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Arlecino [84]
2 years ago
12

if the intensity of a person's voice is 4.6 x 10^-7 w/m^2 at a distance of 2.0 m, how much sound power does that person generate

.
Physics
1 answer:
777dan777 [17]2 years ago
4 0

The sound power the person generated is 2.313 \times 10^{-5} \ W.

<h3>Area of the sound wave</h3>

The area of the sound wave is calculated as follows;

A = 4\pi r^2\\\\A = 4 \pi \times (2)^2\\\\A = 50.272 \ m^2

<h3>Power generated</h3>

The sound power the person generated is calculated as follows;

P = I A\\\\P = 4.6\times 10^{-7} \ W/m^2 \ \ \times \ \ 50.272 \ m^2\\\\P = 2.313 \times 10^{-5} \ W

Learn more about intensity of sound here: brainly.com/question/4431819

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For this problem, we use the conservation of momentum as a solution. Since momentum is mass times velocity, then,

m₁v₁ + m₂v₂ = m₁v₁' + m₂v₂'
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Solving for v₁',
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car traveling on a flat (unbanked), circular track accelerates uniformly from rest with a tangential acceleration of a. The car
Luba_88 [7]

Answer:

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The only force on the car that is not completely in the vertical direction is friction, so let us consider the sums of forces in the tangential and centerward directions.

First the tangential direction

∑Ft =Fft =mat

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Going back to our constant acceleration equations we see that v²t = v²ti +2at∆x = 2at πr/2

So going backwards and plugging in Ffc =m2atπr/ 2r =πmat

Ff = √(F2ft +F2fc)= matp √(1+π²)

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1.70m/s/2 9.80 m/s² x√(1+π²)= 0.572

7 0
3 years ago
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marishachu [46]
  • initial velocity=u=24m/s
  • Acceleration=a=4m/s^2
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  • Final velocity=v

Using 3rd equation of kinematics

\boxed{\Large{\sf v^2-u^2=2as}}

\\ \Large\sf\longmapsto v^2=u^2+2as

\\ \Large\sf\longmapsto v^2=24^2+2(4)(96)

\\ \Large\sf\longmapsto v^2=576+768

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\\ \Large\sf\longmapsto v=\sqrt{1344}

\\ \Large\sf\longmapsto v=36.6m/s

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