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3 years ago

The air in a kitchen has a mass of 60.0kg and a specific heat of 1505J/(kg°C).

Physics

2 answers:

BARSIC [14]3 years ago

8 0

Your answer is 632,100J which is Choice D

AveGali [126]3 years ago

5 0

Answer: The answer is D.

So you have 60kg of air, and the specific heat per kilogram is 1505j/(kg°C), this means you need 1505j to increase 1°C to the temperature of a mass of 1kg of air.

So, if the change in temperature is from 24°C to 31°C, you are increasing 7°C to 60kg of air.

then, for each degree you need 60*1505 j= 90300j

and for 7°c you need 7*90300j = 632100j, wich is answer D.

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NemiM [27]

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$v=f \lambda$

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For the waves in this string we have:

$f=625 Hz$ , since it completes 625 cycles per second

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So the speed of the wave is

$v=(625)(0.0333)=20.6 m/s$

The speed of the waves in a string is related to the tension in the string by

$v=\sqrt{\frac{T}{\mu}}$ (1)

where

T is the tension in the string

$\mu=\frac{m}{L}$ is the linear density

In this problem:

$m=16.5 g = 16.5\cdot 10^{-3} kg$ is the mass of the string

L = 0.75 m is the its length

Solving the equation (1) for T, we find the tension:

$T=\mu v^2 = \frac{m}{L} v^2 = \frac{16.5\cdot 10^{-3}}{0.75}(20.6)^2=9.34 N$

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