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Vikki [24]
2 years ago
9

A small car has a head-on collision with a large truck. Which of the following statements concerning the magnitude of the averag

e force due to the collision is correct? The small car and the truck experience the same average force. The small car experiences the greater average force. The truck experiences the greater average force. It is impossible to tell since the masses are not given. It is impossible to tell since the velocities are not given. 1 points Saved Question 1 of 5 Moving to another question will save this response.
Physics
1 answer:
andrey2020 [161]2 years ago
6 0

Answer:

The small car and the truck experience the same average force.

Explanation:

Here we need to remember two of Newton's laws.

The second one says that:

F = m*a

force equals mass times acceleration.

And the third one says that;

"If an object A exerts a force on object B, then object B must exert a force of equal magnitude and opposite direction back on object A"

From the third law, if the car experiences a force F due to the impact with the truck, then the truck experiences the same force F due to the impact.

But this seems odd, because we would expect to see the car being more affected by the impact, right?

Well, this is explained by the second law.

Suppose that the mass of the car is m, and the mass of the truck is M.

such that M > m

Then for the small car we have:

F = m*a

And for the truck:

F = M*a'

Because the force is the same for both of them, we can write:

m*a = M*a'

a = (M/m)*a'

because M > m, then M/m > 1.

This means that the acceleration that the car experiences is larger than the acceleration for the truck, and this is why we would see that the car seems more affected by the impact, regardless of the fact that both vehicles experience the same force in the impact.

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Vlad1618 [11]

Centripetal force = (mass) x (speed)² / (radius)

                           = (20 kg) x (20 m/s)² / (20 m)

                           = (20 x 20 / 20)  (kg-meter/sec²)

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5 0
3 years ago
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A stuntman with a mass of 80.5 kg swings across a moat from a rope that is 11.5 m. At the bottom of the swing the stuntman's spe
goldenfox [79]

Answer:

  • No
  • 5.49 m/s

Explanation:

The net force required to accelerate the stuntman in a circular arc of radius 11.5 m will be ...

  F = mv²/r . . . . where this m is the mass being accelerated, v is the tangential velocity, and r is the radius.

Here, the net force needs to be ...

  F = (80.5 kg)(8.45 m/s)²/(11.5 m) . . . . . where this m is meters

  ≈ 499.8175 kg·m/s² = 499.8 N

Gravity exerts a force on the stuntman of ...

  F = mg = (80.5 kg)(9.8 m/s²) = 788.9 kg·m/s² = 788.9 N

Then the tension required in the rope/vine is ...

  499.8 N+788.9 N= 1288.7 N

This is more than the capacity of the rope, so we do not expect the stuntman to make it across the moat.

_____

The allowed net force for centripetal acceleration is ...

  1000 N -788.9 N = 211.1 N

Then the allowed velocity is ...

  211.1 = 80.5v²/11.5

  30.16 = v² . . . .  multiply by 11.5/80.5

  5.49 = v . . . . . . take the square root

The maximum speed the stuntman can have is 5.49 m/s.

_____

<em>Comment on crossing the moat</em>

The kinetic energy at the bottom of the swing translates to potential energy at the end of the swing. At the lower speed, the stuntman cannot rise as high, so will traverse a shorter arc. At 8.45 m/s, the moat could be about 16.8 m wide; at 5.49 m/s, it can only be about 11.5 m wide.

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3 years ago
This is my question. I need the answer stat.
Kipish [7]

Answer:

Explanatioyour answers look right, but if there has , has to be another answer its a , but your answers are right

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Volcanic ash and sulfur dioxide spewed out of Mt. Pinatubo in 1991. These materials can reflect incoming solar radiation. Over t
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The suspended ash made for some some spectacular sunsets! Sulfuric acid was spread worldwide, increasing acidity of rain. Ash deflected energy from the sun, causing a slight drop on global temps for a few years.
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Explain how surface and subsurface events are integral parts of the rock cycle.
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Example of surface events are erosion and weathering. Erosion is the carrying of a particle from one place to the other and weathering is the breaking down of particles. These processes help in rock formation because this allows physical changes (grouping together or breaking down) on a certain substance. Subsurface events are those which happened underground such as the flow of underground water which subsequently allow the deposition of minerals, etc. 
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