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almond37 [142]
3 years ago
12

A 150 g egg is dropped from 3.0 meters. The egg is moving at 4.4 m/s right before it hits the ground. The egg comes to a stop in

0.072 seconds.
What is the magnitude of force that the ground exerted on the egg?

0.66 N

9.2 N

13 N

180 N

Answer is 9.2 N
Physics
1 answer:
algol [13]3 years ago
8 0

Answer: Magnitude of the force exerted on the egg by the ground is 9.2N

Explanation:

Given the following :

Mass of egg (m) = 150g = 0.15kg

Height(h) from which egg is dropped = 3m

velocity of egg before hitting the ground (u) = 4.4m/s

Final velocity of egg (V) = 0

Time taken (t) = 0.072s

Magnitude of the force exerted on the egg by the ground can be found by applying Newton's 2nd law:

Momentum = mass × velocity

From Newton's second law:

Force = mass × change in Velocity with time ;

That is

F = m * ΔV / t

Inputting our values

F = 0.15 * (4.4 - 0) / 0.072

F = 0.15 × (4.4 / 0.072)

F = 0.15 × 61.11

F = 9.16N

F = 9.2N

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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

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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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