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Crazy boy [7]
4 years ago
9

A constant force is exerted for a short time interval on a cart that is initially at rest on an air track. This force gives the

cart a certain final speed. Suppose we repeat the experiment but, instead of starting from rest, the cart is already moving with constant speed in the direction of the force at the moment we begin to apply the force.
After we exert the same constant force for the same short time interval, the increase in the cart's speed:

A. is equal to two times its initial speed.
B. is equal to the square of its initial speed.
C. is equal to four times its initial speed.
D. is the same as when it started from rest.
E. cannot be determined from the information provided.
Physics
1 answer:
Ilya [14]4 years ago
5 0

Answer:

d) is the same as when it started from rest

Explanation:

using equation of motion

v = u + at

second law of momentum defines

F = ma

a = F /m

the equation becomes

v = u + (F/m)t

from hear

since v is directly proportional to the force and the force remain the same, the increase in the cart speed will also remain the same.

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An ice cube of volume 15 cm^3 and density 917 kg/m^3 is placed in ethyl alcohol of density 811 kg/m^3. What is the buoyant force
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Answer:

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Explanation:

given,

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density of ethyl alcohol = 811 kg/m³

buoyant force = ?

The density of ice is more than ethyl alcohol hence it will sink.

buoyant  force acting on the ice cube = ρ V g

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A cart traveling at 0.3 m/s collides with stationary object. After the collision, the cart rebounds in the opposite direction. T
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Answer:

Explanation:

First case

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It collide with a stationary body, then after collision the ball rebounds and move in opposite direction. This shows that the ball have a velocity after impulse let say v

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I=m(v-u)

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Second case

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Then, impulse is given as the change in linear momentum of a body

Impulse =m∆v

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Note, momentum is a vector quantity.

I=m(v--u)

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The correct answer is A

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3 years ago
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