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zysi [14]
3 years ago
6

A train pulls the boxcars. What are the vertical forces present? (1 point)

Physics
1 answer:
Greeley [361]3 years ago
5 0

The vertical forces present in a boxcars being pulled by a train are weight and pull of air on the boxcars.

The forces exerted on an object can be resolved into <em>horizontal</em> and <em>vertical components.</em>

<em />

The horizontal components of the forces on an object being pulled include the following;

  • The frictional force on the object; which tends to reduce the motion,
  • force dragging the object forward
  • if the object is inclined to an angle, the horizontal component of the weight

This horizontal force is written as;

\Sigma F_x = 0\\\\F_x - F_k = ma

The vertical component of the force on the object include the following;

  • the weight of the object acting downwards
  • the pull of air on the object acting upwards

The vertical force on the object is written as;

\Sigma F_n = 0\\\\F_n - W = 0

Thus, the vertical forces present in a boxcars being pulled by a train are weight and pull of air on the boxcars.

Learn more here: brainly.com/question/2000189

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Answer:

Option (b) is correct.

Explanation:

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Inelastic collision is defined as a collision where kinetic energy of the system is not conserved whereas the linear momentum is conserved. This loss of kinetic energy may due to the conversion to thermal energy or sound energy or may be due to the deformation of the materials colliding with each other.

As given in the problem, before the collision, total momentum of the system is 2.5~Kg~m~s^{-1} and the kinetic energy is 5~J. After the collision, the total momentum of the system is  2.5~Kg~m~s^{-1}, but the kinetic energy is reduced to 4~J. So some amount of kinetic energy is lost during the collision.

Therefor the situation describes an inelastic collision (and it could NOT be elastic).

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3 years ago
What are the three ways to accelerate?
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3 years ago
Read 2 more answers
A person walking covers 5 m ikn 10 s how fast is the person moving
aniked [119]

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Speed = (5 meters) / (10 seconds)

<em>Speed = 0.5 meter per second</em> .

That's like about 1.1 mile per hour .

Normal walking speed is considered to be around 1.4 m/s ... about 3.1 mph, or 14 meters in 10 seconds.

I've got a grandson who hasn't even turned 1 yet.  He crawls and  doesn't walk, but if you only cover 5m in 10s, he'd leave you in the dust pretty quick.

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3 years ago
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stira [4]
Fortunately, 'force' is a vector.  So if you know the strength and direction
of each force, you can easily addum up and find the 'resultant' (net) force.

When we talk in vectors, one newton forward is the negative of
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the complete solution of the problem.


            (100 N forward) plus (50 N backward)

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        =           50 N forward .

That's it.
Is there any part of the solution that's not clear ?

4 0
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