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3 years ago

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Two cars, one in front of the other, are travelling down the highway at 25 m/s. The car behind sounds its horn, which has a freq

uency of 640 Hz. What is the frequency heard by the driver of the lead car? (Vsound=340 m/s).
The answer choices are:

A) 463 Hz
B) 640 Hz
C)579 Hz
D) 425 Hz
E) 500 Hz

1 answer:

aalyn [17]3 years ago

7 0

Answer:

$f_s = 640 Hz$

Explanation:

For this case we know that the speed of the sound is given by:

$V_s = 340 m/s$

And we have the following info provided:

$v_c = 25 m/s$ represent the car leading

$v_s= 25 m/s$ represent the car behind with the source

$f_o = 640 Hz$ is the frequency for the observer

And we can find the frequency of the source $f_s$ with the following formula:

$f_s = \frac{v-v_o}{v-v_s} f_o$

And replacing we got:

$f_s = \frac{340-25}{340-25} *640 Hz = 640 Hz$

So then the frequency for the source would be the same since the both objects are travelling at the same speed.

$f_s = 640 Hz$

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3 years ago

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Travka [436]

#1

initial height = 25000 ft

as we know that

1 ft = 0.3048 m

so we have

H = 25000(0.3048) = 7620 m

#2

If mass of the man = 90 kg

now the initial potential energy is given as

$U = mgh$

$U = (90)(9.8)(7620)$

$U = 6.72 \times 10^6 J$

#3

Gravitational force that is acting on it is given by

$F = mg$

$F = 90(9.8) = 882 N$

so this is the force by which earth is attracting him towards it as we can see due to this force the man is accelerating towards the earth

#4

As we know that gravitational force is given by

$F = \frac{GMm}{r^2}$

here we know that

r = R + h

R = radius of earth ( $6.37 \times 10^6 m$ )

h = 7620 m

now we have acceleration at that point is

$a = \frac{F}{m}$

$a = \frac{GM}{r^2}$

$a = \frac{(6.67\times 10^{-11})(5.98 \times 10^{24})}{(6.37 \times 10^6 + 7620)^}$

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#5

by energy conservation we have

$KE = PE$

$\frac{1}{2}mv^2 = mgh$

$v^2 = 2gh$

$v^2 = 2(9.8)(7620)$

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#6

final speed due to wind resistance is 150 mph

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1 mile = 1609 meter

so this speed in m/s is given as

$150 mph = 150(\frac{1609 m}{3600 s})$

$v = 67 m/s$

#7

No Luke is not accelerating when his speed is 150 mph

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#8

since at 150 mph the acceleration of Luke is zero

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#9

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#10

Net will reduce the velocity of luke to be zero by taking long time

As we know that force on a system is given by

$F = \frac{\Delta P}{\Delta t}$

so if we increase the time interval here then we will have less force

so Luke will unhurt due to more time to decrease his speed to be zero

5 0

3 years ago