Answer:

To solve this problem we will apply the concepts related to the Doppler effect. The Doppler effect is the change in the perceived frequency of any wave movement when the emitter, or focus of waves, and the receiver, or observer, move relative to each other. Mathematically it can be described as,

Here,

= Frequency of Source

= Speed of sound

f = Frequency heard before slowing down

f' = Frequency heard after slowing down

v = Speed of the train before slowing down

So if the speed of the train after slowing down will be v/2, we can do a system equation of 2x2 at the two moments, then,

The first equation is,

Now the second expression will be,

Dividing the two expression we have,

Solving for v, we have,

Therefore the speed of the train before and after slowing down is 22.12m/s

Answer:
### Final answer:

### Explanation:

### Learn more about Doppler effect here:

The speed of the train can be determined using the Doppler effect formula.

The question involves the Doppler effect, which is the change in frequency or wavelength of a wave as observed by an observer moving relative to the source of the wave. In this case, the train whistle's frequency changes from 300 Hz to 290 Hz as the train approaches the station.

To find the speed of the train before and after slowing down, we can use the formula for the Doppler effect:

**f' = f((v + v_o)/(v - v_s))**

Where:

**f'**is the observed frequency**f**is the source frequency**v**is the speed of sound**v_o**is the speed of the observer (here it is the train)**v_s**is the speed of the source (here it is the speed of sound)

By substituting the given values for observed frequency (290 Hz), source frequency (300 Hz), and the speed of sound (343 m/s), we can solve for the speed of the train before and after slowing down.

#SPJ3

The planet earth orbits around the sun and also spins around its own axis. 33% part (a) calculate the angular momentum of the earth in its orbit around the sun in kg • m2/s

A racing car is travelling at 70 m/s and accelerates at -14 m/s2. What would the car’s speed be after 3 s?

The position of a particular particle as a function of is given by r=(8.5t i+5.6j-2tk )m (a) Determine the particle’s velocity and acceleration as a function of time.(b) Find the instantaneous velocity and acceleration of the particle at t=3.0 s.

What is the density, in Mg/m3, of a substance with a density of 0.14 lb/in3? (3 pts) What is the velocity, in m/sec, of a vehicle traveling 70 mi/hr?

After a package is dropped from the plane, how long will it take for it to reach sea level from the time it is dropped? assume that the package, like the plane, has an initial velocity of 342 km/hour in the horizontal direction. express your answer numerically in seconds. neglect air resistance.

A racing car is travelling at 70 m/s and accelerates at -14 m/s2. What would the car’s speed be after 3 s?

The position of a particular particle as a function of is given by r=(8.5t i+5.6j-2tk )m (a) Determine the particle’s velocity and acceleration as a function of time.(b) Find the instantaneous velocity and acceleration of the particle at t=3.0 s.

What is the density, in Mg/m3, of a substance with a density of 0.14 lb/in3? (3 pts) What is the velocity, in m/sec, of a vehicle traveling 70 mi/hr?

After a package is dropped from the plane, how long will it take for it to reach sea level from the time it is dropped? assume that the package, like the plane, has an initial velocity of 342 km/hour in the horizontal direction. express your answer numerically in seconds. neglect air resistance.

**Answer:**

**Explanation:**

a )

Time period T = 1/3 s

angular velocity = 2π / T

= 2 x 3.14 x 3

ω = 18.84 radian / s

b )

Applying conservation of angular momentum

I₁ ω₁ = I₂ ω₂

I₁ / I₂ = ω₂ / ω₁

2 = ω₂ / ω

ω₂ = 2 ω

c )

(KE)initial = 1/2 I₁ ω²

(KE)final = 1/2 I₂ ω₂²

= 1/2 (I₁ / 2) (2ω)²

= I₁ ω²

c )

Change in rotational kinetic energy

= I₁ ω² - 1/2 I₁ ω²

= + 1/2 I₁ ω²

d )

This energy comes from the work done by centripetal force which is increased to increase the speed of rotation.

**Answer:**

**Explanation:**

Given;

Thickness of the glass plate,

refractive index of the glass plate,

wavelength of light source in vacuum,

distance between the source and the screen,

Distance travelled by the light from source to screen in vacuum:

So the no. of wavelengths in the vacuum:

.......................(1)

__Now we find the wavelength of the light wave in the glass:__

where:

wavelength of light in the medium of glass.

Now the no. of wavelengths in the glass:

............................(2)

From (1) & (2):

- total no. of wavelengths are there between the source and the screen:

The **rate **of change of **angular****displacement **is defined as **angular velocity**. The **angular velocity** will be **22.41rad/s.**

The rate of **change **of **angular displacement **is defined as **angular velocity**. Its unit is **rad/sec.**

ω = θ t

Where,

**θ** is the angle of **rotation**,

tis the **time**

**ω** is the **angular velocity**

The **given **data in the problem is;

**u** is the **initial****velocity**=0

**α** is the **angular****acceleration** = 4.0 rad/s²

**t **is the **time period**=

**n** is the **number** of **revolution** = 10 rev

From **Newton's **second equation of **motion **in terms of **angular velocity**;

Hence the **angular velocity **will be 22.41 rad/s.

To learn more about **angular****velocity **refer to the link

**Answer:**

= 22.41rad/s

**Explanation:**

First, we know that:

a = 4 rad/s^2

S = 10 rev = 62.83 rad

Now we know that:

where is the final angular velocity, the initial angular velocity, a is the angular aceleration and S the radians.

Replacing, we get:

Finally, solving for :

= 22.41rad/s

**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**

(10 Points)

increases

decreases

It decreasessssssssss

**Answer:**

**Explanation:**

The uncertainty in energy is given by

here h is plank's constant which value is and is the time interval which is given as

So using all the parameters the smallest possible uncertainty in electrons energy is