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Butoxors [25]
3 years ago
8

A 1.5kg cart moves along a track at 0.20m/s until it hits a fixed bumper at the end of the track. Find the average force exerted

on the cart by the bumper if the cart is in contact with the bumper for 0.10s and it rebounds back along original direction with speed 0.125m/s
Physics
1 answer:
DochEvi [55]3 years ago
6 0

Answer:

Net force will be 4.875 N

Explanation:

We have given mass of the cart m = 1.5 kg

Initial velocity of the cart  = 0.2 m/sec

And final velocity of the car = - 0.125 m/sec ( Negative direction is due to opposite direction )

Instant of time \Delta t=0.1sec

Change in momentum is given by

\Delta P=1.5\times (0.2-(-0.125))=1.5\times 0.325=0.4875kgm/sec

Now force is given by

F=\frac{\Delta P}{\Delta t}=\frac{0.4875}{0.1}=4.875N

Net force will be 4.875 N

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A 1500 kg car traveling at 15.0 m/s to the south collides with a 4500 kg truck that is at rest at a stopligt. The car comes to a
Burka [1]

Answer:

5 m/s, moving to the South.

Explanation:

Parameters given:

Mass of car, m = 1500 kg

Initial velocity of car, u = 15 m/s

Mass of truck, M = 4500 kg

Initial velocity of truck, v = 0 m/s (Truck is at rest)

Final velocity of car, U = 0 m/s (Car comes to a stop)

Final velocity of truck = V

Because the collision is elastic, we can apply the principle of conservation of momentum, we have that:

Total initial momentum = Total final momentum

m*u + M*U = m*v + M*V

(1500 * 15) + (4500 * 0) = (1500 * 0) + (4500 * V)

22500 + 0 = 0 + 00V

=> V = 22500/4500

V = 5 m/s

The velocity carries a positive sign, hence, it's moving in the same direction as the car was moving initially.

That is, it's moving to the South.

8 0
3 years ago
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docker41 [41]
Air caught in the ball of foil makes the ball less dense than water
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A wire carries a steady current of 2.20
alina1380 [7]
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3 years ago
Question Part Points Submissions Used A car is stopped for a traffic signal. When the light turns green, the car accelerates, in
olya-2409 [2.1K]

(a) 328.6 kg m/s

The linear impulse experienced by the passenger in the car is equal to the change in momentum of the passenger:

I=\Delta p = m\Delta v

where

m = 62.0 kg is the mass of the passenger

\Delta v is the change in velocity of the car (and the passenger), which is

\Delta v = 5.30 m/s - 0 = 5.30 m/s

So, the linear impulse experienced by the passenger is

I=(62.0 kg)(5.30 m/s)=328.6 kg m/s

(b) 404.7 N

The linear impulse experienced by the passenger is also equal to the product between the average force and the time interval:

I=F \Delta t

where in this case

I=328.6 kg m/s is the linear impulse

\Delta t = 0.812 s is the time during which the force is applied

Solving the equation for F, we find the magnitude of the average force experienced by the passenger:

F=\frac{I}{\Delta t}=\frac{328.6 kg m/s}{0.812 s}=404.7 N

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3 years ago
Assume it takes 5.00 minutes to fill a 50.0-gal gasoline tank. (1 U.S. gal = 231 in.3).
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