


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