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Lera25 [3.4K]
3 years ago
12

Why does the frequency of a siren get higher as an ambulance using that siren gets closer

Physics
2 answers:
Fantom [35]3 years ago
6 0

Explanation:

Doppler effect: There is an apparent change in the frequency when there is relative motion between the source and the listener. This phenomenon is Doppler effect.

When the source and the listener are moving towards each other then the frequency of the wave increases.

In the given problem, the frequency of a siren gets higher as an ambulance using that siren gets closer. The frequency of the siren increases due to Doppler effect. The pitch of the sound also becomes louder as the pitch depends on the frequency.

Nadya [2.5K]3 years ago
3 0
The frequency of a wave becomes higher due to the object moving at a fast pace coming towards you with shorter wavelengths  (depending on the speed) aka the Doppler Effect. 
Hope this helps
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An astronaut inside a spacecraft, which protects her from harmful radiation, is orbiting a black hole at a distance of 120 km fr
mestny [16]

An astronaut inside a spacecraft, which protects her from harmful radiation, is orbiting a black hole at a distance of 120 km from its center. The black hole is 5.00 times the mass of the sun and has a Schwarzschild radius of 15.0 km. The astronaut is positioned inside the spaceship such that one of her 0.030 kg ears is 6.0 cm farther from the black hole than the center of mass of the spacecraft and the other ear is 6.0 cm closer.

What is the tension between her ears?

Would the astronaut find it difficult to keep from being torn apart by the gravitational forces?

Answer:

The tension between the ears = 2.07 KN

The astronaut will find it difficult to keep and will eventually be in trouble because the tension is now greater compared to the tension in the human tissues.

Explanation:

Given that:

Orbital radius of the spacecraft (R) = 120 Km = 120 × 10³ m

Mass of the black hole (m) = 5 \ * (M \ _{sun})

where : M_{sun} = 1.99*10^{33} \ kg

Then; we have:

 m = 5*(1.99*10^{30} \ kg ) \\ = 9.95*10^{30} kg

Schwarzchild radius of the black hole

r - 15.0 km

Mass of each ear m_{ear} = 0.030 \ kg

Farther distance between one ear and the black hole (d) = 6.0 cm

= 0.06 m

Closer distance between the other ear and the black home is (d) 6.0 cm

= 0.6 cm

NOW, If we assume that the tension force should be T; then definitely the two ears will posses the same angular velocity .

The net force on the ear closer to the black hole will be:

\frac{GMm_{ear} }{(R-d)}- T = m_{ear} (R -  d) \omega^2

\frac{GMm_{ear} }{(R-d)^2}- \frac{T}{(R-d)} = m_{ear} \omega^2 \ ----> \ (1)

The net force on the ear farther to the black hole is :

\frac{GMm_{ear} }{(R+d)}- T = m_{ear} (R +  d) \omega^2

\frac{GMm_{ear} }{(R+d)^2}- \frac{T}{(R+d)} = m_{ear} \omega^2 \ ----> \ (2)

Equating equation (1) and (2) & therefore making (T) the subject of the formula; we have:

T = \frac{3GMm_{ear}d}{R^3}

T = \frac{3(6.67*10^{-11}N.m^2/kg^2)(1.95*10^{30}kg)(0.03kg)(0.06m)}{(120*10^3m)^3}

T = 2073.9 N\\T = 2.07 KN

The tension between the ears = 2.07 KN

The astronaut will find it difficult to keep and will eventually be in trouble because the tension is now greater compared to the tension in the human tissues.

3 0
4 years ago
A force of 7 N acts on an object. The displacement is, say 8 m, in the direction of the force. Let us take it that the force act
MA_775_DIABLO [31]

Answer:

cos 0 = 1.

Fs = 7×8 = 56 J

Explanation:

7 0
3 years ago
Use the drop-down menu to complete the statement. One difference between prokaryotic cells and eukaryotic cells is that a eukary
Ahat [919]

Answer:

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Explanation: I did the quiz and got it right. Have an amazing day!

8 0
3 years ago
Answer the following questions
kiruha [24]
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4 0
3 years ago
Read 2 more answers
1. Is the sequence geometric? If so, identify the common ratio.
frozen [14]
1) <span>yes;2 6*2=12 12*2=24 24*2=48
2)</span><span>Next Term (or nth term) = ar^n-1 
</span>
a = first term, i.e. 5 
<span>r = common ratio i.e. 3 (as 15/5=3 and 45/15=3 </span>
<span>n = .. </span>
<span>as you already have 1st , 2nd and 3rd terms</span>
<span>substituting now </span>
<span>T4= ar^n-1 </span>
<span>= 5*3^4-1 </span>
<span>= 5*3^3 </span>
<span>= 5*27 </span>
<span>T4 = 135 
</span>T5= ar^n-1 
<span>= 5*3^5-1 </span>
<span>= 5*3^4 </span>
<span>= 5*81 </span>
<span>T5 = 405 </span>


6 0
3 years ago
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