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luda_lava [24]
2 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]2 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]2 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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How do physicists define velocity?
Natalija [7]

Answer:

Velocity, quantity that designates how fast and in what direction a point is moving.

Explanation:

7 0
3 years ago
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What force does the water exert (in addition to that due to atmospheric pressure) on a submarine window of radius 44.0 cm at a d
Butoxors [25]
Calculate the pressure due to sea water as density*depth.
That is, 
pressure = (1025 kg/m^3)*((9400 m)*(9.8 m/s^2) = 94423000 Pa = 94423 kPa

Atmospheric pressure is  101.3 kPa
Total pressure is  94423 + 101.3 = 94524 kPa (approx)

The area of the window is π(0.44 m)^2 = 0.6082 m^2

The force on the window is
(94524 kPa)*(0.6082 m^2) = 57489.7 kN = 57.5 MN approx
3 0
3 years ago
A particle with charge 6 mC moving in a region where only electric forces act on it has a kinetic energy of 1.9000000000000001 J
Vesna [10]

Answer:

16.9000000000000001 J

Explanation:

From the given information:

Let the initial kinetic energy from point A be K_A = 1.9000000000000001 J

and the final kinetic energy from point B be K_B = ???

The charge particle Q = 6 mC = 6 × 10⁻³ C

The change in the electric potential from point B to A;

i.e. V_B - V_A = -2.5 × 10³ V

According to the work-energy theorem:

-Q × ΔV = ΔK

-Q \times ( V_B - V_A) = (K_B - K_A)

-(6\times 10^{-3}\ C) \times ( -2.5 \times 10^3) = (K_B - 1.9000000000000001 \ J)

15 = (K_B - 1.9000000000000001 \ J)

K_B = 15+ 1.9000000000000001 \ J

\mathbf{K_B =1 6.9000000000000001 \ J}

3 0
3 years ago
31. Draw a free body diagram for a 15.5N box that is being pushed to the right with a 18. N force while experiencing 4.30 N of r
posledela

Answer:

See answers below

Explanation:

a.

F = mg,

15.5 N = m(9.8 m/s²)

m = 1.58 kg

b.

Fnet = Applied force - resistance,

Fnet = 18 N - 4.30 N,

Fnet = 13.70 N

Fnet = ma

13.70 N = (1.58 kg)a

a = 8.67 m/s²

For the free body diagram, draw a box with an upward arrow labeled 15.5 N, a downward label labeled 15.5 N, a right label labeled 18 N, and a left label labeled 4.30 N.

7 0
2 years ago
A 47-kg packing crate is pulled with constant speed across a rough floor with a rope that is at an angle of 37 degrees above the
levacccp [35]

Tension in the rope due to applied force will be given as

F = 142 N

angle of applied force with horizontal is 37 degree

displacement along the floor = 6.1 m

so here we can use the formula of work done

W = F d cos\theta

now we can plug in all values above

W = 142 * 6.1 * cos37

W = 691.8 J

So here work done to pull is given by 691.8 J


8 0
2 years ago
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