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zmey [24]
1 year ago
13

If the distance to a point source of sound is doubled, by what multiplicative factor does the intensity change?

Physics
1 answer:
alina1380 [7]1 year ago
8 0

If the distance to a point source of sound is doubled, by a multiplicative factor of 4, the intensity changes.

Intensity of sound is the sound which is perpendicular to sound wave propogation per unit area. It is dependent on the Surface of source sound.

Intensity is the Power per unit area. Its SI unit is Watt/m².

As we move away from a source of sound, the sound starts to diminish. This is due to the decreasing sound intensity with distance.

It can also be understood by the fact that on increasing distance, the Power radiated by the source spreads over a larger area. Hence, the Intensity decreases gradually.

Since, Intensity is proportional to the square of the distance.

Hence, on doubling the distance, Intensity reduces to one fourth of the initial intensity or reduces by a multiplicative factor of 4.

Learn more about Intensity here, brainly.com/question/17583145

#SPJ4

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What change in entropy occurs when a 0.15 kg ice cube at -18 °C is transformed into steam at 120 °c 4.
Studentka2010 [4]

<u>Answer:</u> The change in entropy of the given process is 1324.8 J/K

<u>Explanation:</u>

The processes involved in the given problem are:

1.)H_2O(s)(-18^oC,255K)\rightarrow H_2O(s)(0^oC,273K)\\2.)H_2O(s)(0^oC,273K)\rightarrow H_2O(l)(0^oC,273K)\\3.)H_2O(l)(0^oC,273K)\rightarrow H_2O(l)(100^oC,373K)\\4.)H_2O(l)(100^oC,373K)\rightarrow H_2O(g)(100^oC,373K)\\5.)H_2O(g)(100^oC,373K)\rightarrow H_2O(g)(120^oC,393K)

Pressure is taken as constant.

To calculate the entropy change for same phase at different temperature, we use the equation:

\Delta S=m\times C_{p,m}\times \ln (\frac{T_2}{T_1})      .......(1)

where,

\Delta S = Entropy change

C_{p,m} = specific heat capacity of medium

m = mass of ice = 0.15 kg = 150 g    (Conversion factor: 1 kg = 1000 g)

T_2 = final temperature

T_1 = initial temperature

To calculate the entropy change for different phase at same temperature, we use the equation:

\Delta S=m\times \frac{\Delta H_{f,v}}{T}      .......(2)

where,

\Delta S = Entropy change

m = mass of ice

\Delta H_{f,v} = enthalpy of fusion of vaporization

T = temperature of the system

Calculating the entropy change for each process:

  • <u>For process 1:</u>

We are given:

m=150g\\C_{p,s}=2.06J/gK\\T_1=255K\\T_2=273K

Putting values in equation 1, we get:

\Delta S_1=150g\times 2.06J/g.K\times \ln(\frac{273K}{255K})\\\\\Delta S_1=21.1J/K

  • <u>For process 2:</u>

We are given:

m=150g\\\Delta H_{fusion}=334.16J/g\\T=273K

Putting values in equation 2, we get:

\Delta S_2=\frac{150g\times 334.16J/g}{273K}\\\\\Delta S_2=183.6J/K

  • <u>For process 3:</u>

We are given:

m=150g\\C_{p,l}=4.184J/gK\\T_1=273K\\T_2=373K

Putting values in equation 1, we get:

\Delta S_3=150g\times 4.184J/g.K\times \ln(\frac{373K}{273K})\\\\\Delta S_3=195.9J/K

  • <u>For process 4:</u>

We are given:

m=150g\\\Delta H_{vaporization}=2259J/g\\T=373K

Putting values in equation 2, we get:

\Delta S_2=\frac{150g\times 2259J/g}{373K}\\\\\Delta S_2=908.4J/K

  • <u>For process 5:</u>

We are given:

m=150g\\C_{p,g}=2.02J/gK\\T_1=373K\\T_2=393K

Putting values in equation 1, we get:

\Delta S_5=150g\times 2.02J/g.K\times \ln(\frac{393K}{373K})\\\\\Delta S_5=15.8J/K

Total entropy change for the process = \Delta S_1+\Delta S_2+\Delta S_3+\Delta S_4+\Delta S_5

Total entropy change for the process = [21.1+183.6+195.9+908.4+15.8]J/K=1324.8J/K

Hence, the change in entropy of the given process is 1324.8 J/K

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Explanation:

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Rank these temperatures from hottest to coldest: 32° F,32° C, and 32 K 320 F&gt; 32° C&gt;32 K 32°C 32° F 32 K 32° F 32 K 32° c
Leviafan [203]

Answer:

32 C > 32 F > 32 K

Explanation:

32 F, 32 C, 32 K

Let T1 = 32 F

T2 = 32 C

T3 = 32 K

Convert all the temperatures in degree C

The relation between F and C is given by

(F - 32) / 9 = C / 100

so, (32 - 32) / 9 = C / 100

C = 0

So, T1 = 32 F = 0 C

The relation between c and K is given by

C = K - 273 = 32 - 273 = - 241

So, T3 = 32 K = - 241 C

So, T 1 = 0 C, T2 = 32 c, T3 = - 241 C

Thus, T2 > T1 > T3

32C > 32 F > 32 K

3 0
3 years ago
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