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tino4ka555 [31]
3 years ago
13

An alert physics student stands beside the tracks as a train rolls slowly past. He notes that the frequency of the train whistle

is 452 Hz when the train is approaching him and 442 Hz when the train is receding from him. Using these frequencies, he calculates the speed of the train. What value does he find
Physics
1 answer:
Dovator [93]3 years ago
4 0

Answer:The velocity of the train is 3.84m/s

Explanation:

According to the Doppler effect, if the source is moving towards you then the apparent frequency of the sound emitted by the source is higher and if the source is moving away from you then the apparent frequency of the sound emitted by the source is smaller.

This is given by:

fo = V +-Vo/ V +-Vo × source

Where fo= observed frequency

V= velocity of sound

Vo= vo it of the observer

fsource= frequency the source

Given:

Observed frequency of the approaching train fo1= 452Hz

The observed frequency of train= fo2= 442Hz

Velocity of sound= 334m/s

Velocity of source=?

Train approaching the observer is given by:

fo1= V/(V - Vs)× source ...eq1

Train passes the student is given by:

fo2= V/(V - Vs)×source ...eq2

Divide eq1 by eq2

452/442 = (343+Vs)/(343 - Vs)

1.02 =(343+Vs)/(343 -Vs)

Cross multiply

1.02(343- Vs) = 343 + Vs

350.76 - 1.02Vs = 343 + Vs

Collecting like terms

350.76 -343= 1.02Vs+ Vs

7.76 = 2.02Vs

Vs= 7.76/2.02

Vs= 3.84m/s

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I NEED HELP PLEASE, THANKS! Describe each Newton Law. :)
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Explanation:

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A rectangular block of steel measures 4cm*2cm*1.5cm*. its mass is 93.6g
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3 years ago
lanet R47A is a spherical planet where the gravitational acceleration on the surface is 3.45 m/s2. A satellite orbitsPlanet R47A
qaws [65]

2.6×10^6\:\text{m}

Explanation:

The acceleration due to gravity g is defined as

g = G\dfrac{M}{R^2}

and solving for R, we find that

R = \sqrt{\dfrac{GM}{g}}\:\:\:\:\:\:\:(1)

We need the mass M of the planet first and we can do that by noting that the centripetal acceleration F_c experienced by the satellite is equal to the gravitational force F_G or

F_c = F_G \Rightarrow m\dfrac{v^2}{r} = G\dfrac{mM}{r^2}\:\:\:\:\:(2)

The orbital velocity <em>v</em> is the velocity of the satellite around the planet defined as

v = \dfrac{2\pi r}{T}

where <em>r</em><em> </em>is the radius of the satellite's orbit in meters and <em>T</em> is the period or the time it takes for the satellite to circle the planet in seconds. We can then rewrite Eqn(2) as

\dfrac{4\pi^2 r}{T^2} = G\dfrac{M}{r^2}

Solving for <em>M</em>, we get

M = \dfrac{4\pi^2 r^3}{GT^2}

Putting this expression back into Eqn(1), we get

R = \sqrt{\dfrac{G}{g}\left(\dfrac{4\pi^2 r^3}{GT^2}\right)}

\:\:\:\:=\dfrac{2\pi}{T}\sqrt{\dfrac{r^3}{g}}

\:\:\:\:=\dfrac{2\pi}{(1.44×10^4\:\text{s})}\sqrt{\dfrac{(5×10^6\:\text{m})^3}{(3.45\:\text{m/s}^2)}}

\:\:\:\:= 2.6×10^6\:\text{m}

5 0
3 years ago
34. A train, starting from rest, accelerates along the platform at a uniform rate of 0.6 m/s2. A passenger standing on the platf
ira [324]

Answer:

4.08 s

Explanation:

Let the passenger took "t" time to catch the train

so in this case the total distance moved by the train + 5 m = total distance moved by the passenger

so we will have

distance moved by train is given as

d_1 = \frac{1}{2}(0.6) t^2

also the distance moved by passenger

d_2 = \frac{1}{2}(1.2) t^2

so we will have

d_1 + 5 = d_2

0.3 t^2 + 5 = 0.6 t^2

0.3 t^2 = 5

t = 4.08 s

3 0
3 years ago
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