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Gennadij [26K]
2 years ago
9

Imagine you are waiting for a train to pass at a railroad crossing. Will the train whistle have a higher pitch as the train appr

oaches you or after it has passed you by?
Physics
1 answer:
frosja888 [35]2 years ago
7 0
THE DOPPLER EFFECT. Anyways, it would have a higher whistle as it approaches you, when it gets to you it only gets quieter because it leaves after. Think of a motorcycle going by, its loud coming to you then as it passes it gets quieter.
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Changing the velocity of a moving object will have little effect on its kinetic energy.
Alinara [238K]

Answer:Faulse

Explanation:

5 0
3 years ago
Read 2 more answers
How is it technically correct to say that a car making a u-turn can have a constant speed but cannot have a constant velocity?
saw5 [17]

During the "U" part of the turn, the car would follow an approximately circular path, and if it's moving at a constant speed, it would have to accelerate toward the center of the circle in order to change its direction.

5 0
2 years ago
Select the correct answer
DiKsa [7]

Answer:

C or 3

Explanation:

A: no they are called sources of sound. A is incorrect.

B: It does. Many people attest to this. But this is not a property of physics.

C: A media is required is the correct answer.

D: Dogs might. In general we don't.

I would pick C

7 0
2 years ago
An airplane during departure has a constant acceleration of 3 m / s².
Rama09 [41]

Constant acceleration of plane = 3m/s²

a) Speed of the plane after 4s

Acceleration = speed/time

3m/s² = speed/4s

S = 12m/s

The speed of the plane after 4s is 12m/s.

b) Flight point will be termed as the point the plane got initial speed, u, 20m/s

Find speed after 8s, v

a = 3m/s²

from,

a = <u>v</u><u> </u><u>-</u><u> </u><u>u</u>

t

3 = <u>v</u><u> </u><u>-</u><u> </u><u>2</u><u>0</u>

8

24 = v - 20

v = 44m/s

After 8s the plane would've 44m/s speed.

6 0
2 years ago
(a) Calculate the force (in N) needed to bring a 1100 kg car to rest from a speed of 85.0 km/h in a distance of 125 m (a fairly
nasty-shy [4]

(a) -2451 N

We can start by calculating the acceleration of the car. We have:

u=85.0 km/h = 23.6 m/s is the initial velocity

v = 0 is the final velocity of the car

d = 125 m is the stopping distance

So we can use the following equation

v^2 - u^2 = 2ad

To find the acceleration of the car, a:

a=\frac{v^2-u^2}{2d}=\frac{0-(23.6 m/s)^2}{2(125 m)}=-2.23 m/s^2

Now we can use Newton's second Law:

F = ma

where m = 1100 kg to find the force exerted on the car in order to stop it; we find:

F=(1100 kg)(-2.23 m/s^2)=-2451 N

and the negative sign means the force is in the opposite direction to the motion of the car.

(b) -1.53\cdot 10^5 N

We can use again the equation

v^2 - u^2 = 2ad

To find the acceleration of the car. This time we have

u=85.0 km/h = 23.6 m/s is the initial velocity

v = 0 is the final velocity of the car

d = 2.0 m is the stopping distance

Substituting and solving for a,

a=\frac{v^2-u^2}{2d}=\frac{0-(23.6 m/s)^2}{2(2 m)}=-139.2 m/s^2

So now we can find the force exerted on the car by using again Newton's second law:

F=ma=(1100 kg)(-139.2 m/s^2)=-1.53\cdot 10^5 N

As we can see, the force is much stronger than the force exerted in part a).

8 0
3 years ago
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