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Licemer1 [7]
3 years ago
12

A tank is full of water. Find the work W required to pump the water out of the spout. (Use 9.8 m/s2 for g. Use 1000 kg/m3 as the

weight density of water. Assume that a
Physics
1 answer:
Sergio039 [100]3 years ago
5 0

Answer:

W = 1.06 MJ

Explanation:

- We will use differential calculus to solve this problem.

- Make a differential volume of water in the tank with thickness dx. We see as we traverse up or down the differential volume of water the side length is always constant, hence, its always 8.

- As for the width of the part w we see that it varies as we move up and down the differential element. We will draw a rectangle whose base axis is x and vertical axis is y. we will find the equation of the slant line that comes out to be y = 0.5*x. And the width spans towards both of the sides its going to be 2*y = x.

- Now develop and expression of Force required:

                                             F = p*V*g

                                             F = 1000*(2*0.5*x*8*dx)*g

                                             F = 78480*x*dx

- Now, the work done is given by:

                                             W = F.s

- Where, s is the distance from top of hose to the differential volume:

                                             s = (5 - x)

- We have the work as follows:

                                            dW = 78400*x*(5-x)dx

- Now integrate the following express from 0 to 3 till the tank is empty:

                                           W = 78400*(2.5*x^2 - (1/3)*x^3)

                                           W = 78400*(2.5*3^2 - (1/3)*3^3)

                                           W = 78400*13.5 = 1058400 J

 

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2 years ago
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alukav5142 [94]

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8 0
3 years ago
1pt A cannon fires a 5-kg ball horizontally from a
Klio2033 [76]

Answer: Both cannonballs will hit the ground at the same time.

Explanation:

Suppose that a given object is on the air. The only force acting on the object (if we ignore air friction and such) will be the gravitational force.

then the acceleration equation is only on the vertical axis, and can be written as:

a(t) = -(9.8 m/s^2)

Now, to get the vertical velocity equation, we need to integrate over time.

v(t) = -(9.8 m/s^2)*t + v0

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if the object is dropped (or it only has initial velocity on the horizontal axis) then v0 = 0m/s

and:

v(t) = -(9.8 m/s^2)*t

Now, if two objects are initially at the same height (both cannonballs start 1 m above the ground)

And both objects have the same vertical velocity, we can conclude that both objects will hit the ground at the same time.

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7 0
3 years ago
A pitcher exerts a force on a baseball that is 30 times the balls weight. How fast is the pitcher accelerating the ball?
iVinArrow [24]

Answer:

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Explanation:

  Force applied on baseball = 30 times weight of the ball.

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  We have force applied is also equal to product of mass and acceleration.

                            F = ma = 30 x mg

                                 a = 30g

   So, pitcher is accelerating the ball at 30 times of acceleration due to gravity = 294 m/s²

8 0
3 years ago
How do I find the approximate velocity of "the object", on the graph at 5 seconds?
MakcuM [25]

First, foremost, and most critically, you must look at the graph, and critically
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Without that ability ... since the graph is nowhere to be found ... I am hardly
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7 0
3 years ago
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