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nekit [7.7K]
3 years ago
5

A 2.00-kilogram mud ball drops from rest at a height of 17.0 m. If the impact between the ball and the ground lasts 0.46 s, what

is the magnitude of the average force exerted by the ball on the ground
Physics
1 answer:
Softa [21]3 years ago
5 0

Answer: 79.35 N

Explanation: according to the impulse momentum theorem,

Impulse = change in momentum

Where impulse = force × time and change in momentum = m ( v - u).

The object was initially at rest, hence it initial velocity is zero.

To get the final velocity, we use the formulae below

v² = u² + 2gh

Where h = height of the cliff = 17m

v² = 2 × 9.8 × 17

v² = 333.2

v = √333.2

v = 18.25 m/s

At t = 0.46s and v = 18.25 m/s, we can get the average force of impact

F×0.46 = 2 (18.25 - 0)

F × 0.46 = 2 (18.25)

F × 0.46 = 36.5

F = 36.5 /0.46 = 79.35 N

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Answer:

-4.72005 m/s

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Displacement is the minimum distance between the initial and final points of the journey.

Here, the displacement of the whole episode is 0 as the initial and final point is zero.

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Consider this situation: Four ropes, each attached to the end
faust18 [17]

The forces acting on the elevator are:

Gravity force

Tension force

Air resistance

Explanation:

Let's go through each of the forces listed and see which ones are acting on the elevator.

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  • Applied force: NO. Here there is no applied force, since there is nobody "pushing" or "pulling" the elevator.
  • Friction force: NO. As we are considering the forces on the elevator, and the elevator is not sliding against any surfaces, there is no force of friction. (The force of friction acts whenever there are two surfaces sliding against each other, which is not the case here)
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5 0
3 years ago
A box is pushed 40 m by a mover. The amount of work done was 2,240 j. How much force was exerted on the box
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The force exerted on the box is 56 N

Explanation:

The work done by a force on an object is given by

W=Fd cos \theta

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F is the magnitude of the force

d is the displacement of the object

\theta is the angle between the direction of the force and of the displacement

For the box in this problem, we have:

W = 2240 J is the work done

d = 40 m is the displacement of the box

Assuming that the  force is parallel to the displacement, \theta=0

Solving the equation for F, we find the force exerted on the box:

F=\frac{W}{d cos \theta}=\frac{2240}{(40)(cos 0)}=56 N

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3 0
3 years ago
A 3.0kg mass tied to a string
dem82 [27]

Answer:

\boxed{\sf Tension \ in \ the \ string \ (T) = 3 \ kN}

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Tension in the string (T)

Explanation:

Tension (T) is the string will be equal to centripetal force (\sf F_c).

\boxed{ \bold{ T = F_c  =  \frac{m {v}^{2} }{r} }}

Substituting value of m, v & r in the equation:

\sf \implies T =  \frac{3 \times  {20}^{2} }{0.4}  \\  \\  \sf \implies T = \frac{3 \times 400}{0.4}  \\  \\  \sf \implies T =3 \times 1000 \\  \\  \sf \implies T =3000 \: N \\  \\ \sf \implies T =3 \: kN

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Tension in the string (T) = 3 kN

5 0
3 years ago
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