All categories

3 years ago

You can hear someone around the corner before you can see them due to reflection, diffraction, refraction, or absorption?

Physics

1 answer:

Irina-Kira [14]3 years ago

7 0

The answer to this is reflection, because the sound waves are reflected off of the walls

You might be interested in

HELP ASAP

Scorpion4ik [409]

The answer is B. helium-filled balloon, since sound travels the slowest through gases.

8 0

4 years ago

Read 2 more answers

Is a process that modifies light waves so they vibrate in a single plane

madam [21]

The process you're fishing for is "polarization", but that's a

misleading description.

Polarization doesn't do anything to change the light waves.

It simply filters out (absorbs, as with a polarizing filter) the

light waves that aren't vibrating in the desired plane, and

allows only those that are to pass.

The intensity of a light beam is always reduced after

polarizing it, because much (most) of the original light

has been removed.

A laser light source may be thought of as an exception,

since everything coming out of the laser is polarized.

8 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.

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