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astra-53 [7]
2 years ago
13

In an inelastic collision, as compared to an elastic collision, what is to be expected?.

Physics
1 answer:
motikmotik2 years ago
3 0
The loss or conservation of kinetic energy is the difference between an elastic and an inelastic collision. Kinetic energy is not preserved in an inelastic collision, and it will change forms into sound, heat, radiation, or another form. The kinetic energy in an elastic collision is preserved and does not change forms.
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A ball is released from rest at a height of 10 m and falls freely to the
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Answer:

The new kinetic energy would be 16 times greater than before.

Explanation:

Kinetic energy is found using this formula:

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  • where KE = kinetic energy (J), m = mass (kg), and v = velocity (m/s)

We can see that kinetic energy is directly proportional to the square of the velocity, meaning that if the speed was increased by 4 times, then the kinetic energy would get increased by a factor of 16.

The velocity just before the ball hits the ground can be found by the equation:

  • √(2gh)

Let's substitute h = 10 m and h = 40 m into this formula.

  • √(2g(10))
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We can see that the velocity increases by a factor of 4 (10 m → 40 m).

Therefore, this means that the kinetic energy would also be increased by a factor of (4)² = 16. Thus, the answer is D) The new kinetic energy would be 16 times greater than before.

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A physicist found that a force of 0.68 N was measured between two charged spheres. The distance between the spheres was 1.0m. Ca
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Ask Your Teacher A circular wire loop whose radius is 10.0 cm carries a current of 3.60 A. It is placed so that the normal to it
e-lub [12.9K]

Answer:

M=0.113\ Am^2

Explanation:

Given that,

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Uniform magnetic field, B = 12 T

To find,

The magnetic dipole moment of the loop.

Solution,

Let M is the magnitude of magnetic dipole moment of the loop. We know that the product of current flowing and the area of cross section. Its formula is given by :

M=I\times A

A is the area of circular wire

M=I\times \pi r^2

M=3.6\ A \times \pi (0.1\ m)^2

M=0.113\ Am^2

Therefore, the magnetic dipole moment of the loop is 0.113\ Am^2. Hence, this is the required solution.

7 0
3 years ago
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