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2 years ago

12

Wave A has a high frequency and a short wavelength. Wave B has a lower frequency and a longer wavelength. Which wave carries gre

ater energy?

1 answer:

dusya [7]2 years ago

4 0

Answer:

wave A

Explanation:

shortest wavelength carry the most energy

more energy in a wave, the higher its frequency. The lower the frequency is, the less energy in the wave.

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3 years ago

A popular classroom demonstration is to place a gas can on a burner and boil water in it. Left unchecked this has the potential

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D) shrivel up, since the atmosphere exerts more force on the can as it cools.

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3 years ago

Compare the frequency, wavelength, and radiant energy of yellow visible light to that of purple visible light.

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Answer:

Violet has a higher frequency (approximately 7.5×1014 Hz 7.5 × 10 14 Hz ) than red light (approximately 4.3×1014 Hz 4.3 × 10 14 Hz ). Since the speed of both waves is the same, we infer that violet has a shorter wavelength (400 nm ) than red (700 nm ).

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4 0

1 year ago

The two speakers at S1 and S2 are adjusted so that the observer at O hears an intensity of 6 W/m² when either S1 or S2 is sounde

Zanzabum

Answer:

The minimum frequency is 702.22 Hz

Explanation:

The two speakers are adjusted as attached in the figure. From the given data we know that

$S_1 S_2$ =3m

$S_1 O$ =4m

By Pythagoras theorem

$S_2O=\sqrt{(S_1S_2)^2+(S_1O)^2}\\S_2O=\sqrt{(3)^2+(4)^2}\\S_2O=\sqrt{9+16}\\S_2O=\sqrt{25}\\S_2O=5m$

Now

The intensity at O when both speakers are on is given by

$I=4I_1 cos^2(\pi \frac{\delta}{\lambda})$

Here

I is the intensity at O when both speakers are on which is given as 6 $W/m^2$

I1 is the intensity of one speaker on which is 6 $W/m^2$

δ is the Path difference which is given as

$\delta=S_2O-S_1O\\\delta=5-4\\\delta=1 m$

λ is wavelength which is given as

$\lambda=\frac{v}{f}$

Here

v is the speed of sound which is 320 m/s.

f is the frequency of the sound which is to be calculated.

$16=4\times 6 \times cos^2(\pi \frac{1 \times f}{320})\\16/24= cos^2(\pi \frac{1f}{320})\\0.667= cos^2(\pi \frac{f}{320})\\cos(\pi \frac{f}{320})=\pm0.8165\\\pi \frac{f}{320}=\frac{7 \pi}{36}+2k\pi \\ \frac{f}{320}=\frac{7 }{36}+2k \\\\ {f}=320 \times (\frac{7 }{36}+2k )$

where k=0,1,2

for minimum frequency $f_1$ , k=1

${f}=320 \times (\frac{7 }{36}+2 \times 1 )\\\\{f}=320 \times (\frac{79 }{36} )\\\\ f=702.22 Hz$

So the minimum frequency is 702.22 Hz

5 0

2 years ago