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bagirrra123 [75]
4 years ago
13

According to the Doppler Effect, which statement is true about a sound as it approaches you and then moves away.

Physics
2 answers:
Anarel [89]4 years ago
8 0

A) As the source approaches you, the sound waves are compressed so the pitch of the noise increases.

Fiesta28 [93]4 years ago
6 0
As the source approaches you, the sound waves are compressed, so
the pitch of the sound is higher than what the source is actually emitting.

Then, after it passes you and begins moving away, the sound waves
are stretched, so the pitch of the sound is lower than what the source
is actually emitting.
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A stone is thrown horizontally with an initial speed of 10 m/s from the edge of a cliff. A stopwatch measures the stone's trajec
olga2289 [7]

Answer:

The height of the cliff is 90.60 meters.

Explanation:

It is given that,

Initial horizontal speed of the stone, u = 10 m/s

Initial vertical speed of the stone, u' = 0 (as there is no motion in vertical direction)

The time taken by the stone from the top of the cliff to the bottom to be 4.3 s, t = 4.3 s

Let h is the height of the cliff. Using the second equation of motion in vertical direction to find it. It is given by :

h=u't+\dfrac{1}{2}gt^2

h=\dfrac{1}{2}gt^2

h=\dfrac{1}{2}\times 9.8\times (4.3)^2

h = 90.60 meters

So, the height of the cliff is 90.60 meters. Hence, this is the required solution.

5 0
3 years ago
OK brainly if you want hard here here are sixty cups on a table. If one falls down, then how many remain
vivado [14]

Answer:

If one cup falls down then there will be 59 cups left.

5 0
4 years ago
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A sailboat moves north for a distance of 10.00 km when blown by a wind from the exact south with a force of 5.00 x 10^4 n. how m
Nataliya [291]
Work is defined as the force times the distance which is mathematically expressed W = Fxd. The given force is 5x10^4 and the distance is 10000 m (the distance is converted as meter because Nm = J) the work done by the wind is W = 5x10^4 N (10000) = 500 x 10^6 Joules. I hope it answered your question
8 0
3 years ago
If two normal distributions each have the same mean, but one has a standard deviation of 1 and the other has a standard deviatio
likoan [24]

In comparison to a distribution with a standard deviation of 5, one has a curve with a higher peak. The variability will be greater the larger the standard deviation. It denotes increased variability in a distribution with a standard deviation of 5.

<h3>What do you mean by the term standard deviation?</h3>

The term "standard deviation" (or "") refers to a measurement of the data's dispersion from the mean. A low standard deviation implies that the data are grouped around the mean, whereas a large standard deviation shows that the data are more dispersed. In contrast, a high or low standard deviation indicates that the data points are, respectively, above or below the mean. A standard deviation that is close to zero implies that the data points are close to the mean. the curve at the top is more dispersed and has a greater standard deviation than the curve at the bottom, which is more concentrated around the mean and has a lower standard deviation.

To learn more about standard deviation, Visit:

brainly.com/question/14650840

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8 0
1 year ago
A railroad freight car rolls on a track at 3.50 m/s toward two identical coupled freight cars, which are rolling in the same dir
ladessa [460]

Answer:

Explanation:

Hello! To solve this problem we must be clear about the concept of energy conservation, and kinetic energy with the following sentence

The kinetic energy of the two cars (v = 1.2m / S) plus the kinetic energy of the third car (v = 3.5m / S) must be equal to the kinetic energy of the three cars together.

The kinetic energy is calculated by the following equation.

E=0.5mV^2

m= mass of the cars=26500kg

V=speed

E=kinetic energy

taking into account the above, the following equation is inferred

1=  the cars are separated

2= the cars are togheter

E1=E2

E1=0.5mV1^2+0.5mV1^2+0.5m(Va)^2

where

m= mass of each car

V1= 1.2m/s

Va=3.5,m/S

E2=0.5(3)(m)V^2

m= mass of each car

V=speed (in m/s) of the three coupled cars after the first couples with the other two

Solving

0.5mV1^2+0.5mV1^2+0.5m(Va)^2=0.5(3)(m)V^2

V1^2+V1^2+(Va)^2=(3)V^2.\\2V1^2+(Va)^2=(3)V^2\\V^2=\frac{2V1^2+(Va)^2}{3} \\

V=\sqrt{\frac{2V1^2+(Va)^2}{3}} \\V=\sqrt{\frac{2(1.2)^2+(3.5)^2}{3}} \\\\V=2.245m/s

the speed  of the three coupled cars after the first couples with the other two is 2.245m/s

7 0
4 years ago
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