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raketka [301]
3 years ago
10

If a sound with frequency fs is produced by a source traveling along a line with speed vs. If an observer is traveling with spee

d vo along the same line from the opposite direction toward the source, then the frequency of the sound heard by the observer is fo = c + vo c − vs fs where c is the speed of sound, about 332 m/s. (This is the Doppler effect.) Suppose that, at a particular moment, you are in a train traveling at 32 m/s and accelerating at 1.3 m/s2. A train is approaching you from the opposite direction on the other track at 48 m/s, accelerating at 1.9 m/s2, and sounds its whistle, which has a frequency of 439 Hz. At that instant, what is the perceived frequency that you hear? (Round your answer to one decimal place.) Hz
Physics
1 answer:
Tcecarenko [31]3 years ago
5 0

Answer: The frequency heard is 562.7 Hz.

Explanation: <u>Doppler</u> <u>Effect</u> happens when there is shift in frequency during a realtive motion between a source and the observer of that source.

It can be calculated as:

f_{o} = f_{s}(\frac{c+v_{o}}{c-v_{s}} )

where:

c is the speed of light (c = 332m/s)

all the subscripted s is related to the Source (frequency, velocity);

all the subscripted o is related to the Observer (frequency, velocity);

As the source is moving towards the observer and the observer is moving towards the source, the velocities of each are opposite related to direction.

So, the frequency perceived by the observer:

f_{o} = 439(\frac{332+32}{332-48} )

f_{o} = 439(\frac{364}{284} )

f_{o} = 439(1.282 )

f_{o} = 562.7 Hz

At this condition, the observer hears the train's horn in a perceived frequency of 562.7 Hz

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A person observes a firework display for A safe distance of .750 km. Assuming that sound travels at 340 m/s in air what is the t
WINSTONCH [101]

Answer:

t = 2.2 s

Explanation:

Given that,

A person observes a firework display for A safe distance of 0.750 km.

d = 750 m

The speed of sound in air, v = 340 m/s

We need to find the between the person see and hear a firework explosion. let it is t. So, using the formula of speed.

v=\dfrac{d}{t}\\\\t=\dfrac{d}{v}\\\\t=\dfrac{750\ m}{340\ m/s}\\\\t=2.2\ s

So, the required time is 2.2 seconds.

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(1.08×10-3)(9.3×10-4)
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Which causes the thermocline?
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The downward force produced when air flows over the winglike spoiler on a race car is an example of ___ principle
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P_1 + \frac{1}{2}\rho v_1^2 = P_2 + \frac{1}{2}\rho v_2^2

here

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P_2 = pressure downwards

v_1 = velocity of air upwards

v_2 = velocity of air downwards

now from this equation we can say that the pressure difference will be

P_1 - P_2 = \frac{1}{2}\rho v_2^2 - \frac{1}{2}\rho v_1^2

now the force due to this pressure difference will be

F = (P_1 - P_2)A

so this is the above force which is given above

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3 years ago
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Suppose the ball is thrown from the same height as in the PRACTICE IT problem at an angle of 32.0°below the horizontal. If it st
scoray [572]

The figure of the problem is missing: find in attachment.

(a) 1.64 s

The ball follows a projectile motion path. The horizontal displacement is given by

x(t) = v_0 cos \theta t

where

v_0 is the initial speed

t is the time

\theta=32.0^{\circ} is the angle below the horizontal

We can rewrite this equation as

t=\frac{x(t)}{v_0 cos \theta} (1)

The vertical displacement instead is given by

y(t) = -v_0 sin \theta t - \frac{1}{2}gt^2 (2)

where

g=9.8 m/s^2 is the acceleration of gravity

Substituting (1) into (2),

y(t) = -x(t) tan \theta - \frac{1}{2}gt^2

We know that for t = time of flight, the horizontal displacement is

x(t) =50.8 m

We also know that the vertical displacement is

y(t) = -45 m

Substituting everything into the equation, we can find the time of flight:

\frac{1}{2}gt^2=-y -x tan \theta\\t=\sqrt{\frac{2(-y-xtan \theta)}{g}}=\sqrt{\frac{2(-(-45)-50.8 tan 32.0^{\circ})}{9.8}}=1.64 s

(b) 36.5 m/s

We can now find the initial speed directly by using the equation for the horizontal displacement:

x(t) = v_0 cos \theta t

where we have

x = 50.8 m

\theta=32.0^{\circ}

Substituting the time of flight,

t = 1.64 s

We find:

v_0 = \frac{x}{t cos \theta}=\frac{50.8}{(1.64)(cos 32.0^{\circ})}=36.5 m/s

(c) 47.1 m/s at 48.8 degrees below the horizontal

As the ball follows a projectile motion, its horizontal velocity does not change, so its value remains equal to

v_x = v_0 cos \theta = (36.5)(cos 32.0^{\circ})=31.0 m/s

The initial vertical velocity is instead

u_y = -v_0 sin \theta = -(36.5)(sin 32.0^{\circ})=-19.3 m/s

And it changes according to the equation

v_y = u_y -gt

So at t = 1.64 s (when the ball hits the ground),

v_y = -19.3 - (9.8)(1.64)=-35.4 m/s

So the impact speed is:

v=\sqrt{v_x^2+v_y^2}=\sqrt{(31.0)^2+(-35.4)^2}=47.1 m/s

While the direction is:

\theta=tan^{-1}(\frac{v_y}{v_x})=tan^{-1}(\frac{-35.4}{31.0})=-48.8^{\circ}

8 0
3 years ago
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