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Andrew [12]
3 years ago
5

You are trying to overhear a most interesting conversation, but from your distance of 10.0 m , it sounds like only an average wh

isper of 20.0 dB . So you decide to move closer to give the conversation a sound level of 60.0 dB instead. How close should you come
Physics
1 answer:
Alexandra [31]3 years ago
4 0

Answer:

r₂=0.1 m

Explanation:

Given that

r₁= 10 m  , β₁ = 20 dB

At r₂ ,β₂= 60 dB

As we know that intensity level of sound given as

\beta =10\ log\dfrac{I}{10^{-12}}

\beta _1=10\ log\dfrac{I_1}{10^{-12}}

20=10\ log\dfrac{I_1}{10^{-12}}

10² x 10⁻¹² = I₁

I₁=10⁻¹⁰ W/m²

\beta _2=10\ log\dfrac{I_2}{10^{-12}}

60=10\ log\dfrac{I_1}{10^{-12}}

10⁶ x  10⁻¹² = I₂

I₂ = 10⁻⁶ W/m²

I₁=10⁻¹⁰ W/m²

P = I A

P=Power ,I =Intensity  ,A=Area

\dfrac{I_1}{I_2}=\dfrac{r^2_2}{r^2_1}

\dfrac{10^{-10}}{10^{-6}}=\dfrac{r^2_2}{10^2}

r₂=0.1 m

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Answer:

L1/L2 = 6.47

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In order to calculate the ratio of the lengths of the wires you use the following formula for the resistivity of a wire:

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\rho_1=\frac{\pi r_1^2R_1}{L_1}=1.70*10^{-8}\Omega.m\\\\\rho_2=\frac{\pi r_2^2R_2}{L_2}=11.0*10^{10^{-8}}\Omega.m

The resistance and radius of the wires are the same, that is, R1 = R2 = R and r1 = r2 = r. By taking into account this last and dive the equation for the wire 2 into the wire 1, you obtain:

\frac{\rho_2}{\rho_1}=\frac{11.0*10^{-8}\Omega.m}{1.70*10^{-8}\Omega.m}=\frac{L_1}{L_2}\\\\\frac{L_1}{L_2}=6.47

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3 years ago
What happens when a light ray passes through the focal point, and then reflects off of a convex mirror?
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A convex mirror is sometimes referred to as a diverging mirror due to the fact that incident light originating from the same point and will reflect off the mirror surface and diverge. ... After reflection, the light rays diverge; subsequently they will never intersect on the object side of the mirror.

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Explanation:

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From Kinematics we have v^2= v_0^2+2a(y_f-y_i).

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When the rock touches the ground:

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