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wolverine [178]
3 years ago
8

In which situation can you be at rest and moving at then same time

Physics
1 answer:
anyanavicka [17]3 years ago
4 0

If you are stationary, but in/on a moving vehicle/object you can be at rest and moving at then same time.

<u>Explanation</u>:

  • A particle, when viewed from a given frame of reference, cannot be both at rest and in motion. However, in one frame of reference, a particle can be in motion whereas in another frame of reference the particle is in motion.
  • For example, if you are seated in a plane, the plane is stationary in that reference frame and the Earth moves under it, but in the reference frame of the Earth, the plane is moving concerning the Earth. When you are standing still on Earth, in your frame of reference, the Earth is stationary, and the Sun and stars move around the Earth.
  • However, in the frame of reference of the center of our solar system, the Earth orbits the Sun and the Sun are perturb slightly by the rest of the planets, but the rest of the galaxy orbits our solar system. Of course, in rest from our Galaxy, our solar system orbits a giant black hole at its center.
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What is an alpha particle composed of?
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Explanation:

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3 years ago
An object is thrown straight up into the air with an initial speed of 8 m/s, and reaches a greatest height of 15 m before it fal
Ugo [173]

Answer:

The value is T_t =  2.5659 \  s    

Explanation:

From the we are told that  

         The initial speed of the object is  u =  8 \  m/s

         The greatest height it reached is  h  =  15 \  m

Generally from kinematic equation we have that

      v^2 =  u^2 + 2gH

At maximum height v  =  0 m/s

So

      0^2 =  8^2 + 2 *  - 9.8 *  H

=>    H  =  3.27 \  m

Here H is the height from the initial height to the maximum height

So the initial height is mathematically represented as  

      s =  h - H

=>    s =  15 - 3.27

=>    s =  11.73 \  m

Generally the time taken for the object to reach maximum height is mathematically evaluated using kinematic equation as follows

            v  =  u + (-g) t

At maximum height v  =  0 m/s

           0 = 8 - 9.8t

=>         t = 0.8163 \  s

Generally the time taken for the object to move from the maximum height to the ground is mathematically using kinematic equation as follows

       h  =  ut_1 + \frac{1}{2}  gt_1^2

Here the initial velocity is  0 m/s given that its the velocity at maximum height

Also  g is positive because we are moving in the direction of gravity  

So

       15  =  0* t  +  4.9 t^2

=>      t_1  =  1.7496

Generally the total time taken is mathematically represented as

          T_t =  t + t_1

=>        T_t =  0.8163  +   1.7496

=>        T_t =  2.5659 \  s            

 

6 0
3 years ago
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Answer:

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To get the acceleration (magnitude and direction) of the man we apply

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