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2 years ago

Which situations describe an elastic collision?

Physics

2 answers:

Solnce55 [7]2 years ago

6 0

Answer:

Option (1)

Explanation:

There are two types of collision.

1. Elastic collisions

2. Inelastic collisions

The conditions for elastic collisions are given below:

a. the momentum is conserved.

b. kinetic energy is conserved.

c. all the forces during the collision are conservative in nature.

The conditions for inelastic collisions are given below:

a. the momentum is conserved.

b. the mechanical energy is conserved.

c. all or some of the forces are non conservative in nature.

When the two glass marbles strikes and bounce off each other, the momentum and kinetic energy both are conserved so it is the example of elastic collision.

V125BC [204]2 years ago

5 0

Elastic collisions are collisions in which both momentum and kinetic energy are conserved. I think the correct answer from the choices listed above is option A. Two <span>glass marbles bounce off each other is an example of an elastic collision. Hope this answers the question.

I hope this helps! <3
</span>

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oksano4ka [1.4K]

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2 years ago

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Marysya12 [62]

Answer:

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3 years ago

What do you think would happen to the force of attraction of two interacting charges if their distance apart is halved?

sweet [91]

Answer:

The new force becomes 4 times the initial force.

Explanation:

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2 years ago

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qwelly [4]

Answer:

Explanation:

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2 years ago

A stone is thrown vertically upward with a speed of 15.5 m/s from the edge of a cliff 75.0 m high .

rjkz [21]

a) 2.64 s

We can solve this part of the problem by using the following SUVAT equation:

$s=ut+\frac{1}{2}at^2$

where

s is the displacement of the stone

u is the initial velocity

t is the time

a is the acceleration

We must be careful to the signs of s, u and a. Taking upward as positive direction, we have:

- s (displacement) negative, since it is downward: so s = -75.0 m

- u (initial velocity) positive, since it is upward: +15.5 m/s

- a (acceleration) negative, since it is downward: so a= g = -9.8 m/s^2 (acceleration of gravity)

Substituting into the equation,

$-75.0 = 15.5 t -4.9t^2\\4.9t^2-15.5t-75.0 = 0$

Solving the equation, we have two solutions: t = -5.80 s and t = 2.84 s. Since the negative solution has no physical meaning, the stone reaches the bottom of the cliff 2.64 s later.

b) 10.4 m/s

The speed of the stone when it reaches the bottom of the cliff can be calculated by using the equation:

$v=u+at$

where again, we must be careful to the signs of the various quantities:

- u (initial velocity) positive, since it is upward: +15.5 m/s

- a (acceleration) negative, since it is downward: so a = g = -9.8 m/s^2

Substituting t = 2.64 s, we find the final velocity of the stone:

$v = 15.5 +(-9.8)(2.64)=-10.4 m/s$

where the negative sign means that the velocity is downward: so the speed is 10.4 m/s.

c) 4.11 s

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$s=ut+\frac{1}{2}at^2$

where

s is the displacement of the package

u is the initial velocity

t is the time

a is the acceleration

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s = -105 m (vertical displacement of the package, downward so negative)

u = +5.40 m/s (initial velocity of the package, which is the same as the helicopter, upward so positive)

a = g = -9.8 m/s^2

Substituting into the equation,

$-105 = 5.40 t -4.9t^2\\4.9t^2 -5.40 t-105=0$

Which gives two solutions: t = -5.21 s and t = 4.11 s. Again, we discard the first solution since it is negative, so the package reaches the ground after

t = 4.11 seconds.

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2 years ago

Read 2 more answers