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Anon25 [30]
2 years ago
10

How does a push or pull affect motion?

Physics
1 answer:
STatiana [176]2 years ago
8 0

Answer:

The motion of an object acted on by a force depends partly on the strength of the push or pull. The stronger the push or pull, the faster the object will move.

Explanation:

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3000 N is exerted for 4.0 seconds on a 9500 kg object.<br><br> What is the change in momentum?
WARRIOR [948]
Force is the change in momentum over a specific time. The change of momentum is therefore the force multiplied by the time that the force acts, so 3000x4.0=12000 N s=12000 kg m/s
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2 years ago
A roller coaster has a mass of 275 kg. It sits at the top of a hill with height 85 m. If it drops from this hill, how fast is it
Minchanka [31]

kinematic equation

v squared = u squared + 2 a x s

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i thimk

4 0
2 years ago
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Four point charges have the same magnitude of 2.4×10^−12C and are fixed to the corners of a square that is 4.0 cm on a side. Thr
Gwar [14]

Answer:

7.2N/C

Explanation:

Pls see attached file

6 0
3 years ago
A 0.750 kg block is attached to a spring with spring constant 17.5 N/m. While the block is sitting at rest, a student hits it wi
Dmitriy789 [7]

Answer:

a

 A =  0.081 \  m

b

The value is  u =  0.2569 \  m/s

Explanation:

From the question we are told that

   The mass is  m  =  0.750 \ kg

   The spring constant is  k  =  17.5 \  N/m

    The instantaneous speed is  v  =  39.0 \  cm/s= 0.39 \  m/s

    The position consider is  x =  0.750A  meters from equilibrium point

   

Generally from the law of  energy conservation we have that

        The kinetic energy induced by the hammer  =  The energy stored in the spring

So

          \frac{1}{2} *  m * v^2  =  \frac{1}{2}  *  k  *  A^2

Here a is the amplitude of the subsequent oscillations

=>      A =  \sqrt{\frac{m *  v^ 2 }{ k} }

=>      A =  \sqrt{\frac{0.750 *  0.39 ^ 2 }{17.5} }

=>       A =  0.081 \  m

Generally from the law of  energy conservation we have that

The kinetic energy  by the hammer  =  The energy stored in the spring at the point considered   +   The kinetic energy at the considered point

             \frac{1}{2}  * m *  v^2 = \frac{1}{2}  * k x^2 + \frac{1}{2}  * m *  u^2

=>          \frac{1}{2}  * 0.750 *  0.39^2 = \frac{1}{2}  * 17.5* 0.750(0.081 )^2 + \frac{1}{2}  * 0.750 *  u^2

=>          u =  0.2569 \  m/s

3 0
2 years ago
A ball is tossed straight up from the surface of a small, spherical asteroid with no atmosphere. The ball rises to a height equa
alukav5142 [94]

Answer:

d. equal to one-fourth the acceleration at the surface of the asteroid.

Explanation:

The explanation is attached as a picture with this answer

Newton's law of universal gravitation is being used to compare the accelerations at the surface and at the top of the ball's path.

as it can be seen in the explanation that the proportional form of the equation is used because we do not need to necessarily use to final form with "G" for comparison calculations.

As per the given scenario only difference between the two points in the gravitational field is the distance from center of the spherical asteroid, i.e. r.

It is  taken 2r for the top is the path. hence we obtain (1/4)g as our answer.

4 0
3 years ago
Read 2 more answers
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