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KATRIN_1 [288]
1 year ago
15

an open tank has the shape of a right circular cone (see figure). the tank is 8 feet across the top and 6 feet high. how much wo

rk is done in emptying the tank by pumping the water over the top edge? (the weight-density of water is 62.4 pounds per cubic foot.) ft-lb
Physics
1 answer:
Liono4ka [1.6K]1 year ago
7 0

The amount of work done in emptying the tank by pumping the water over the top edge is 163.01* 10³ ft-lbs.

Given that, the tank is 8 feet across the top and 6 feet high

By the property of similar triangles, 4/6 = r/y

6r = 4y

r = 4/6*y = 2/3*y

Each disc is a circle with area, A = π(2/3*y)² = 4π/9*y²

The weight of each disc is m = ρw* A

m = 62.4* 4π/9*y² = 87.08*y²

The distance pumped is 6-y.

The work done in pumping the tank by pumping the water over the top edge is

W = 87.08 ∫(6-y)y² dy

W = 87.08 ∫(6y³ - y²) dy

W =  87.08 [6y⁴/4 - y³/3]

W =  87.08 [3y⁴/2- y³/3]

The limits are from 0 to 6.

W =  87.08 [3*6⁴/2 - 6³/3] = 87.08* [9*6³ - 2*36] = 87.08(1872) = 163013.76 ft-lbs

The amount of work done in emptying the tank by pumping the water over the top edge is 163013.76 ft-lbs.

To know more about work done:

brainly.com/question/16650139

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A 1000 kg elevator is accelerated upward at a rate of 0.70 m/s2. What is the tension in the cable pulling the elevator upward wh
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Answer:

Tension is also known as Force...

and Force is mass× acceleration.

so....1000×0.70=700N

5 0
3 years ago
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A 50.0 kg object is moving at 18.2 m/s when a 200 N force
gayaneshka [121]

Answer:

distance = 21.56 m

Explanation:

given data

mass = 50 kg

initial velocity  = 18.2 m/s

force = -200 N ( here force applied to opposite direction )

final velocity = 12.6 m/s

solution

we know here acceleration will be as

acceleration a  = force ÷ mass

a = \frac{-200}{50}   =  -4 m/s²

we get here now required time that is

required time = \frac{V_{(final)} - V_{(initial)}}{a}     ...............1

put here value

required time = \frac{12.6-18.2}{-4}  

so distance will be

distance = \frac{V_{(final)}^2 - V_{(initial)}^2}{2a}    ........2

distance = \frac{12.6}^2 -{18.2}^2}{2\times (-4)}  

distance = 21.56 m

7 0
2 years ago
A 5-kg ball collides inelastically head-on with a 10-kg ball, which is initially stationary. Which of the following statements i
NARA [144]

Answer:

The magnitude of the change of velocity the 5-kg ball experiences is less than that of the 10-kg ball.

Explanation:

In inelastic collision, the total momentum is always conserved after collision but the kinetic energy is reduced.

Momentum is Mass X velocity.

5 kg ball is in motion, while 10 kg ball is stationary; that is zero velocity.

The momentum of 10 kg ball before collision is zero while the momentum of 5 kg ball before collision is more than zero. Therefore, the magnitude of change in momentum will not be equal.

Next possible options are in kinetic Energy

Initial Kinetic energy = \frac{1}{2}mu^2

Final kinetic energy =\frac{1}{2}mv^2

Change in kinetic energy = Final Kinetic Energy - Initial Kinetic Energy

Change in kinetic energy of 5kg ball = \frac{1}{2}mv^2 -\frac{1}{2}mu^2 = \frac{1}{2}m(v-u)^2

Since the 5-kg ball has initial velocity (u), the magnitude of the change in velocity will be reduced.

Change in kinetic energy of 10kg ball:

the ball is initially at rest, therefore the initial velocity (u) will be zero (0)

Δ K.E = \frac{1}{2}mv^2 -\frac{1}{2}mu^2 = \frac{1}{2}m(v-u)^2 = \frac{1}{2}m(v-0)^2 = \frac{1}{2}mv^2

From the solution above, the magnitude of the change in velocity experienced by 10 kg ball is higher than 5 kg ball.

Hence, The magnitude of the change of velocity the 5-kg ball experiences is less than that of the 10-kg ball

4 0
3 years ago
4. Johnny exerts a 3.55 N rightward force on a 0.200-kg box to accelerate it across a low-friction track. If the total resistanc
Anon25 [30]

a) 15.2 m/s^2

b) 1.96 N

c) 1.96 N

Explanation:

a)

To find the box's acceleration, we have to find first the net force acting on the box in the horizontal direction.

We have:

- The forward force of 3.55 N

- The backward, resistive force of 0.52 N

So, the net force forward is

\sum F=3.55-0.52=3.03 N

Now we can find the acceleration by using Newton's second law of motion, which states that:

\sum F=ma

where

m = 0.200 kg is the mass of the box

a is its acceleration

And solving for a, we find the acceleration:

a=\frac{\sum F}{m}=\frac{3.03}{0.200}=15.2 m/s^2

b)

The gravitational force on an object is the force with which the object is pulled towards the ground by the Earth.

It is given by

W=mg

where

m is the mass of the object

g is the gravitational field strength

In this problem we have

m = 0.200 kg is the mass of the box

g=9.8 m/s^2 is the gravitational field strength

So, the gravitational force on the box is

W=(0.200)(9.8)=1.96 N

c)

The normal force is the reaction force exerted by the floor on the box, in the upward direction.

In order to find the magnitude of this force, we apply Newton's second law of motion along the vertical direction.

We have two forces in this direction:

- The gravitational force, W, downward

- The normal force, N, upward

So the net force is

\sum F=N-W

According to Newton's second law,

\sum F=ma

However, the box is at rest in the vertical direction, so the vertical acceleration is zero:

a=0

This means that the net force is zero:

\sum F=0

And so, we can find the normal force:

N-W=0\\N=W=1.96 N

4 0
2 years ago
How does a model differ from a theory?
yuradex [85]

Answer:

Explained

Explanation:

A model is practical representation of reality with help of various tools called model. It is more purposeful, appealing and provide greater understanding of the specific phenomenon called modeling.

Theory on the other hand are set of written statement which is more generalized and aimed to explain the phenomenon. Theory is developed through the process of abstraction, experimentation, and deduction.

A model is often is used to describe an application of a theory for a specific case.

4 0
3 years ago
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