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saw5 [17]
3 years ago
9

What is the momentum of a 950 kg car moving at 10.0 m/s?

Physics
2 answers:
KonstantinChe [14]3 years ago
6 0
What's 950 x 10 bro?
<span>
9500 kg m/s
</span>
OverLord2011 [107]3 years ago
6 0

Answer:

Momentum of the car, p = 9500 kg-m/s.

Explanation:

It is given that,

Mass of the car, m = 950 kg

Velocity of the car, v = 10 m/s

We need to find the momentum of the car. It is given by the product of mass and velocity. It is given by :

p=m\times v

p=950\ kg\times 10\ m/s

p = 9500 kg-m/s

So, the momentum of the car is 9500 kg-m/s. Hence, this is the required solution.

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A ball of mass M is suspended by a thin string (of negligible mass) from the ceiling of an elevator.uploaded image
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Answer:

(a) The elevator is traveling upward and its upward velocity is decreasing as it nears a stop at a higher floor.  T > mg

(b) The elevator is traveling upward and its upward velocity is increasing as it begins its journey towards a higher floor. T > mg

(c) The elevator is traveling downward and its downward velocity is decreasing as it nears a stop at a lower floor. T < mg

(d) The elevator is traveling downward at a constant velocity. T = mg

(e) The elevator is traveling downward and its downward velocity is increasing. T < mg

(f) The elevator is stationary and remains at rest. T = mg

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To answer this question, consider all the forces acting on the elevator.

The mass of the ball acting downwards due to gravity = mg

The tension on the string depends on upward or downwards force on the ball. T = m(a+g)

where a is acceleration and increase in velocity causes increase in acceleration, and vice versa. (a = v/t)

(a) The elevator is traveling upward and its upward velocity is decreasing as it nears a stop at a higher floor.

If the upward velocity is decreasing, its acceleration is also decreasing, and acceleration is not equal to Zero

T = m(a+g) > mg

(b) The elevator is traveling upward and its upward velocity is increasing as it begins its journey towards a higher floor.

If the upward velocity is increasing, its acceleration is also increasing.

Then, T = m(a+g) > mg

(c) The elevator is traveling downward and its downward velocity is decreasing as it nears a stop at a lower floor.

If the downward velocity is decreasing, its acceleration is also decreasing, and acceleration is not equal to Zero

T = m(a-g) < mg

(d) The elevator is traveling downward at a constant velocity

At constant velocity, acceleration is zero, because acceleration is the rate of change of velocity.

T = m(0+g) = mg

(e) The elevator is traveling downward and its downward velocity is increasing

If the downward velocity is increasing, its acceleration is also increasing

T = m(a-g) < mg

(f) The elevator is stationary and remains at rest.

if the elevator is at rest, its acceleration is zero

T = m(0+g) = mg

6 0
3 years ago
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