A train whistle is heard at 300 Hz as the train approaches town. The train cuts its speed in half as it nears the station, and the sound of the whistle is then 290 Hz. What is the speed of the train before and after slowing down?

Answers

Answer 1

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,

$f = f_0 ((v_0)/(v_0-v))$

Here,

$f_0$ = Frequency of Source

$v_s$ = 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,

$f = f_0 ((v_0)/(v_0-v))$

$300 = f_0 ((343)/(343-v))$

$(300*343) - 300v = 343f_0$

Now the second expression will be,

$f' = f_0 ((v_0)/(v_0-v/2))$

$290 = (343)((v_0)/(343-v/2))$

$290*343-145v = 343f_0$

Dividing the two expression we have,

$((300*343) - 300v)/(290*343-145v) = 1$

Solving for v, we have,

$v = 22.12m/s$

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

Answer 2

Answer:

Final answer:

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

Explanation:

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.

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Answers

Answer: 1.11 x 10⁸ Pa

Explanation:

At any deep, the absolute pressure is the same for all points located at the same level, and can be expressed as follows:

p = p₀ + δ. g . h, where p₀ = atmospheric pressure = 101, 325 Pa

Replacing by the values, we get:

p= 101,325 Pa + 1025 Kg/m³ . 9.8 m/s². 11,033 m = 1.11 x 10⁸ Pa.

Calculate the work done by a 50.0 N force on an object as it moves 9.00 m, if the force is oriented at an angle of 135° from the direction of travel. Explain what your answer means in terms of the object’s energy.

Answers

Answer:

Work done, W = -318.19 Joules

Explanation:

It is given that,

Force acting on the object, F = 50 N

Distance covered by the force, d = 9 m

Angle between the force and the distance traveled, $\theta=135^(\circ)$

The work done by an object is equal to the product of force and distance traveled. It is equal to the dot product of force and the distance. Mathematically, it is given by :

$W=Fd\ cos\theta$

$W=50* 9* \ cos(135)$

W = -318.19 Joules

So, the work done by the force is 318.19 Joules. The work is done in opposite to the direction of motion. Hence, this is the required solution.

Which two processes allow water to enter the atmosphere

Answers

The two processes that allow water to enter the atmosphere are: Evaporation and Transpiration.

1. **Evaporation:** Evaporation is the process by which liquid water on the Earth's surface (such as oceans, lakes, rivers, and even moist soil) is heated by the sun and turns into water vapor. This water vapor then rises into the atmosphere. Evaporation is a key component of the water cycle, where water constantly moves between the surface of the Earth and the atmosphere.

2. **Transpiration:** Transpiration is the process by which water is released into the atmosphere from plants. Plants take up water from the soil through their roots, and this water travels through the plant and eventually evaporates from small openings called stomata on the leaves. Transpiration serves various functions in plants, including cooling the plant and transporting nutrients.

Together, evaporation and transpiration contribute to the overall movement of water from the Earth's surface into the atmosphere, where it eventually condenses to form clouds and participates in various atmospheric processes before returning to the surface as precipitation through processes like rain, snow, sleet, or hail.

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Answer:

sublimation and transpiration

Explanation:

The water can enter the atmosphere from snow and ice with the process of sublimation where they also make water vapors. Last water can get in the atmosphere from plants through transpiration which means that the water is evaporated through the pores of the leaves

What operation do you apply to the position function of a particle to compute the particles velocity

Answers

Answer:

the derivative with respect to time

Explanation:

This is an exercise in kinematics, where the velocity is defined as a function of the position of a body of the form

v = dx/dt

where v is the velocity of the body, x is the position that we assume is a continuous and differentiable function.

The function written in the equation is the derivative with respect to time

A hydrogen atom contains a single electron that moves in a circular orbit about a single proton. Assume the proton is stationary, and the electron has a speed of 7.5 105 m/s. Find the radius between the stationary proton and the electron orbit within the hydrogen atom.

Answers

Answer:

450 pm

Explanation:

The electron is held in orbit by an electric force, this works as the centripetal force. The equation for the centripetal acceleration is:

a = v^2 / r

The equation for the electric force is:

F = q1 * q2 / (4 * π * e0 * r^2)

Where

q1, q2: the electric charges, the charge of the electron is -1.6*10^-19 C

e0: electric constant (8.85*10^-12 F/m)

If we divide this force by the mass of the electron we get the acceleration

me = 9.1*10^-31 kg

a = q1 * q2 / (4 * π * e0 * me * r^2)

v^2 / r = q1 * q2 / (4 * π * e0 * me * r^2)

We can simplify r

v^2 = q1 * q2 / (4 * π * e0 * me * r)

Rearranging:

r = q1 * q2 / (4 * π * e0 * me * v^2)

r = 1.6*10^-19 * 1.6*10^-19 / (4 * π * 8.85*10^-12 * 9.1*10^-31 * (7.5*10^5)^2) = 4.5*10^-10 m = 450 pm

A 2.4-kg ball falling vertically hits the floor with a speed of 2.5 m/s and rebounds with a speed of 1.5 m/s. What is the magnitude of the impulse exerted on the ball by the floor