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devlian [24]
4 years ago
7

A moderate force will break an egg. However, an egg droppedon the road usually breaks while one dropped on the grass usuallydoes

nt break. THis is because the egg dropped on thegrass............???
A. has a greater change in momentum
B. has a lesser change in momentum
C. the time interval for stoppin is greater
D. the time interval for stoppin is less
Physics
1 answer:
Fed [463]4 years ago
6 0

Answer:

C. the time interval for stopping is greater.

Explanation:

As the egg falls onto the grass, it takes a a greater amount of time for it to stop, and thus the force that is being applied to it is in increments; there is never enough force applied on the egg for it to break. That's why the egg doesn't break when it lands on the grass.

In contrast, when the egg is dropped on the road, <em>all of the force that is being applied by the road on the egg is in the tiny interval when the egg touches the road</em>, That force is large enough to break the egg because it is being applied in a tiny amount of time. That's why the egg dropped on the road breaks.

<em>So here is the rule of thumb: if you don't want to break your things but still want to drop them, drop them such that it takes  some for them to stop—because force will applied to them gradually. </em>

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Answer:

Approximately 2.1\; \rm km, assuming that g = -9.8\; \rm m \cdot s^{-2}.

Explanation:

Let t denote the time required for the package to reach the ground. Let h(\text{initial}) and h(\text{final}) denote the initial and final height of this package.

\displaystyle h(\text{final}) = \frac{1}{2}\, g\, t^2 + h(\text{initial}).

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\displaystyle t^{2} = \frac{2\, (h(\text{final}) - h(\text{initial}))}{g}.

\begin{aligned} t &= \sqrt{\frac{2\, (h(\text{final}) - h(\text{initial}))}{g} \\ &\approx \sqrt{\frac{2\times (0\; \rm m - 2500\; \rm m)}{(-9.8\; \rm m \cdot s^{-2})}} \approx 22.588\; \rm s\end{aligned}.

Assume that the air resistance on this package is negligible. The horizontal ("forward") velocity of this package would be constant (supposedly at 95\; \rm m \cdot s^{-1}.) From calculations above, the package would travel forward at that speed for about 22.588\; \rm s. That corresponds to approximately:95\; \rm m \cdot s^{-1} \times 22.588\; \rm s \approx 2.1 \times 10^{3}\; \rm m = 2.1\; \rm km.

Hence, the package would land approximately 2.1\; \rm km in front of where the plane released the package.

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