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Vanyuwa [196]
3 years ago
8

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. Nine times higher. B. Nine times lower. C. Three times higher. D. Three times lower.
Physics
1 answer:
Eva8 [605]3 years ago
4 0

The intensity of the sound will be B. Nine times lower

Explanation:

The intensity of a sound follows an inverse square law, which means that it is inversely proportional to the square of the distance from the source:

I\propto \frac{1}{r^2}

where

I is the intensity

r is the distance from the source

In this problem, the siund has an intensity of I when the receiver is placed at a distance r from the source.

Later, the receiver is placed three times farther away, so the new distance is

r' = 3r

Therefore, the new intensity of the sound will be:

I'\propto \frac{1}{r'^2}=\frac{1}{(3r)^2}= \frac{1}{9} (\frac{1}{r^2})= \frac{1}{9}I

Therefore, the intensity of the sound received will be nine times lower.

#LearnwithBrainly

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3 years ago
Is the relationship between velocity and centripetal force a direct, linear or nonlinear square relationship?
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Answer:

non linear square relationship

Explanation:

formula for centripetal force is given as

a = mv^2/r

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a = constant × v^2

a α v^2

hence non linear square relationship

5 0
3 years ago
A 7.5-cmcm-diameter horizontal pipe gradually narrows to 4.5 cmcm . When water flows through this pipe at a certain rate, the ga
tino4ka555 [31]

Answer

given,

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               d₂ = 4.5 cm

P₁ = 32 kPa

P₂ = 25 kPa

Assuming, we have calculation of flow in the pipe

using continuity equation

 A₁ v₁ = A₂ v₂

 π r₁² v₁ = π r₂² v₂

 v_1= \dfrac{r_2^2}{r_1^2} v_2

 v_1= \dfrac{2.25^2}{3.75^2} v_2

 v_1= 0.36 v_2

Applying Bernoulli's equation

 \Delta P = \dfrac{1}{2}\rho (v_2^2-v_1^2)

 P_1-P_2 = \dfrac{1}{2}\rho (v_2^2-(0.36 v_2)^2)

 32-25 = \dfrac{1}{2}1000\times v_2^2 (1 - 0.1269)

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 v_2=\sqrt{16.084}

       v₂ = 4.01 m/s

fluid flow rate

Q = A₂ V₂

Q = π (0.0225)²  x 4.01

Q = 6.38 x 10⁻³ m³/s

flow in the pipe is equal to 6.38 x 10⁻³ m³/s

4 0
3 years ago
A system gains 1500 J of heat, while the internal energy of the system increases by 4500 J and the volume decreases by . Assume
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Answer:

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Explanation:

Given data

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We know work done is

W= Q- ΔU

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The change in the volume at constant pressure is

ΔV= W/P

there fore P = W/ΔV= -3000/-0.01= 3×10^5

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3 0
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Answer:

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