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3 years ago

A crowd in a stadium is heard by a man standing 823.2 m away. The man hears the noise

Physics

1 answer:

Reil [10]3 years ago

8 0

Answer: 329.28 m/s

Explanation:

Given that:

Distance of sound = 823.2 m

Time taken for sound = 2.5 seconds

The speed of traveling sound = ?

Speed is obtained by dividing the distance travelled by the time taken for the travel.

Speed = Distance / time

Speed = 823.2m/2.5 seconds

Speed = 329.28 m/s

Thus, the sound traveling as fast as 329.28 m/s

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$T=51.64^\circ F$

$t=180.10s$

Explanation:

The Newton's law in this case is:

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Here, $T_m$ is the air temperture, C and k are constants.

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$70^\circ F$ in $t=0$

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$T(0)=70^\circ F\\T(0)=10^\circ F+Ce^{k(0)}\\70^\circ F=10^\circ F+C\\C=70^\circ F-10^\circ F=60^\circ F$

And we have $60^\circ F$ in $t=30 s$ , So:

$T(30)=60^\circ F\\T(30)=10^\circ F+(60^\circ F)e^{k(30)}\\60^\circ F=10^\circ F+(60^\circ F)e^{k(30)}\\50^\circ F=(60^\circ F)e^{k(30)}\\e^{k(30)}=\frac{50^\circ F}{60^\circ F}\\(30)k=ln(\frac{50}{60})\\k=\frac{ln(\frac{50}{60})}{30}=-0.0061$

Now, we have:

$T=10^\circ F+(60^\circ F)e^{-0.0061t}(1)$

Applying (1) for $t=1 min=60s$ :

$T=10^\circ F+(60^\circ F)e^{-0.0061*60}\\T=10^\circ F+(60^\circ F)0.694\\T=10^\circ F+41.64^\circ F\\T=51.64^\circ F$

Applying (1) for $T=30^\circ F$ :

$30^\circ F=10^\circ F+(60^\circ F)e^{-0.0061t}\\30^\circ F-10^\circ F=(60^\circ F)e^{-0.0061t}\\-0.0061t=ln(\frac{20}{60})\\t=\frac{ln(\frac{20}{60})}{-0.0061}=180.10s$

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