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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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Three masses (3 kg, 5 kg, and 7 kg) are located in the xy-plane at the origin, (2.3 m, 0), and (0, 1.5 m), respectively.
Artist 52 [7]

Answer:

a) C.M =(\bar x, \bar y)=(0.767,0.7)m

b) (x_4,y_4)=(-1.917,-1.75)m

Explanation:

The center of mass "represent the unique point in an object or system which can be used to describe the system's response to external forces and torques"

The center of mass on a two dimensional plane is defined with the following formulas:

\bar x =\frac{\sum_{i=1}^N m_i x_i}{M}

\bar y =\frac{\sum_{i=1}^N m_i y_i}{M}

Where M represent the sum of all the masses on the system.

And the center of mass C.M =(\bar x, \bar y)

Part a

m_1= 3 kg, m_2=5kg,m_3=7kg represent the masses.

(x_1,y_1)=(0,0),(x_2,y_2)=(2.3,0),(x_3,y_3)=(0,1.5) represent the coordinates for the masses with the units on meters.

So we have everything in order to find the center of mass, if we begin with the x coordinate we have:

\bar x =\frac{(3kg*0m)+(5kg*2.3m)+(7kg*0m)}{3kg+5kg+7kg}=0.767m

\bar y =\frac{(3kg*0m)+(5kg*0m)+(7kg*1.5m)}{3kg+5kg+7kg}=0.7m

C.M =(\bar x, \bar y)=(0.767,0.7)m

Part b

For this case we have an additional mass m_4=6kg and we know that the resulting new center of mass it at the origin C.M =(\bar x, \bar y)=(0,0)m and we want to find the location for this new particle. Let the coordinates for this new particle given by (a,b)

\bar x =\frac{(3kg*0m)+(5kg*2.3m)+(7kg*0m)+(6kg*a)}{3kg+5kg+7kg+6kg}=0m

If we solve for a we got:

(3kg*0m)+(5kg*2.3m)+(7kg*0m)+(6kg*a)=0

a=-\frac{(5kg*2.3m)}{6kg}=-1.917m

\bar y =\frac{(3kg*0m)+(5kg*0m)+(7kg*1.5m)+(6kg*b)}{3kg+5kg+7kg+6kg}=0m

(3kg*0m)+(5kg*0m)+(7kg*1.5m)+(6kg*b)=0

And solving for b we got:

b=-\frac{(7kg*1.5m)}{6kg}=-1.75m

So the coordinates for this new particle are:

(x_4,y_4)=(-1.917,-1.75)m

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A person is hauling their taco stand and it takes 3,500 Joules of work to stop the taco stand. What force was exerted on the tac
Lapatulllka [165]

Given the work done, the force exerted on the taco stand through the given distance is 2.3 × 10³ Newtons.

<h3>What is Work done?</h3>

Work done is simply defined as the energy transfer that takes place when an object is either pushed or pulled over a certain distance by an external force. It is expressed as;

W = f × d

Where f is force applied and d is distance travelled.

Given that;

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  • Distance covered d = 1.5m
  • Force applied F = ?

W = f × d

3500kgm²/s² = f × 1.5m

f = 3500kgm²/s² ÷ 1.5m

f = 2.3 × 10³ kgm/s²

f = 2.3 × 10³ N

Given the work done, the force exerted on the taco stand through the given distance is 2.3 × 10³ Newtons.

Learn more about work done here: brainly.com/question/26115962

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