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larisa86 [58]
3 years ago
11

Two drag cars race. They line up at the starting line at rest. The winning car accelerates at a constant rate a and reaches the

finish line with a final velocity v. The losing car accelerates at a constant rate a/4. How far had the losing car traveled, d, when the winning car crossed the finish line?
Physics
1 answer:
Mila [183]3 years ago
5 0

Answer:d=\frac{v^2}{8a}

Explanation:

Given

winning car accelerates with a and its final velocity is v

considering they both start from rest

time taken by winning car is

v=u+at

where u=initial velocity

a=acceleration

t=time

v=at

t=\frac{v}{a}

Now loosing car is accelerating with \frac{a}{4}

Distance traveled by loosing car in time t

s_1=ut+\frac{at^2}{2\cdot 4}

s_1=0+\frac{a}{8}\times (\frac{v}{a})^2

s_1=\frac{v^2}{8a}

Thus distance d traveled by loosing car is given by d=\frac{v^2}{8a}

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Answer:

Explanation:

(A)

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(C)

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When n=2 , this is the second longest wavelength mode.

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When n=3, this is the third longest wavelength mode.

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f_i=\frac{v}{\lambda_i}

Here, f_i is the frequency of the i^{th} normal mode, v is wave speed, and \lambda_i is the wavelength of i^{th} normal mode.

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