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garri49 [273]
3 years ago
12

g PRACTICE ANOTHER Two trains approach each other on separate but adjacent tracks. Train 1 is traveling at a speed of 30.3 m/s a

nd train 2 at a speed of 22.5 m/s. If the engineer of train 1 sounds his horn which has a frequency of 520 Hz, determine the frequency of the sound heard by the engineer of train 2. (Use 343 m/s as the speed of sound. Enter your answer to the nearest Hz.)
Physics
1 answer:
e-lub [12.9K]3 years ago
6 0

Answer:

  f'= 607.8 Hz

Explanation:

This is a Doppler effect exercise due to the relative velocity of the sound source and the observer.

By the time the source and the observer are getting closer the expression is

         f ’=f_o \ \frac{ v+ v_o}{v - v_s}

where vs is the speed of the source, vo is the speed of the observer, if the bodies move away the signs are exchanged

in this case, train 1 emits sound, so its speed is v_s = 30.3 m / s and train 2 is the receiver of the sound v₀ = 22.5 m / s

let's calculate

          f ’= 520 \ ( \frac{343 + 22.5}{343 - 30.3} )520 (343+ 22.5 / 343 - 30.3)

          f'= 607.8 Hz

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3 0
3 years ago
Jason has 13720 J of gravitational potential energy standing at the top of a cliff over the lake. If he jumps off the cliff and
Anna [14]

The conservation of energy and Newton's second law allows us to find the results about Jason's falling motion are;

  • The energy when reaching the water is K = 13720 J
  • The average force of the water to stop it is: F = 2744 N

<h3>Energy conservation.</h3><h3> </h3>

The conservation of energy is one of the most important principles of physics, stable that if there is no friction force, mechanical energy is conserved at all points.

Mechanical energy is the sum of kinetic energy plus potential energy.

Let's look for the energy at two points

Starting point. Get higher.

         Em₀ = U = 13720 J

Final point. Lower down.

         Em_f = K

Friction in the air is negligible, so energy is conserved.

          Em_o= Em_f

          K = 13720J

<h3>Kinematics and Newton's law.</h3><h3> </h3>

They indicate that it stops 5m under the water, if we assume that the water acts with a constant force, we can use kinematics and Newton's second law to find this force.

The kinematics expression to find the acceleration is

            v² =v₀² – 2ay

When it stops the speed is zero.

            a = \frac{v_o^2}{2y}  

 

Newton's second law is:

           F = ma

           F = m ( \frac{v_o^2}{2y} )

The expression for the kinetic energy is:

          K = ½ m v₀²

          v_o^2 = \frac{2K}{m}  

Let's substitute.

           F = m (\frac{2K}{m}) \frac{1}{2y}  

           F= \frac{K}{y}  

Let's calculate.

           F= \frac{13720}{5}  

           F = 2744N

In conclusion using conservation of energy and Newton's second law we can find the results about Jason's falling motion are;

  • The energy when reaching the water is K = 13720 J
  • The average force of the water to stop it is: F = 2744 N

Learn more about energy here:  brainly.com/question/14274074

6 0
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Your pet hamster sits on a record player whose angular speed is constant. If he moves to a point twice as far from the center, t
baherus [9]

Answer:

doubles

Explanation:

The relationship between angular speed and linear speed in case of a circular motion is the following:

v=\omega r

where

v is the linear speed

\omega is the angular speed

r is the distance from the centre of the circular path

In this proble, the pet hamster moves to a point twice as far from the center; this means that its distance from the centre, r', is doubled:

r'=2r

While the angular speed \omega remains the same. Therefore, the new linear speed is

v'=\omega r' = \omega (2r)=2 \omega r=2v

So, the correct answer is

the linear speed doubles

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4 years ago
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