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solmaris [256]
3 years ago
13

Sam is walking through the park. He hears a police car coming down the street toward him. What happens to the sound of the siren

as the police car approaches and then passes Sam?
Sound waves from the siren remain constant. As the police car drives past Sam, the sound waves decrease in frequency.
The frequency of sound waves from the siren increase. As the police car continues past Sam, the sound waves decrease in frequency.
The frequency of sound waves from the siren decrease. As the police car drives past Sam, the sound waves increase in frequency.
The frequency of the sounds waves from the siren increase and continues to increase as the police car passes and drives away from Sam.
Physics
2 answers:
geniusboy [140]3 years ago
8 0
I believe it is the first one.
This is because of the Doppler Effect. 
The Doppler Effect is when the frequency of the wave changes for an observer moving relative to the wave source.
Basically, as the police car moves away from you, the distance between you and the car increases, which makes the sound waves spread out. 
Damm [24]3 years ago
5 0

Answer:

The correct answer is "The frequency of sound waves from the siren increase. As the police car continues past Sam, the sound waves decrease in frequency."

Explanation:

The correct answer is "The frequency of sound waves from the siren increase. As the police car continues past Sam, the sound waves decrease in frequency."

This is due to the effect called "Doppler Effect".

The Doppler effect is defined as the change in apparent frequency of a wave produced by the relative movement of the source with respect to its observer. In other words, this effect is the change in the perceived frequency of any wave movement when the sender and the receiver, or observer, move relative to each other.

According to this effect, the frequency perceived by the receiver will increase when the receiver and sender increase their separation distance and will decrease whenever the separation distance between them is reduced. As mentioned, then when the police car approaches Sam, the wave "compresses", that is, the wavelength is short, and the frequency is high. On the other hand, when the police car moves away, the wave "decompresses", that is, the wavelength is long and the frequency low.

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You are pushing a 60 kg block of ice across the ground. You exert a constant force of 9 N on the block of ice. You let go after
kotykmax [81]

Answer: d = 33 cm or 0.33 m

Explanation: In physics, Work is the amount of energy transferred to an object to make it move. It can be expressed by:

W = F.d.cosθ

F is the force applied to the object, d is the displacement and θ is the angle formed between the force and the displacement.

For the ice block, the angle is 0, i.e., force and distance are at the same direction, so:

W = F.d.cos(0)

W = F.d

To determine d:

d = \frac{W}{F}

d = \frac{3}{9}

d = 0.33 m

The distance d the block ice moved is 33 cm.

7 0
3 years ago
A crane lifts a 1,750 kg mass using a steel cable whose mass per unit length is 0.88 kg/m. What is the speed of transverse waves
Sauron [17]

Answer:

139.6m/s

Explanation:

Calculate the tension first, T=m*g

mass(m): 1750kg, gravity(g): 9.8m/s^2

T= 1750*9.8

 =17150N

Then calculate the wave speed using the equation v = √ (T/μ)

v= √(17150N)/(0.88kg/m)

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4 0
2 years ago
This is physics class, plzzz help ! 86.72 m to cm
Vaselesa [24]

Answer:

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5 0
3 years ago
.A hard rubber ball, released at chest height, falls to the pavement and bounces back to nearly the same height. When it is in c
ohaa [14]

Answer:

 a = 1.1 10⁵ m / s²

Explanation:

This is a momentum exercise, where we use the relationship between momentum and momentum

          I = ∫ F dt = Δp

= p_f - p₀

as they indicate that the ball bounces at the same height, we can assume that the moment when it reaches the ground is equal to the moment when it bounces, but in the opposite direction

        F t = 2 (m v)

therefore the average force is

         F = 2 m v / t

where in general the mass of the ball unknown, the velocity of the ball can be calculated using the conservation of energy

starting point. Done the ball is released with zero initial velocity

        Em₀ = U = mgh

final point. Upon reaching the ground, just before the deformation begins

        Em_f = K = ½ m v²

energy is conserved in this system

        Em₀ = Em_f

        m g h = ½ m v²

        v = √ (2gh)

This is the velocity of the body when it reaches the ground, so the force remains

        F = 2m √(2gh)   /t

where the height of the person's chest is known and the time that the impact with the floor lasts must be estimated in general is of the order of milli seconds

knowing this force let's use Newton's second law

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          a = 2 √(2gh) / t

We can estimate the order of magnitude of this acceleration, assuming the person's chest height of h = 1.5 m and a collision time of t = 1 10⁻³ s

         a = 2 √ (2 9.8 1.5) / 10⁻³

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Flura [38]
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