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Viktor [21]
3 years ago
8

A high speed train is traveling at a speed of 44.7 m/s (100 mph) when the engineer sounds the 415 Hz warning horn. The speed of

sound is 343 m/s. What are the frequency and wavelength of the sound, as perceived by a person standing at a crossing when the train is (a) approaching and (b) leaving the crossing
Physics
1 answer:
Zarrin [17]3 years ago
8 0

Answer:

a) When the train is approaching the frequency is 477.19 Hz and the wavelenght is 0.719 m

b) When the train is leaving the crossing the frequency is 367.152 Hz and the wavelenght is 0.934 m

Explanation:

Given data:

v = speed of the train = 44.7 m/s

V = speed of the sound = 343 m/s

f = frequency of sound = 415 Hz

a) If the train is approaching, the frequency is:

f_{ap} =f(\frac{V}{V-v} )=415(\frac{343}{343-44.7} )=477.19Hz

The wavelength is:

w=\frac{V}{f_{ap} } =\frac{343}{477.19} =0.719m

b) If the train leaves the crossing, the frequency is:

f_{le} =f(\frac{V}{V+v} )=415(\frac{343}{343+44.7} )=367.152Hz

The wavelength is:

w=\frac{V}{f_{le} } =\frac{343}{367.152} =0.934m

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4 0
2 years ago
A block of mass 3.6 kg, sliding on a horizontal plane, is released with a velocity of 1.7 m/s. The block slides and stops at a d
pentagon [3]

Answer:

The block would have slid <u>12.6 m</u> if its initial velocity were increased by a factor of 2.8.

Explanation:

Given:

Mass of the block (m) = 3.6 kg

Initial velocity (u) = 1.7 m/s

Final velocity (v) = 0 m/s

Displacement (S) = 1.6 m

First we will find the acceleration of the block.

Using the equation of motion, we have:

v^2=u^2+2aS\\\\a=\dfrac{v^2-u^2}{2S}

Now, plug in the given values and solve for 'a'. This gives,

a=\frac{0-1.7^2}{2\times 1.6}\\\\a=\frac{-2.89}{3.2}=0.903\ m/s^2

The acceleration is negative as it is resisting the motion.

Now, the initial velocity is increased by a factor of 2.8. So,

New initial velocity = 2.8 × 1.7 = 4.76 m/s

Again using the same equation of motion and expressing the result in terms of 'S'. This gives,

v^2=u^2+2aS\\\\S=\dfrac{v^2-u^2}{2a}

Now, plug in the given values and solve for 'S'. This gives,

S=\frac{0-4.76^2}{2\times -0.903}\\\\S=\frac{-22.6576}{-1.806}=12.6\ m

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6 0
3 years ago
Help me with this plzzz
Marysya12 [62]

Answer:

yeah I'm Pretty sure it's b

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Answer:

1m/s^2

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Mass of block=10 kg

Applied horizontal force =F=20 N

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We have to find the acceleration of block.

Net force=Applied horizontal force-friction force

ma=F-f

Where F= Horizontal force

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