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Olenka [21]
4 years ago
10

Your amplifier is playing a 200 Hz tone. When you double the amplitude(loudness), what happens to the speed of the sound (v) and

the frequency (f)? a. v doubles, f stays the same b. v doubles, f also doubles c. v stays the same, and f doubles d. v and f both stay the same e. v and f both decrease
Chemistry
1 answer:
kherson [118]4 years ago
8 0
Its d. Speed (v) and frequency(f) both stay the same.

<span>Because Amplitude does not affect wave frequency. When you put more energy into the wave (larger amplitude) it does not affect how often you repeat a cycle. 
</span>Also speed can be altered when the medium changes through which it travels.
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Answer:

(a) ΔU = 7.2x10²

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From the definition of power (p), we have:

p = \frac {\Delta W}{\Delta t} = \frac {\Delta U}{\Delta t} (1)

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\Delta U = p \cdot \Delta t = 100 \frac{J}{s} \cdot 2.0\cdot 10^{-2} h \cdot \frac{3600s}{1h} = 7.2 \cdot 10^{2} J  

(b) The work is related to pressure and volume by:

\Delta W = -p \Delta V

<em>where p: pressure and ΔV: change in volume = V final - V initial      </em>

\Delta W = - p \cdot (V_{fin} - V_{ini}) = - 1.0 atm (5.88L - 0.85L) = - 5.03 L \cdot atm \cdot \frac{101.33J}{1 L\cdot atm} = -5.1 \cdot 10^{2} J

(c) By the definition of Energy, we can calculate q:

\Delta U = \Delta W + \Delta q

<em>where Δq: is the heat transfer </em>

\Delta q = \Delta U - \Delta W = 7.2 J - (-5.1 \cdot 10^{2} J) = 5.2 \cdot 10^{2} J    

I hope it helps you!  

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Read 2 more answers
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