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Bingel [31]
3 years ago
15

I have a question that's confusing, Will a large can of solids or empty can reach end of the ramp faster first in a can race? Al

so will a small can of solids or empty can reach end of ramp faster first? Explain briefly why Thank You so much
Physics
1 answer:
I am Lyosha [343]3 years ago
4 0
The large can of solids will reach before the empty can reason being is it has more additional weight its like dropping a 1 pound weight compared to a 4 pound the 4 pound reaches faster because of the gravitational pull the size of the pull depends on the masses of the objects 
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Answer:

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A proton moves along the x-axis in the laboratory with velocity uy = 0.6c. An observer moves with a velocity of v=0.8c along the
Dima020 [189]

Explanation:

Given that,

Velocity of the proton in lab frame u_{x} = 0.6c

Velocity of the observer v= 0.8c

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Using formula of velocity

u'=\dfrac{v-u_{x}}{1-\dfrac{u_{x}v}{c^2}}

u'=\dfrac{0.8c-0.6c}{1-\dfrac{0.6c\times0.8c}{c^2}}

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(a). We need to calculate the total energy of the proton in the lab frame

Using formula of kinetic energy

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Put the value into the formula

K.E=938.28\times(\dfrac{1}{\sqrt{1-\dfrac{(0.6c)^2}{c^2}}}-1)

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Using formula of kinetic energy

K.E=m_{0}c^2(\dfrac{1}{\sqrt{1-\dfrac{u'^2}{c^2}}}-1)

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K.E=938.28\times(\dfrac{1}{\sqrt{1-(0.385)^2}}-1)

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Using formula of momentum

P_{obs}=\dfrac{m_{0}u'^2}{\sqrt{1-\dfrac{u'^2}{c^2}}}

We know that,

Proton mass energy = m₀c²

m_{0}=\dfrac{938.28}{c^2}

P_{obs}=\dfrac{\dfrac{938.28}{c^2}\times(0.385c)^2}{\sqrt{1-\dfrac{(0.385c)^2}{c^2}}}

P_{obs}=\dfrac{938.28\times(0.385)^2}{\sqrt{1-0.385^2}}

P_{obs}=150.69\ MeV/c

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3 0
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Answer:

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==================================

In the real world, a dropped stone would only take  3.03  seconds
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Alternatively, a stone that fell for  4.6 seconds from rest would fall
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Answer:

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