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devlian [24]
3 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]3 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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A particle of charge 2.0 x 10^-8C experiences an upward force of magnitude 4.0 x10^-6 when it is placed in a particular point in
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Answer:

a) The electric field at that point is 2.0\times 10^{2} newtons per coulomb.

b) The electric force is 2.0\times 10^{-6} newtons.

Explanation:

a) Let suppose that electric field is uniform, then the following electric field can be applied:

E = \frac{F_{e}}{q} (1)

Where:

E - Electric field, measured in newtons per coulomb.

F_{e} - Electric force, measured in newtons.

q - Electric charge, measured in coulombs.

If we know that F_{e} = 4.0\times 10^{-6}\,N and q = 2.0\times 10^{-8}\,C, then the electric field at that point is:

E = \frac{4.0\times 10^{-6}\,N}{2.0\times 10^{-8}\,C}

E = 2.0\times 10^{2}\,\frac{N}{C}

The electric field at that point is 2.0\times 10^{2} newtons per coulomb.

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F_{e} = E\cdot q

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F_{e} = 2.0\times 10^{-6}\,N

The electric force is 2.0\times 10^{-6} newtons.

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3 years ago
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