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luda_lava [24]
3 years ago
15

A surface receiving sound is moved from its original position to a position three times farther away from the source of the soun

d. the intensity of the received sound thus becomes
a. three times as high.
b. three times as low.
c. nine times as high.
d. nine times as low.
Physics
2 answers:
murzikaleks [220]3 years ago
5 0
As long as all the waves stay in the same medium, the intensity of
any waves ... electromagnetic or mechanical ... decrease in proportion
to the square of the distance.

If the distance increases to 3 x the original distance, then the intensity
changes to  1/3² or 1/9 of the original intensity.

I suppose choice-'d' is the correct one, but I have to tell you that
the phrase "nine times as low" is mathematically meaningless,
and it really grinds my gears.
shepuryov [24]3 years ago
5 0

Correct answer choice is :


D) Nine times as low.


Explanation:


Sound intensity level also identified as acoustic intensity is described as the energy taken by sound waves per unit area in a path perpendicular to that area. The SI unit of intensity, which involves sound intensity, is the watt per square meter (W/m2). The loudness of a sound describes the intensity of any given sound to the intensity at the start of sound. It is marked in decibels (dB).

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A cylinder contains a mixture of helium and argon gas in equilibrium at 160°C.
Artist 52 [7]

Answer:

a. E=8.97*10^{-21} J

b. v-He2 = 1.161 km/s

v-Ar2 = 369.74 m/s

Explanation:

(a) The average kinetic energy is given by

E_k=\frac{1}{2}k_BT

where KB is the Boltzman's constant and T is the temperature. For each molecule we have:

E_k=\frac{3}{2}(1.38066*10^{-23}\frac{J}{K})(433.15K)=8.97*10^{-21}J

is the same for both type of molecules because is independent of the mass

(b)

v_{rms}=\sqrt{\frac{3RT}{M}}

where R is the constant of ideal gases, and M is the mass of the molecule. BY replacing for each type of molecule we obtain:

v_{rms-He_{2}}=\sqrt{\frac{3(8.311434Jmol^{-1}K^{-1})(433.15K)}{0.008kg\ mol^{-1}}}=1161.9\frac{m}{s}\\\\v_{rms-Ar_{2}}=\sqrt{\frac{3(8.311434Jmol^{-1}K^{-1})(433.15K)}{0.079kg\ mol^{-1}}}=369.74\frac{m}{s}

v-He2 = 1.161 km/s

v-Ar2 = 369.74 m/s

hope this helps!!

8 0
3 years ago
the density of gas is 1.775kg/m^3 at 27 degree celcius and pressure of 10^-5n/m^2. if the specific heat capacity at constant pre
MrRissso [65]

Answer:

76.52

Explanation:

5 0
2 years ago
A sailboat starts from rest and accelerates at a rate of 0.21 m/s^2 over a distance of 280 m. find the magnitude of the boat's f
sasho [114]

We use the kinematic equations,

v=u+at                                          (A)

S= ut + \frac{1}{2} at^2                  (B)

Here, u is initial velocity, v is final velocity, a is acceleration and t is time.

Given,  u=0, a=0.21 \ m/s^2 and s= 280 m.

Substituting these values in equation (B), we get

280 \ m = 0 +\frac{1}{2} (0.21 m/s^2) t^2 \\\\ t^2 = \frac{280 \times 2}{0.21 } \\\\ t= 51.63 \ s.

Therefore from equation (A),

v = 0 + (0.21) \times (51.63 s)= 10.84 \ m/s

Thus, the magnitude of the boat's final velocity is 10.84 m/s and the time taken by boat to travel the distance 280 m is 51.63 s



8 0
3 years ago
With a force of 5 newtons, Amanda pushes the stack of books to the right. At the same time, Jeremy, her little brother, pushes t
Vaselesa [24]
Its letter C. 5N to the left. Since Jeremy's force in Newtons are higher than Amanda's (in newtons), and since Jeremy's force directs to the left, then the direction of the force will be to the LEFT. Then subtract the higher one to the lower one so that would be: 10N-5N=5N. So it is C. 5N to the left.
5 0
3 years ago
Read 2 more answers
PLEASE HELP ME I REALLY NEED HELP
allochka39001 [22]

Answer:

The answer is A.

Explanation:

The graph would show the unit rate otherwise known as a ratio

3 0
3 years ago
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