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3 years ago

What happens in terms of energy when a moving car hits a parked car, causing the parked car to move?

Physics

2 answers:

Helen [10]3 years ago

8 0

the answer to this question is c

meriva3 years ago

5 0

This is an example Newton's Third Law. All the kinectic energy from the moving car transferred the potential energy of the parked car. This potential is not much since the brakes are on (hopefully) and it's not in a non-moving position.

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A car travels at a constant velocity of 20.0 meters/second for 15 seconds. What is the power of the car if the initial force app

Zielflug [23.3K]

Work done by the force = Force x displacement. Power = work done/time = F.s/t = F.u.t/t = F.u = 95 x 20 = 1900J. {S=ut because acceleration is zero since car is moving at constant velocity}.

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3 years ago

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Two cars are traveling along a straight road. Car A maintains a constant speed of 95 km/h and car B maintains a constant speed o

natulia [17]

Answer

given,

Speed of car A = 95 Km/h

= 95 x 0.278 = 26.41 m/s

Speed of Car B = 121 Km/h

= 121 x 0.278 = 33.64 m/s

Distance between Car A and B at t=0 = 41 Km

a) Distance travel by car B

d = 26.41 t + 41000

speed of the car A = 33.64 m/s

distance = s x t

26.41 t + 41000 = 33.64 x t

7.23 t = 41000

t = 5670.82 s

time taken by Car B to cross Car A is equal to t = 5670.82 s

distance traveled by car A

D = s x t = 26.41 x 5670.82 = 149766.25 m = 149.76 Km

b) distance travel by the car B in 30 s after overtaking car A

D' = s x t = 33.64 x 30 = 1009.2 m = 1 Km

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3 years ago

Which kind of cloud touches the ground making it hard to see

luda_lava [24]

A funnel cloud is a funnel-shaped cloud of condensed water droplets. They usually appear with a rotating column of air. These extend from the bottom of a cloud that does not touch the ground or a water surface.

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3 years ago

A child is riding a merry-go-round that is turning at 7.18 rpm. if the child is standing 4.65 m from the center of the merry-go-

Alisiya [41]

First we need to convert the angular speed from rpm to rad/s. Keeping in mind that
$1 rev= 2 \pi rad$
$1 min = 60 s$
the angular speed is
$\omega = 7.18 \frac{rev}{min} \cdot \frac{2 \pi}{60} = 0.75 rad/s$

And so now we can calculate the tangential speed of the child, which is the angular speed times the distance of the child from the center of the motion:
$v= \omega r = (0.75 rad/s)(4.65 m)=3.50 m/s$

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3 years ago

PLEASE HELP ASAP WILL GIVE BRAINLIEST!!!!

nydimaria [60]

Explanation:

Please mark me as the brainliest answer

7 0

2 years ago

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