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3 years ago

Which statement describes the reason for the change in the velocity of waves as temperature decreases?

Physics

1 answer:

DIA [1.3K]3 years ago

7 0

Answer:

O The particles of the medium move more slowly and there are fewer chances to transfer energy.

Explanation:

Various media are made up of particles. These particles are in constant motion according to the kinetic theory of matter. Recall that temperature has been defined as the average kinetic energy of the particles in a medium. Hence, for any given medium, the velocity of particle motion increases or decreases linearly with temperature.

The speed of particles in any medium increases or decreases as the temperature of the medium increases or decreases as emphasised above. Hence, at low temperature, the velocity of waves set up by the motion of particles in a medium decreases and transfer the wave energy to neighbouring particles occurs more slowly than at high temperatures.

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Answer:

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Explanation:

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3 years ago

You are driving along a highway at 35.0 m/s when you hear the siren of a police car approaching you from behind and you perceive

Irina-Kira [14]

Answer:

Explanation:

In the first case you can use the expression for the Doppler effect when the source is getting closer and getting away

$f'=f(\frac{v}{v-v_{s}})$ ( 1 )

$f''=f(\frac{v}{v+v_s})$ ( 2 )

f' = perceived frequency when the source is getting closer

f'' = perceived frequency when the source is getting away

f = source frequency

v = relative speed

vs = sound speed

by dividing (1) and (2) you have

$\frac{f'}{f''}=\frac{f}{f}\frac{\frac{v}{v-v_s}}{\frac{v}{v+v_s}}=\frac{v+v_s}{v-v_s}\\\\f'v-f'v_s=f''v+f''v_s\\\\v(f'-f'')=v_s(f''+f')\\\\v=v_s\frac{f''+f'}{f'-f''}=(340\frac{m}{s})\frac{1370Hz+1330Hz}{1370Hz-1330Hz}=67.5\frac{m}{s}$

but this is the relative velocity, you have that

$v=v_{sir}-v_{car}\\v_{sir}=v+v_{car}=67.5\frac{m}{s}+35\frac{m}{s}=102.5\frac{m}{s}$

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3 years ago

The heat flux for a given wall is in the x-direction and given as q^n = 11 W/m^2, the walls thermal conductivity is 1.7 W/mK and

MrMuchimi

Answer:

$\frac{dT}{dx} = 6.47 ^oC/m$

Also as we can see the equation that heat flux directly depends on the temperature gradient so more is the temperature gradient then more will be the heat flux.

For positive temperature gradient the heat will flow outwards while for negative temperature gradient the heat will flow inwards

Explanation:

As we know that heat flux is given by the formula

$q^n = K\frac{dT}{dx}$

here we know that

K = thermal conductivity

$\frac{dT}{dx}$ = temperature gradient

now we know that

$q^n = 11 W/m^2$

also we know that

K = 1.7 W/mK

now we have

$11 = 1.7 \frac{dT}{dx}$

so temperature gradient is given as

$\frac{dT}{dx} = \frac{11}{1.7} = 6.47 K/m$

also in other unit it will be same

$\frac{dT}{dx} = 6.47 ^oC/m$

Also as we can see the equation that heat flux directly depends on the temperature gradient so more is the temperature gradient then more will be the heat flux.

For positive temperature gradient the heat will flow outwards while for negative temperature gradient the heat will flow inwards

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3 years ago

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stepan [7]

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