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yaroslaw [1]
3 years ago
15

A car is stopped at a traffic light. When the light turns green, the car starts from rest with an acceleration of 2.5 m/s2 . Als

o, as the light turns green, a truck traveling with a constant velocity of 10 m/s passes the car. The truck’s horn is stuck, emitting sound at a frequency of 440Hz. What frequency of sound is heard by the driver of the car, (a) 5 seconds before the light turned green? (b) 5 seconds after the light turned green? (c) 10 seconds after the light turned green?
Physics
1 answer:
erma4kov [3.2K]3 years ago
8 0

Answer:

a) f = 453.3Hz

b) f = 443.1Hz

c) f = 420Hz

Explanation:

First of all we need to know the positions and velocities of both vehicles.

For the car:

Xc(-5) = 0m;   Vc(-5) = 0m/s

Xc(5) = 2.5*5^2/2=31.25m;  Vc(5)=2.5*5 = 12.5m/s

Xc(10)=2.5*10^2/2=125m;    Vc(10)=2.5*10=25m/s

For the truck:

Xt(-5)=10*(-5) = -50m;   Vt(5)=10m/s

Xt(5)=10*5 = 50m;       Vt(5)=10m/s

Xt(10)=10*10=100m;     Vt(10)=10m/s

Now for part a) t=-5s. The truck is behind the car, so:

f=\frac{C}{C-Vt}*fo = 453.3Hz

Now for part b) t=5s. The car is behind the truck, so:

f=\frac{C+Vc}{C+Vt}*fo = 443.1Hz

Now for part b) t=10s. The truck is behind the car, so:

f=\frac{C-Vc}{C-Vt}*fo = 420Hz

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asambeis [7]

Complete Question:

Check the file attached to get the complete question

Answer:

In the film Ice word Revenge, vehicle 2 did not fall of the cliff because, Weight_{vehicle 1} < Weight_{vehicle 2} but in Claire's test, vehicle 2 off the cliff because Weight_{vehicle 1} \geq Weight_{vehicle 2}

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You shoot an arrow into the air. Two seconds later (2.00 s) the arrow has gone straight upward to a height of 35.0 m above its l
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This question can be solved by using the equations of motion.

a) The initial speed of the arrow is was "9.81 m/s".

b) It took the arrow "1.13 s" to reach a height of 17.5 m.

a)

We will use the second equation of motion to find out the initial speed of the arrow.

h= v_it + \frac{1}{2}gt^2\\

where,

vi = initial speed = ?

h = height = 35 m

t = time interval = 2 s

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Therefore,

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<u>vi =  9.81 m/s</u>

b)

To find the time taken by the arrow to reach 17.5 m, we will use the second equation of motion again.

h= v_it + \frac{1}{2}gt^2\\

where,

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t = time = ?

Therefore,

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solving this quadratic equation using the quadratic formula, we get:

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Therefore,

<u>t = 1.13 s</u>

Learn more about equations of motion here:

brainly.com/question/20594939?referrer=searchResults

The attached picture shows the equations of motion in the horizontal and vertical directions.

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Answer:

Explanation:

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