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anyanavicka [17]
3 years ago
14

A 60-kg skier is stationary at the top of a hill. She then pushes off and heads down the hill with an initial speed of 4.0 m/s.

Air resistance and the friction between the skis and the snow are both negligible. How fast will she be moving after she is at the bottom of the hill, which is 10 m in elevation lower than the hilltop
Physics
2 answers:
GREYUIT [131]3 years ago
6 0

Answer:

The velocity is  v = 8.85 m/s

Explanation:

From the question we are told that

    The mass of the skier is m_s =  60 \ kg

      The initial speed is u =  4.0 \ m/s

       The height is  h =  10 \ m

According to the law of energy conservation

     PE_t  + KE_t  =  KE_b + PE_b

Where PE_t is the potential energy at the top which is mathematically evaluated as

       PE_t =  mg h

substituting values

       PE_t  =  60 * 4*9.8

      PE_t  =  2352 \ J

And  KE_t is the kinetic energy at the top which equal to zero due to the fact that velocity is zero at the top of the hill

And  KE_b is the kinetic energy at the bottom of the hill which is mathematically represented as

         KE_b  =  0.5 * m * v^2

  substituting  values

         KE_b  =  0.5 * 60 * v^2

=>     KE_b  = 30 v^2

Where v is the velocity at the bottom

   And PE_b is the potential  energy at the bottom which equal to zero due to the fact that height  is zero at the bottom of the hill

So  

        30 v^2 =  2352

=>      v^2 =   \frac{2352}{30}

=>       v =  \sqrt{ \frac{2352}{30}}

        v = 8.85 m/s

         

Romashka [77]3 years ago
5 0

Answer:

The Skier's velocity at the bottom of the hill will be 18m/s

Explanation:

This is simply the case of energy conversion between potential and kinetic energy. Her potential energy at the top of the hill gets converted to the kinetic energy she experiences at the bottom.

That is

mgh = 0.5 mv^{2}

solving for velocity, we will have

v= \sqrt{2gh}

hence her velocity will be

v=\sqrt{2 \times 9.81 \times 10}=14.00m/s

This is the velocity she gains from the slope.

Recall that she already has an initial velocity of 4m/s. It is important to note that since velocities are vector quantities, they can easily be added algebraically. Hence, her velocity at the bottom of the hill is 4 + 14 = 18m/s

The Skier's velocity at the bottom of the hill will be 18m/s

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A ball of mass 0.160 kg is dropped from a height of 2.25 m. When it hits the ground it compresses 0.087 m.
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A) 6.64 m/s downward

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A)

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So we can write:

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The motion of the ballduring the collision is a uniformly accelerated motion (= with constant acceleration), so the time of impact can be found by using a suvat equation:

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