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kotykmax [81]
2 years ago
10

A 6,600 kg train car moving at +2.0 m/s bumps into and locks together with one of mass 5,400 kg moving at -3.0 m/s.

Physics
1 answer:
Katarina [22]2 years ago
5 0
Option B would be right one
according to momentum conservation
6600*2 = 13200kgm/s
5400*3 = 16200kgm/s
16200-13200 = 3000
now 6600-5400 = 1200 kg
thus 3000/1200 = 2.5 v
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A person throws a pumpkin at a horizontal speed of 4.0 — off a cliff. The pumpkin travels 9.5 m horizontally
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Complete Question

A person throws a pumpkin at a horizontal speed of   4.0 m/s off a cliff. The pumpkin travels 9.5m horizontally before it hits the ground. We can ignore air resistance.What is the pumpkin's vertical displacement during the throw? What is the pumpkin's vertical velocity when it hits the ground?

Answer:

The  pumpkin's vertical displacement  is  H = 27 .67 \ m

The  pumpkin's vertical velocity when it hits the ground is  v_v__{f}} = 23.298 \  m/s

Explanation:

From the question we are told that

   The  horizontal speed is  v_h  =  4 m/s

    The horizontal distance traveled is  d =  9.5 \ m

The horizontal distance traveled is mathematically represented as

           S =  v_h * t

Where t is the time taken

substituting values

          9.5 =  4 * t

   =>     t =  \frac{9.5}{4}

            t = 2.38 \ sec

Now the vertical displacement is mathematically represented as

        H  =  v_v t  +  \frac{1}{2} a_v t^2

now the vertical velocity before the throw is  zero

    So

          H =  0 +  \frac{1}{2} (9.8) * (2.375)^2

          H = 27 .67 \ m

Now the final vertical velocity  is mathematically represented as

          v_v__{f}} =  v_v + at

  substituting values

             v_v__{f}} =  0 + (9.8)* (2.375)

            v_v__{f}} = 23.298 \  m/s

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3 years ago
Hooke’s law states that the distance that a spring is stretched by hanging object varies directly as the mass of the object. If
trasher [3.6K]

Answer:

d_{2} = 33.33 cm

Explanation:

Given:

When mass, m_{1} =21 kg

          distance travelled is  d_{1}  = 140 cm

When mass, m_{2} =5 kg

         distance travelled is  d_{2}  = ?

Hooke's law state that within elastic limit, when an external force is applied to a body, the body gets deformed and when the force is released the gets back to its original form.

Therefore according to the question,

\frac{d_{1}}{m_{1}}=\frac{d_{2}}{m_{2}}

\frac{140}{21}=\frac{d_{2}}{5}

d_{2} = 33.33 cm

Distance travelled is 33.33 cm when mass is 5 kg.

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