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3 years ago

What is the difference between sounds that have the same pitch and loudness?

Physics

2 answers:

amm18123 years ago

8 0

Answer:

Option C

Sound quality

Explanation:

Sounds with the same pitch and loudness means they share natural frequency where frequency here implies the the number of vibrations that an individual particle makes per unit time (seconds). Additionally, when the pitch and loudness are the sane, the resonance and standing waves of these sounds will be similar. However, the quality of the sounds will vary. Therefore, option C is the correct one.

LuckyWell [14K]3 years ago

7 0

Answer:

"Sounds with the same pitch and loudness means they share natural frequency where frequency here implies the the number of vibrations that an individual particle makes per unit time (seconds). Additionally, when the pitch and loudness are the sane, the resonance and standing waves of these sounds will be similar. However, the quality of the sounds will vary. Therefore, option C is the correct one."

Explanation:

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Answer:

The amount of energy required is $152.68\times 10^{3}Joules$

Explanation:

The energy required to convert the ice to steam is the sum of:

1) Energy required to raise the temperature of the ice from -20 to 0 degree Celsius.

2) Latent heat required to convert the ice into water.

3) Energy required to raise the temperature of water from 0 degrees to 100 degrees

4) Latent heat required to convert the water at 100 degrees to steam.

The amount of energy required in each process is as under

1) $Q_1=mass\times S.heat_{ice}\times \Delta T\\\\Q_1=50\times 2.05\times 20=2050Joules$

where

$'S.heat_{ice}$ ' is specific heat of ice = $2.05J/^{o}C\cdot gm$

2) Amount of heat required in phase 2 equals

$Q_2=L.heat\times mass\\\\\therefore Q_{2}=334\times 50=16700Joules$

3) The amount of heat required to raise the temperature of water from 0 to 100 degrees centigrade equals

$Q_3=mass\times S.heat\times \Delta T\\\\Q_1=50\times 4.186\times 100=20930Joules$

where

$'S.heat_{water}$ ' is specific heat of water= $4.186J/^{o}C\cdot gm$

4) Amount of heat required in phase 4 equals

$Q_4=L.heat\times mass\\\\\therefore Q_{4}=2260\times 50=113000Joules\\\\$ $\\\\\\\\$ Thus the total heat required equals $Q=Q_{1}+Q_{2}+Q_{3}+Q_{4}\\\\Q=152.68\times 10^{3}Joules$

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