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kykrilka [37]
2 years ago
11

a sports car accelerates from a standing start to 65 mi\h in 4.61s how far can it travel in that time

Physics
1 answer:
igor_vitrenko [27]2 years ago
3 0

The car can travel 220.27 ft in 4.61 s of time.

Given that a sports car accelerates from a standing start to 65 mi/hr.

So the initial velocity = 0 ft/s

The final velocity = 65 mi/hr.

We convert the final velocity from mi/hr. to ft/s.

1 mi/hr = 5280/3600 ft/s

          = 1.47 ft/s

Therefore final velocity = 65 x 1.47 = 95.55 ft/s

The time taken by the car = 4.61 s

Now the average acceleration of the car = final velocity - initial velocity / time taken = 95.55 - 0/ 4.61 s = 20.73 ft/s^2

The displacement of the car , s= ut +\frac{1}{2} at^2 ,

where u is the initial velocity of the car, t, the time taken by the car and a, the average acceleration.

So, displacement = 0 x 4.61 + 1/2 x 20.73 x 4.61^2

                             = 220.27 ft

Hence the distance travelled by the sports car in 4.61s is 220.27 ft.

Learn more about the displacement at brainly.com/question/17006213

#SPJ4

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Explanation:

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bentar mo numpang cerita nih

jadi gini

eist yg takut cerita horor Jan di baca

awokawokawok

ok lanjut ya dek ke cerita nya :)

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dah lah sampe disini aja dulu ya dek

byee byee

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6 0
3 years ago
Suppose a moving car has 2000 J of kinetic energy. If the carʹs speed doubles, how much kinetic energy would it then have?
Bond [772]

Answer:

option A

Explanation:

given,

Kinetic energy of the car = 2000 J

speed of the car is doubled

we know,

KE_1 = \dfrac{1}{2}mv^2

2000= \dfrac{1}{2}mv^2........(1)

now, speed of the car is doubled

v' = 2 v

KE_2 = \dfrac{1}{2}mv'^2

KE_2 = \dfrac{1}{2}m(2v)^2

from equation (1)

KE_2 = 4\times \dfrac{1}{2}m(v)^2

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The correct answer is option A.

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Explanation:

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