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saw5 [17]
3 years ago
10

Why does sound travel slower at a cold temperature?

Physics
2 answers:
tatyana61 [14]3 years ago
7 0
<h2>Answer:</h2><h2>It is due to a refractment of light.</h2>

Sound moves faster in warmer air than colder air the way bends away from the warm air and back towards of air.

natulia [17]3 years ago
3 0

Answer:

Sound travels slower in cold temperatures than in hot temperatures because the particles in cold air are more closely compacted together and move slower compared to the particles of hot air. Thus the sound, vibrations of particles in the atmosphere, will travel slower.

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PLEASE ANSWER QUESTION 25! NOT 24. Brainliest and 25 points!
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"The 4-wire setup is inherently safer and better able to prevent electrical shock, which in the case of a 220/240-volt circuit can be fatal."

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A wave has a wavelength of 5m and frequency of 5 Hz. What is the speed of the wave?​
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If you multiply m (the unit for wavelength) with 1s (the unit for frequency), you will get m/s, the unit for speed. Now multiply! 25 m/s is your final answer!
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3 years ago
efrigerant-134a is expanded isentropically from 600 kPa and 70°C at the inlet of a steady-flow turbine to 100 kPa at the outlet.
PolarNik [594]

Answer:

Inlet : v_i=0.0646\frac{m}{s}

Outlet:  v_o=0.171\frac{m}{s}

Explanation:

1) Notation and important concepts

Flow of mass represent "the mass of a substance which passes per unit of time".

Flow rate represent "a measure of the volume of liquid that moves in a certain amount of time"

Specific volume is "the ratio of the substance's volume to its mass. It is the reciprocal of density."

Isentropic process is a "thermodynamic process, in which the entropy of the fluid or gas remains constant".

We know that the flow of mass is given by the following expression

\dot{m}=\frac{\dot{V}}{\upsilon}, where \dot{V} represent the flow rate and \upsilon the specific volume at the pressure and temperature given.

A_i=0.5m^2 is the inlet area

P_i=600Kpa pressure at the inlet area

T_i=70C temperature at the inlet area

A_o=1m^2 is the outlet area

P_o=100Kpa pressure at the outlet area

T_o=C temperature at the outlet area

\dot{m}=0.75\frac{kg}{s} represent the flow of mass

If we look at the first figure attached Table A-13 we see that the specific volume for the inlet condition is

\upsilon_i =0.04304\frac{kg}{m^3} and the entropy is h_i=1.0645\frac{KJ}{KgK}=h_o

With the value of entropy and the outlet pressure of 100 Kpa we can find we specific volume at the outlet condition since w ehave the entropy h_o=1.0645\frac{KJ}{KgK}

Since on the table we don't have the exact value we need to interpolate between these two values (see the second figure attached)

h_1=1.0531\frac{KJ}{KgK} , \upsilon_1=0.22473\frac{kg}{m^3}

h_2=1.0829\frac{KJ}{KgK} , \upsilon_2=0.23349\frac{kg}{m^3}

Our interest value would be given using interpolation like this:

\upsilon=0.22473+\frac{(0.23349-0.22473)}{(1.0829-1.0531)}(1.0645-1.0531)=0.228\frac{kg}{m^3}

2) Solution to the problem

Now since we have all the info required to solve the problem we can find the velocities on this way.

We know from the definition of flow of mass that \dot{m}=\frac{\dot{V}}{\upsilon}, but since \dot{V}=Av we have this:

\dot{m}=\frac{Av}{\upsilon}

If we solve from the velocity v we have this:

v=\frac{\upsilon \dot{m}}{A}   (*)

And now we just need to replace the values into equation (*)

For the inlet case:

v_i=\frac{\upsilon_i \dot{m}}{A_i}=\frac{0.043069\frac{kg}{m^3}(0.75\frac{kg}{s})}{0.5m^2}=0.0646\frac{m}{s}

For the oulet case:

v_o=\frac{\upsilon_o \dot{m}}{A_o}=\frac{0.228\frac{kg}{m^3}(0.75\frac{kg}{s})}{1m^2}=0.171\frac{m}{s}

7 0
3 years ago
A vertically polarized beam of light of intensity 100 W/m2 passes through two ideal polarizers. The transmission axis of the fir
TEA [102]

To solve the problem it is necessary to apply the Malus Law. Malus's law indicates that the intensity of a linearly polarized beam of light, which passes through a perfect analyzer with a vertical optical axis is equivalent to:

I=I_0 cos^2\theta

Where,

I_ {0} indicates the intensity of the light before passing through the polarizer,

I is the resulting intensity, and

\theta indicates the angle between the axis of the analyzer and the polarization axis of the incident light.

Since we have two objects the law would be,

I=I_0cos^2\theta_1*cos^2(\theta_2-\theta_1)

Replacing the values,

I=100*cos^2(20)*cos^2(40-20)

I=100*cos^4(20)

I=77.91W/m^2

Therefore the intesity of the light after it has passes through both polarizers is 77.91W/m^2

7 0
3 years ago
6. A girl pushes her little brother on his sled with a force of 300 N for 750 m.
Oduvanchick [21]

w=225000 J  

E=225000 J  

P=9000 W  

P=22500 W

Ew=Fd=E  

P=E/t

5 0
3 years ago
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