Question

A large storage tank, open to the atmosphere at the top and filled with water, develops a small hole in its side at a point 19.1 m below the water level. If the rate of flow from the leak is 2.00 × 10^(-3) m^3/min, determine the following. (a) Determine the speed at which the water leaves the hole in m/s. (b) Determine the diameter of the hole.

          A large storage tank, open to the atmosphere at the top and filled with water, develops a small hole in its side at a point 19.1 m below the water level. If the rate of flow from the leak is 2.00 × 10^(-3) m^3/min, determine the following.
(a) Determine the speed at which the water leaves the hole in m/s.
(b) Determine the diameter of the hole.

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Added by Patrick A.

University Physics with Modern Physics

Hugh D. Young 14th Edition

Instant Answer

Solved by Expert Adi S

06/20/2022

Step 1

Given: H = 19.1 m, g = 9.8 m/s^2 Using the equation v^2 = 2gh, where v is the speed at which water leaves the hole, we can calculate: v = sqrt(2 * 9.8 * 19.1) v = sqrt(376.36) v = 19.39 m/s Show more…

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A large storage tank, open to the atmosphere at the top and filled with water, develops a small hole in its side at a point 19.1 m below the water level. If the rate of flow from the leak is 2.00 × 10^(-3) m^3/min, determine the following. (a) Determine the speed at which the water leaves the hole in m/s. (b) Determine the diameter of the hole.

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Transcript

00:01 In this problem we have been given a large storage tank open to the atmosphere at the top and filled with water.

00:08 Now it develops a small hole somewhere at a side 19 .1 meter below the water level.

00:14 So let us see how we can solve this problem.

00:16 So the height given in the problem is 19 .1 meters.

00:21 Also the speed which is asked in the first part of the problem is under root 2gh.

00:28 Now, how we arrive at this particular relation? we arrive at this particular relation because v square minus u square is equal to 2gh.

00:38 And here, this is the equation from third law of motion.

00:43 Now here the u value is zero.

00:46 So, v square is equals to 2gh and hence we have this particular relationship.

00:52 Now here, g is gravity definitely.

00:54 Now, if we plug in the value of h in this particular equation, we get the value of v as 2 multiplied 9 .8 multiplied 19 .1.

01:06 Finally, the v value arrives as 374 .36 meter per second.

01:15 So this is the answer for the first part of the problem.

01:18 Now in the second part, let us see what values we have been given.

01:21 In the second part, we have to find out the value of diameter.

01:25 So, diameter value is asked.

01:28 Let us see how to solve it.

01:31 So a is my area of cross section.

01:33 So area formula will be pi d square over 4 in terms of d.

01:40 Now we can write the value of v that is rate of flow as 0 .0022 over 60.

01:48 So this will be equals to 3 .67 multiplied 10 raise to the power minus 5.

01:55 So, this is the value of rate of flow...

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