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3 years ago

Why do sound engineers use spongy materials in the walls? (1 point)

Physics

1 answer:

mihalych1998 [28]3 years ago

8 0

Answer:

o to increase the frequency of sound waves. It increases the sound waves to a level of frequency that humans cannot hear so you won't be able to hear many things though the wall other then low noises like pounding.

Explanation:

I am in construction class as well as a student teacher for other construction type programs trust me :D

Brainiest would be appreciated

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1. Which letter represents the location of the battery in this diagram?

My name is Ann [436]

Answer:

location of battery in this diagram is at A and location of switch is at B.

4 0

3 years ago

Read 2 more answers

Two force vectors are oriented such that the angle between their directions is 38 degrees and they have the same magnitude. If t

Usimov [2.4K]

Answer:

Sum of the forces will be equal to 3.479 N

Explanation:

We have given two same forces are oriented at an angle of 38°

Magnitude of each force is given $F_1=F_2=1.84$

We have to find the sum of the forces

Sum of the forces will be equal to

$F=\sqrt{F_1^2+F_2^2+2F_1F_2cos\Theta }$

So sum of the forces will be equal to $F=\sqrt{1.84^2+1.84^2+2\times 1.84\times 1.84\times cos38^{\circ} }=3.479$

So sum of the force will be equal to 3.479 N

6 0

3 years ago

A student measured the specific heat of water to be 4.29 J/g.Co. The

leonid [27]

Answer:

2.63 %.

Explanation:

Given that,

The calculated value of the specific heat of water is 4.29 J/g.C

Original value of specific heat of water is 4.18 J/g.C.

We need to find the student's percent error. The percentage error in any quantity is given by :

$P=\dfrac{|\text{original value-calculated value}|}{\text{original value}}\times 100\\\\P=\dfrac{4.29-4.18}{4.18}\times 100\\\\P=2.63\%$

So, the student's percent error is 2.63 %.

7 0

2 years ago

Find the acceleration a body whose velocity increases from 11m/s to 33m/s in 10 seconds

solong [7]

Answer:

1.1 m/(s)^2

Explanation:

u=11 m/s

v=33 m/s

t=10s

v=u+at

=> 33=22+(a)(10)

=> 33-22=10a

=> 10a=11

=> a=11/10=1.1 m/(s)^2

7 0

2 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

5 0

2 years ago