A train station bell gives off a fundamental tone of 505 Hz as the train approaches the station at a speed of 27.6 m/s. If the speed of sound in air on that day is 339 m/s, what will be the apparent frequency of the bell to an observer riding the train


Answer 1


Apparent frequency of the bell to the observer is 546.12 Hz


The frequency of train bell (frequency of source) = 505 Hz

The speed of train (observer) = 27.6 m/s

The speed of sound in the air is (velocity of sound) = 339 m/s

The apparent frequency of the bell to the observer is calculated as follows:

Apparent frequency of bell to the observer.

= \text{frequency of source} * (Observer + velocity \ of \ sound )/( velocity \ of \ sound ) \n= 505 * (27.6 + 339)/(339) \n= 546.12 Hz

Related Questions

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Bailey wants to find out which frozen solid melts the fastest: soda, ice, or orange juice. She pours each of the three liquids into the empty cubes of an ice tray, and then places the ice tray in the freezer overnight. The next day, she pulls the ice tray out and sets each cube on its own plate. She then waits and watches for them to melt. When the last part of the frozen liquid melts, she records the time.
An airplane is traveling 835 km/h in a direction 41.5 ∘ west of north. Find the components of the velocity vector in the northerly and westerly directions. How far north and how far west has the plane traveled after 2.20 h ?

A tuning fork vibrates at 15,660 oscillations every minute. What is the period (in seconds) of one back and forth vibration of the tuning fork?


We are given:

The tuning fork vibrates at 15660 oscillations per minute

Period of one back-and forth movement:

the given data can be rewritten as:

1 minute / 15660 oscillations

60 seconds / 15660 oscillations          (1 minute  = 60 seconds)

dividing the values

0.0038 seconds / Oscillation

Therefore, one back and forth vibration takes 0.0038 seconds

A rubber ball is shot straight up from the ground with speed v0. Simultaneously, a second rubber ball at height h directly above the first ball is dropped from rest. a) At what height above the ground do the balls collide? Your answer will be a symbolic expression in terms of v0, h, and g .b) What is the maximum value of h for which a collision occurs before the first ball falls back to the ground?
Express your answer in terms of the variable v0 and appropriate constants.
c) For what value of h does the collision occur at the instant when the first ball is at its highest point?
Express your answer in terms of the variable v0 and appropriate constants.


In this exercise we have to have knowledge about the horizontal launch, so we have to use the known formulas to find that:

a) h*(1 - 1/2 g * h/v_0^2)

b)h = v_0^2/ g

c)h = v_0^2/ g

So we have to remember some famous equations like the position and velocity of an object moving in a constant, like this:

y = y_0 + v_0*t + 1/2 * a * t^2\nv = v_0 + a * t


  • y = height at time t
  • y0 = initial height
  • v0 = initial velocity
  • a = acceleration
  • t = time
  • v = velocity

a) When the balls collide, h1 = h2. Then,

h_1 = h_2\nv_0 * t - 1/2 g * t^2 = h - 1/2 * g * t^2\nv_0 * t = h\nt = h / v_0

Replacing in the equation of the height of the first ball:

h_1 = v_0 * h/v_0 - 1/2g * h^2/v_0^2\nh_1 = h - 1/2 g * h^2/ v_0^2\nh_1 = h*(1 - 1/2 g * h/v_0^2)

b)  that the balls collide at t = h/v0. Then:

h/ v_0 = -v_0/-g\nh = v_0^2/ g

c) The value of h for which the collision occurs at the instant when the first ball is at its highest point is the maximum value,Then:

h = v_0^2/ g

See more about velocity at brainly.com/question/862972

A 1200-kg cannon suddenly fires a 100-kg cannonball at 35 m/s. what is the recoil speed of the cannon? assume that frictional forces are negligible and the cannon is fired horizontally.



 Recoil velocity of cannon = 2.92 m/s


By law of conservation of momentum, we have momentum of cannon = momentum of cannonball.

 Mass of cannon = 1200 kg

Mass of cannon ball = 100 kg

Velocity of cannon ball = 35 m/s

 We have, Momentum of cannon = momentum of cannon ball

                  1200 x v = 100 x 35

                            v =3500/1200 = 2.92 m/s

 Recoil velocity of cannon = 2.92 m/s

Final answer:

The recoil speed of the cannon is 2.92 m/s.


To find the recoil speed of the cannon, we can use the conservation of momentum. The initial momentum of the cannon and cannonball system is zero since the cannon is at rest before firing. The final momentum is the sum of the momenta of the cannon and cannonball after firing. Using the equation:

Initial momentum = Final momentum

(mass of cannon) x (recoil speed of cannon) = (mass of cannonball) x (velocity of cannonball)

Plugging in the given values:

(1200 kg) x (recoil speed of cannon) = (100 kg) x (35 m/s)

Solving for the recoil speed of the cannon:

recoil speed of cannon = (100 kg x 35 m/s) / 1200 kg = 2.92 m/s

Learn more about recoil speed of cannon here:



What is the motion of the particles in this kind of wave? A hand holds the left end of a set of waves. The waves themselves make a larger set of waves in the same direction as that of the smaller waves. A label Wave motion is above the series of waves and an arrow next to the label points right. The particles will move up and down over large areas. The particles will move up and down over small areas. The particles will move side to side over small areas. The particles will move side to side over large areas.





D its incorrect in edge




The particles will move side to side over large areas

A space probe has two engines. Each generates the same amount of force when fired, and the directions of these forces can be in- dependently adjusted. When the engines are fired simultaneously and each applies its force in the same direction, the probe, starting from rest, takes 28 s to travel a certain distance. How long does it take to travel the same distance, again starting from rest, if the engines are fired simultaneously and the forces that they apply to the probe are perpendicular



t = 39.60 s


Let's take a careful look at this interesting exercise.

In the first case the two motors apply the force in the same direction

            F = m a₀          

           a₀ = F / m

with this acceleration it takes t = 28s to travel a distance, starting from rest

           x = v₀ t + ½ a t²

           x = ½ a₀ t²

           t² = 2x / a₀

           28² = 2x /a₀          (1)

in a second case the two motors apply perpendicular forces

we can analyze this situation as two independent movements, one in each direction


in the direction of axis a, there is a motor so its force is F/2


the acceleration on this axis is

          a = F/2m

          a = a₀ / 2

so if we use the distance equation

             x = v₀ t + ½ a t²

as part of rest v₀ = 0

             x = ½ (a₀ / 2) t²


let's clear the time

             t² = (2x / a₀)  2

we substitute the let of equation 1

             t² = 28² 2

             t = 28 √2

             t = 39.60 s

Calculate the linear momentum per photon,energy per photon, and the energy per mole of photons for radiation of wavelength; (a) 600 nm (red), (b) 550 nm (yellow), (c) 400 nm (violet), (d) 200 nm (ultraviolet), (e) 150 pm (X-ray), and (f ) 1.0 cm (microwave).



The detailed explanations is attached below


What is applied is the De brogile equation and the equation showing a relationship between Energy, speed of light and wavelength.

The explanation is as attached below.