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aleksklad [387]
3 years ago
15

A train station bell gives off a fundamental tone of 500 Hz as the train approaches the station at a speed of 30.0 m/s. If the s

peed of sound in air is 339 m/s, what will be the apparent frequency of the bell to an observer riding the train?
Physics
1 answer:
TiliK225 [7]3 years ago
3 0

Answer:

548.5 Hz

Explanation:

We are given that

Frequency of source=500 Hz

Speed of train=30m/s

Speed of sound in air=339m/s

We have to find the frequency of the bell to an observer riding the train.

The apparent frequency for the listener

f=\frac{v_0}{v_0-v_s}f_1

Where f_1=Frequency of source

v_0=Speed of sound

v_s=Speed of source

Using the formula

The apparent frequency of the bell to an observer riding the train=\frac{339}{339-30}\times 500=548.5 Hz

Hence, the apparent frequency of the bell to an observer riding the train=548.5 Hz

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A building made with a steel structure is 565 m high on a winter day when the temperature is 0◦F. How much taller is the buildin
anyanavicka [17]

To solve this problem we apply the thermodynamic equations of linear expansion in bodies.

Mathematically the change in the length of a body is subject to the mathematical expression

\Delta L = L_0 \alpha \Delta T

Where,

L_0 = Initial Length

\alpha = Thermal expansion coefficient

\Delta T = Change in temperature

Since we have values in different units we proceed to transform the temperature to degrees Celsius so

0\°F \Rightarrow (0-32)*\frac{5}{9} = -17.77\°C

103\°F \Rightarrow (103-32)*(\frac{5}{9})= 39.44\°C

The coefficient of thermal expansion given is

\alpha = 1.1*10^{-5}/\°C

The initial length would be,

L_0 = 565m

Replacing we have to,

\Delta L = L_0 \alpha \Delta T

\Delta L = (565)(1.1*10^{-5})(39.44-(-17.77))

\Delta L = (565)(1.1*10^{-5})(39.44-(-17.77))

\Delta L = 0.355m

This means that the building will be 35.5cm taller

3 0
3 years ago
During destructive interference, two waves moving through the same medium will
ddd [48]
They will amplify eachother.
6 0
2 years ago
Read 2 more answers
If a dog walks north for 10 meters and then east for 10 meters, what is the direction of its displacement?
Katyanochek1 [597]

The direction of its displacement wil be

c.northeast

In fact, the dog walks north for 10 meters and east for another 10 meters. The path of the dog can be represented with two vectors, A pointing north (of magnitude 10 meters) and B pointing east (of magnitude 10 meters). The direction of the resultant vector (due to east) will be given by

tan \theta =\frac{A}{B}=\frac{10}{10}=1

\theta=tan^{-1} (1)=45^{\circ}

and the direction will be north-east.

5 0
3 years ago
Riding in a car, you suddenly put on the brakes. As you experience it inside the car, do Newton's law apply? Do they apply as se
alisha [4.7K]

Answer with Explanation:

Newton's laws are applicable for inertial frames of reference which is a frame which is not accelerating when seen from the observer standing on earth.

For the person as he presses the brakes his frame is a decelerating frame of reference hence he cannot apply the newtons laws of motion as they are in their original form but if he analyses the motion he has to apply a correction known as  pseudo-force on the object he is analyzing. Pseudo Force has no basis in newton's laws but are a correction that needs to be applied if he wishes to analyse the motion from non inertial frame of reference

While as a person standing on earth outside the car since his frame is an inertial frame of reference he can apply newton's laws of motion without any correction.  

3 0
3 years ago
What is the passenger's apparent weight at t=1.0s?
Fittoniya [83]

Answer:

For the complete question provided in explanation, if the elevator moves upward, then the apparent weight will be 1035 N. While for downward motion the apparent weight will be 435 N.

Explanation:

The question is incomplete. The complete question contains a velocity graph provided in the attachment. This is the velocity graph for an elevator having a passenger of 75 kg.

From the slope of graph it is clear that acceleration at t = 1 sec is given as:

Acceleration = a = (4-0)m/s / (1-0)s = 4 m/s^2

Now, there are two cases:

1- Elevator moving up

2- Elevator moving down

For upward motion:

Apparent Weight =  m(g + a)

Apparent Weight = (75 kg)(9.8 + 4)m/s^2

<u>Apparent Weight = 1035 N</u>

For downward motion:

Apparent Weight =  m(g - a)

Apparent Weight = (75 kg)(9.8 - 4)m/s^2

<u>Apparent Weight = 435 N</u>

4 0
3 years ago
Read 2 more answers
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