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aev [14]
3 years ago
7

A stationary siren emits sound of frequency 1000 Hz and wavelength 0.343 m. An observer who is moving toward the siren will meas

ure a frequency f and wavelength ? for this sound such that
Answer







f > 1000 Hz and ? > 0.343 m.








f > 1000 Hz and ? = 0.343 m.








f > 1000 Hz and ? < 0.343 m.








f = 1000 Hz and ? < 0.343 m
Physics
1 answer:
Ainat [17]3 years ago
4 0

Answer:

f>1000Hz and wavelength=0.343 m

Explanation:

We are given that

Frequency of stationary siren,f=1000 Hz

Wavelength of stationary sound,\lambda=0.343 m

When a observer is moving towards the siren then the frequency increases.

Therefore,an observer who is moving towards the siren measure a frequency >1000 Hz.

The wavelength depends upon the speed of source.

But we are given that siren is stationary.

Therefore, source is not moving and then the wavelength remains same.

f>1000Hz and wavelength=0.343 m

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A train, traveling at a constant speed of 22.0 m/s, comes to an incline with a constant slope. While going up the incline, the t
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Answer:

123.30 m

Explanation:

Given

Speed, u = 22 m/s

acceleration, a = 1.40 m/s²

time, t = 7.30 s

From equation of motion,

                       v = u + at

where,

v is the final velocity

u is the initial velocity

a is the acceleration

t is time  

                       V = at + U

using equation  v - u = at to get line equation for the graph of the motion of the train on the incline plane

                       V_{x} = mt + V_{o}      where m is the slope

Comparing equation (1) and (2)

V = V_{x}

a = m    

U = V_{o}

Since the train slows down with a constant acceleration of magnitude 1.40 m/s² when going up the incline plane. This implies the train is decelerating. Therefore, the train is experiencing negative acceleration.

          a = -  1.40 m/s²

Sunstituting a = -  1.40 m/s² and  u = 22 m/s

                        V_{x} = -1.40t + 22

                            V_{x} = -1.40(7.30) + 22

                             V_{x} = -10.22 + 22

                             V_{x} = 11. 78 m/s

The speed of the train at 7.30 s is 11.78 m/s.

The distance traveled after 7.30 sec on the incline is the area cover on the incline under the specific interval.

           Area of triangle +  Area of rectangle

          [\frac{1}{2} * (22 - 11.78) * (7.30)]  + [(11.78 - 0) * (7.30)]

                           = 37.303 + 85.994

                           = 123. 297 m

                           ≈ 123. 30 m

                 

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