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mr Goodwill [35]
2 years ago
15

I NEED HELP ASAP!!!

Physics
2 answers:
Tamiku [17]2 years ago
7 0
I think option C is correct in my opinion
Savatey [412]2 years ago
3 0
I think option C is correct..hope it helps
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A car accelerates at a constant rate from 12 m/s to 27 m/s while it travels 125 m. How long does it take to achieve this speed?
MArishka [77]

Answer:

6.41 s

Explanation:

Under constant acceleration we know that

average velocity × time taken = displacement

s=\frac{u+v}{2}t

125=\frac{27+12}{2}t

    t = 6.41 s

The proof of used equation is given in the attachment.

3 0
3 years ago
For a body falling freely from rest​ (disregarding air​ resistance), the distance the body falls varies directly as the square o
jasenka [17]

Answer:

The answer to the question is

The object would fall 57.625 m in the first 5 seconds

Explanation:

To solve the question, we note that

the height of fall = 490 ft = ‪149.352‬ m

Time to touch the ground = 7 seconds

We are required to find out how far the object falls in the first 5 seconds

We apply the relation

S = u·t + 0.5×g·t ² = We then have

‪149.352‬ = U×7+0.5*9.81*49 From where u = -13 m/s

Therefore to find how far it falls in the first 5 seconds, we have

-13*5 + 0.5*9.81*25 = 57.625 m

5 0
3 years ago
Read 2 more answers
A race car accelerates from 16.5 m/s to 45.1 m/s in 2.27 seconds. Determine the acceleration of the car.
ioda

Answer:

12.6

Explanation:

4 0
3 years ago
Can someone help? Please?
zubka84 [21]

Answer:

A. Speed

Explanation:

Speed is the magnitude of velocity, which is given in the question. Velocity is a vector quantity and therefore has both a magnitude and a direction. Only the former is implied in the question.

5 0
3 years ago
Read 2 more answers
Two masses, each weighing 1.0 × 103 kilograms and moving with the same speed of 12.5 meters/second, are approaching each other.
juin [17]
A perfectly elastic<span> collision is defined as one in which there is no loss of </span>kinetic energy<span> in the collision. Therefore, we just add the kinetic energies of each system. We calculate as follows:

KE = 0.5(</span>1.0 × 10^3)(12.5 )^2 + 0.5(1.0 × 10^3)(12.5 )^2
KE = 156250 J = 1.6 x 10^5 J -------> OPTION A
5 0
3 years ago
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