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3 years ago

7

Two person A and B of weight 60kg and 40kg respectively stand facing each other, and pull on a light rope streched between them.

Find the ratio of distance covered by them when they meet if their intial separation was 5 metre

1 answer:

goblinko [34]3 years ago

6 0

To be able to successfully pull the rope with balance, the work done by person A should be just equal to the work done by person B.

<span>workA = workB ----> 1</span>

Where: work = force * distance

In this case, force = mass * gravity

So work = mass * gravity * distance

Rewriting equation 1:

<span>massA * gravity * distanceA = massB * gravity * distanceB ---->2</span>

We know that distanceA + distanceB = 5 m, therefore distanceA = 5 – distanceB. Substituting this to equation 2 and cancelling constant g:

massA * distanceA = massB * distanceB

60 kg * (5 – distanceB) = 40 kg * distanceB

distanceB = 3 m

distanceA = 5 – distanceB = 2 m

<span>ratio is: distanceA / distanceB = 2/3 = 0.67</span>

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n = any integer which represent the number of repetition of the spectrum (n)1,2,3...)(Number of harmonic)

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How much work is required to accelerate a proton from rest up to a speed of 0.993 c? Express your answer with the appropriate un

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Answer:

(a). The work done is 7001 MeV.

(b). The momentum of this proton is $4.20\times10^{8}\ kg-m/s$ .

Explanation:

Given that,

Speed = 0.993 c

We need to calculate the work done

Using work energy theorem

The work done is equal to the kinetic energy relative to the proton

$W=K.E$

$W=\dfrac{m_{p}c^2}{\sqrt{1-\dfrac{v^2}{c^2}}}-m_{p}c^2$

Put the value into the formula

$W=\dfrac{1.67\times10^{-27}\times(3\times10^{8})^2}{\sqrt{1-(\dfrac{0.993c}{c})^2}}-1.67\times10^{-27}\times(3\times10^{8})^2$

$W=\dfrac{1.67\times10^{-27}\times(3\times10^{8})^2}{\sqrt{1-(0.993)^2}}-1.67\times10^{-27}\times(3\times10^{8})^2$

$W=1.122\times10^{-9}\ J$

$W=7001\ MeV$

(b). We need to calculate the momentum of this proton

Using formula of momentum

$p=\dfrac{m_{0}v}{\sqrt{1-\dfrac{v^2}{c^2}}}$

Put the value into the formula

$p=\dfrac{1.67\times10^{-27}\times0.993c}{\sqrt{1-(\dfrac{0.993c}{c})^2}}$

$p=\dfrac{1.67\times10^{-27}\times0.993c}{\sqrt{1-(0.993)^2}}$

$p=1.404\times10^{-26}c$

$p=4.20\times10^{8}\ kg-m/s$

Hence, (a). The work done is 7001 MeV.

(b). The momentum of this proton is $4.20\times10^{8}\ kg-m/s$ .

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