PHYSICS COLLEGE

A train is traveling at 30.0 m/sm/s relative to the ground in still air. The frequency of the note emitted by the train whistle is 262 HzHz. The speed of sound in air should be taken as 344 m/sm/s.A. What frequency fapproach is heard by a passenger on a train moving at a speed of 18.0 m/s relative to the ground in a direction opposite to the first train and approaching it?B. What frequency frecede is heard by a passenger on a train moving at a speed of 18.0 m/s relative to the ground in a direction opposite to the first train and receding from it?

Answers

Answer 1

Answer:

Answer

given,

speed of sound = 344 m/s

speed of train = 30 m/s

frequency emitted by the train = 262 Hz

Doppler's effect

$f_L = (v + v_L)/(v + v_s)\ f_S$

f_L is the frequency of listener

f_S is the frequency of the source of the sound

v is the speed of the sound

v_L is the speed of listener.

v_S is the speed of the source

a) Speed of the passenger in another train , v = 18 m/s

another train is moving in opposite direction and approaching

v_L is positive as the listener is moving forward.

v_S is negative at the source is moving toward the listener.

$f_L = (344 + 18)/(344 - 30)* 262$

$f_L = 302\ Hz$

b) Speed of the passenger in another train , v = 18 m/s

another train is moving in opposite direction and receding

v_L is negative as the listener is moving away from source.

v_S is positive at the source is moving away the listener.

$f_L = (344 - 18)/(344 + 30)* 262$

$f_L = 228.37\ Hz$

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Recall your experimental setup from Lab 05A: a constant force was applied to a disc by attaching a mass to a light string wrapped around a mass-less pulley and hanging the mass over the edge of the apparatus. In the lab, you used energy conservation arguments to derive an expression for the angular velocity of the disc after the mass had fallen a distance x . Your goal now is to use kinematics and dynamics to confirm your expression. Use the following symbols throughout this question: m is the mass of the hanging mass, M is the mass of the disc, r is the radius of the pulley, R is the radius of the disc, x is the distance the mass has fallen, and g is the acceleration due to gravity. What is the linear acceleration of the mass after it has fallen a distance x

Answers

Answer:

w = $\sqrt{(2gy)/(r^2 + (1)/(2) R^2 ) }$

Explanation:

For this exercise let's start by applying Newton's second law to the mass with the string

W - T = m a

In this case, as the system is going down, we will assume the vertical directional down as positive.

T = W - m a

Now we apply Newton's second law for rotational motion to the pulley of radius r. We will assume the positive counterclockwise rotations

∑ τ = I α

T r = I α

the moment of inertia of the disk is

I = ½ M R²

angular and linear acceleration are related

a = α r

we substitute

T r = (½ m R²) (a / r)

T = ½ m ( $(R)/(r)$ )² a

we write our two equations

T = W - m a

T = ½ m ( $(R)/(r)$ )² a

we solve the system of equations

W - m a = ½ m (\frac{R}{r} )² a

m g = m a [ 1 + ½ (\frac{R}{r} )² ]

a = $(g)/( 1 + (1)/(2) ((R)/(r))^2 )$

this acceleration is constant throughout the trajectory, so with the angular and lineal kinematics relations

w² = w₀² + 2 α θ

v² = v₀² + 2 a y

as the system is released its initial angular velocity is zero

w² = 0 + 2 α θ

v² = 0 + 2 a y

we look for the angular acceleration

a =α r

α = a / r

α = $(g)/(r (1 + (1)/(2) ((R)/(r))^2 )$

we look for the angle, remember that they must be measured in radians

θ = s / r

in this case we approximate the arc to the distance

s = y

θ = y / r

we substitute

w = $\sqrt{2 (g)/( r( (1)/(2) ((R)/(r))^2 ) (y)/(r) }$

w = $\sqrt{(2gy)/(r^2 + (1)/(2) R^2 ) }$

for the simple case where r = R

w = $\sqrt{ (2gy)/( (3)/(2) R^2 ) }$

w = $\sqrt{ (4)/(3) (gy)/(R^2) }$

Boyle's Law states that when a sample of gas is compressed at a constant temperature, the pressure P and the volume V satisfy the equation PV=C, where C is a constant. Suppose that at a certain instant the volume is 600cm3, the pressure is 150 kPa, and the pressure is increasing at a rate of 20 kPa/min. At what rate is the volume decreasing at this instant?

Answers

Answer: The volume is decreasing at a rate of 80 cm3/min

Explanation: Please see the attachments below

Answer: 80 cm³/min

Explanation:

Just solved it

An ordinary drinking glass is filled to the brim with water (268.4 mL) at 2.0 ° C and placed on the sunny pool deck for a swimmer to enjoy. If the temperature of the water rises to 32.0 ° C before the swimmer reaches for the glass, how much water will have spilled over the top of the glass? Assume the glass does not expand.

Answers

Answer:

$\Delta V=1.667*10^(-3)$

Explanation:

Given the initial temperature T_i=2° C

final temperature T_f= 32° C

The original volume of water Vo=268.8 mL= 0.2688 L

we need to calculate the change in the volume

As we know that volume expansion is given by

$(\Delta V)/(V_0)= \beta\Delta T$

ΔV= change in Volume

β= expansion coefficient = $207*10^(-6) K^(-1)$

therefore,

$\Delta V= \beta\Delta T V_0$

plugging values we get

$\Delta V=207*10^(-6) K^(-1) (32-2)*0.2688$

$\Delta V=1.667*10^(-3)$

A 2.4 kg toy oscillates on a spring completes a cycle every 0.56 s. What is the frequency of this oscillation?

Answers

Answer:

$f=1.79Hz$

Explanation:

The period is defined as the time taken by an object to complete a cycle in a simple harmonic motion. As the frequency of the motion increases, the period decreases. Therefore, they are inversely proportional. The frequency does not depend on the mass of the object.

$f=(1)/(T)\nf=(1)/(0.56 s)\nf=1.79Hz$

The word acid comes from the Latin word

Answers

Hi :)

The word acid comes from the Latin word acere, which means sour

Hope this helps!

It is acere but for future reference you can search of definition press more and google tells you its origin

When a hammer thrower releases her ball, she is aiming to maximize the distance from the starting ring. Assume she releases the ball at an angle of 54.6 degrees above horizontal, and the ball travels a total horizontal distance of 30.1 m. What angular velocity must she have achieved (in radians/s) at the moment of the throw, assuming the ball is 1.15 m from the axis of rotation during the spin?