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malfutka [58]
2 years ago
8

From what height should you drop a ball so that it hits the ground at 40 m/s? How long will it take to hit the ground?

Physics
1 answer:
Ksenya-84 [330]2 years ago
7 0

Explanation:

It is given that,

A ball hits the ground at 40 m/s

Let h is the height from where you drop a ball. It is based on the conservation of energy as :

mgh=\dfrac{1}{2}mv^2\\\\h=\dfrac{v^2}{2g}\\\\\text{Putting all the values in above formula}\\\\h=\dfrac{40^2}{2\times 9.8}\\\\h=81.63\ m

It is dropped from a height of 81.63 m.

Let it take t seconds to hit the ground. When it hits the ground, its final speed, v = 0 and u = 40 m/s. So,

h=ut+\dfrac{1}{2}gt^2\\\\h=\dfrac{1}{2}gt^2\\\\t=(\dfrac{2h}{g})^{1/2}\\\\t=(\dfrac{2\times 81.63}{9.8})^{1/2}\\\\t=4.08\ s

So, it will take 4.08 seconds to hit the ground.

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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.

  • Normal force: NO. The normal force is a force exerted by a surface whenever there is another object "pushing" on it. For instance, when a box is at rest on a table, the box is "pushing" on the table (due to its weight), and the table "pushes back" on the box, upward, in order to balance its weight: this is the normal force. In this case, the elevator is lifted, so it is not pushing on anything, therefore there is no normal force.
  • Gravity force: YES. The force of gravity acts on every object located in the gravitational field of the Earth; it pulls downward, and its magnitude is mg, where m is the mass of the object and g is the acceleration of gravity.
  • 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)
  • Tension force: YES. The tension force is the force exerted by a rope or a string when pulling an object. In this case, there are four ropes pulling the elevator, therefore there are 4 forces of tension acting on the elevator, upward.
  • Air resistance: YES. As the elevator is moving through the air, the interaction between the molecules of air with the surface of the elevator produces a force (called air resistance) that "resists" the motion of the elevator, therefore pushing downward. However, the magnitude of this force is negligible in this case.

Learn more about forces:

brainly.com/question/8459017

brainly.com/question/11292757

brainly.com/question/12978926

#LearnwithBrainly

5 0
3 years ago
A man with a mass of 65.0 kg skis down a frictionless hill that is 5.00 m high. At the bottom of the hill the terrain levels out
anzhelika [568]

Answer:

The horizontal distance is 4.823 m

Solution:

As per the question:

Mass of man, m = 65.0 kg

Height of the hill, H = 5.00 m

Mass of the backpack, m' = 20.0 kg

Height of ledge, h = 2 m

Now,

To calculate the horizontal distance from the edge of the ledge:

Making use of the principle of conservation of energy both at the top and bottom of the hill (frictionless), the total mechanical energy will remain conserved.

Now,

KE_{initial} + PE_{initial} = KE_{final} + PE_{final}

where

KE = Kinetic energy

PE = Potential energy

Initially, the man starts, form rest thus the velocity at start will be zero and hence the initial Kinetic energy will also be zero.

Also, the initial potential energy will be converted into the kinetic energy thus the final potential energy will be zero.

Therefore,

0 + mgH = \frac{1}{2}mv^{2} + 0

2gH = v^{2}

v = \sqrt{2\times 9.8\times 5} = 9.89\ m/s

where

v = velocity at the hill's bottom

Now,

Making use of the principle of conservation of momentum in order to calculate the velocity after the inclusion, v' of the backpack:

mv = (m + m')v'

65.0\times 9.89 = (65.0 + 20.0)v'

v' = 7.56\ m/s

Now, time taken for the fall:

h = \frac{1}{2}gt^{2}

t = \sqrt{\frac{2h}{g}}

t = \sqrt{\frac{2\times 2}{9.8} = 0.638\ s

Now, the horizontal distance is given by:

x = v't = 7.56\times 0.638 = 4.823\ m

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3 years ago
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Answer:

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Answer:

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Answer:

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