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ollegr [7]
2 years ago
7

A feather and a minivan are dropped vertically downward from a height of twenty meters and both are subject to air drag as they

fall. The minivan reaches the ground much faster than the feather. Which one of the following statements concerning this situation is true, if any?
a) The minivan has a larger terminal velocity than the feather because it experiences less air resistance than the feather.

b) The minivan encounters a smaller force of air resistance than the feather and falls faster.

c) Each object experiences the same amount of air drag, but the minivan experiences the greatest force of gravity.

d) The feather experiences more air drag than the minivan and has a smaller terminal velocity.
Physics
1 answer:
Luda [366]2 years ago
6 0

Answer:

maybe C i'm not 100% sure

I'm sorry if this didn't help

Explanation:

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Explanation:

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A positive kaon (K+) has a rest mass of 494 MeV/c² , whereas a proton has a rest mass of 938 MeV/c². If a kaon has a total energ
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Answer:

<em>0.85c </em>

Explanation:

Rest mass of Kaon M_{0K} = 494 MeV/c²

Rest mass of proton M_{0P}  = 938 MeV/c²

The rest energy is gotten by multiplying the rest mass by the square of the speed of light c²

for the kaon, rest energy E_{0K} = 494c² MeV

for the proton, rest energy E_{0P} = 938c² MeV

Recall that the rest energy, and the total energy are related by..

E = γE_{0}

which can be written in this case as

E_{K} = γE_{0K} ...... equ 1

where E = total energy of the kaon, and

E_{0} = rest energy of the kaon

γ = relativistic factor = \frac{1}{\sqrt{1 - \beta ^{2} } }

where \beta = \frac{v}{c}

But, it is stated that the total energy of the kaon is equal to the rest mass of the proton or its equivalent rest energy, therefore...

E_{K} = E_{0P} ......equ 2

where E_{K} is the total energy of the kaon, and

E_{0P} is the rest energy of the proton.

From E_{K} = E_{0P} = 938c²    

equ 1 becomes

938c² = γ494c²

γ = 938c²/494c² = 1.89

γ = \frac{1}{\sqrt{1 - \beta ^{2} } } = 1.89

1.89\sqrt{1 - \beta ^{2} } = 1

squaring both sides, we get

3.57( 1 - \beta^{2}) = 1

3.57 - 3.57\beta^{2} = 1

2.57 = 3.57\beta^{2}

\beta^{2} = 2.57/3.57 = 0.72

\beta = \sqrt{0.72} = 0.85

but, \beta = \frac{v}{c}

v/c = 0.85

v = <em>0.85c </em>

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[7 pts] 1. Suppose that a car starts from rest, its engine providing an acceleration of 4 ft/s2 while air resistance provides 0.
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Answer:

a) Initial Value Problem

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Explanations:

The complete explanations of each of the sections contained in the question are in the files attached to this solution.

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