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sesenic [268]
2 years ago
9

Which statement is best supported by the information in the chart? Wave X and Wave Y are mechanical waves, and Wave Z is an elec

tromagnetic wave. Wave X and Wave Y are electromagnetic waves, and Wave Z is a mechanical wave. Wave X and Wave Z are electromagnetic waves, and Wave Y is a mechanical wave. Wave X and Wave Z are mechanical waves, and Wave Y is an electromagnetic wave.
Physics
1 answer:
cluponka [151]2 years ago
5 0

Answer:

what??????????????????????

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Why didn’t early Portuguese and Dutch explorers decide to create settlements in Australia
aleksandrvk [35]
I think it is because most of the settlements in Australia are under the Spanish rule. Aside from that, Australia was part of the Spanish territory because of the papal bull mandated by Pope Alexander VI. The papal bull divided the world between Spain and Portugal. Spain has full control of the west while Portugal owns the east side of the world

Portuguese explorers were forced to study Australia in secret. 
3 0
3 years 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

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What is a learned behavior of a puppy
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Which statements describe a physical property of matter? Check all that apply. A cumulus cloud is puffy and white. Vinegar has a
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3 years ago
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What type of clouds usually accompany cold fronts?
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Answer: Cumulus

Explanation: Most large cloud fronts are made up of cumulus clouds, large storm clouds are cumulonimbus clouds.

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3 years ago
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