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barxatty [35]
3 years ago
8

A bystander observes the musicians heading toward each other. When musician #1 is 100 m away, the intensity is 1.24 x 10-8 W/m^2

. What is the intensity AND intensity level when musician #1 is 25 m away?
Physics
1 answer:
777dan777 [17]3 years ago
7 0

Explanation:

Given that,

Distance 1, r = 100 m

Intensity, I_1=1.24\times 10^{-8}\ W/m^2

If distance 2, r' = 25 m

We need to find the intensity and the intensity level at 25 meters. Intensity and a distance r is given by :

I=\dfrac{P}{4\pi r^2}.........(1)

Let I' is the intensity at r'. So,

I'=\dfrac{P}{4\pi r'^2}............(2)

From equation (1) and (2) :

I'=\dfrac{Ir}{r'^2}

I'=\dfrac{1.24\times 10^{-8}\times 100}{25^2}

I'=1.98\times 10^{-9}\ W/m^2

Intensity level is given by :

dB=10\ log(\dfrac{I'}{I_o}), I_o=10^{-12}\ W/m^2

dB=10\ log(\dfrac{1.98\times 10^{-9}}{10^{-12}})

dB = 32.96 dB

Hence, this is the required solution.

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Which car has the most kinetic energy? <br><br> (PLEASE HELP NEED ANSWER ASAP)
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6 0
3 years ago
A jogger accelerates from rest to 3.0 m/s in 2.0 s. A car accelerates from 38.0 to 41.0 m/s also in 2.0 s. (a) Find the accelera
iren [92.7K]

Answer:

(a)  a₁:  jogger  acceleration= 1.5 m/s²

(b)  a₂:  car  acceleration = 1.5 m/s²

(b)  d= 76m : the car travels 76 meters longer than the jogger during the 2 seconds

Explanation:

we apply uniformly accelerated motion formulas:

vf= v₀+at Formula (1)

vf²=v₀²+2*a*d Formula (2)

d= v₀t+ (1/2)*a*t² Formula (3)

Where:  

d:displacement in meters (m)  

t : time in seconds (s)

v₀: initial speed in m/s  

vf: final speed in m/s  

a: acceleration in m/s²

Nomenclature

d₁:  jogger displacement   

t₁ :  jogger time

v₀₁:  jogger initial speed

vf₁:  jogger  final speed

a₁:  jogger  acceleration

d₂: car displacement   

t₂ : car  time

v₀₂: car  initial speed

vf₂:  car  final speed

a₂:  car  acceleration

Data

v₀₁ = 0

vf₁ = 3 m/s

t₁ =2.0 s

v₀₂ = 38.0m/s

vf₂ = 41.0 m/s

t₂ = 2.0 s

Problem development

(a) Find the acceleration (magnitude only) of the jogger.

We apply the formula (1) for calculate acceleration :

vf₁= v₀₁+a₁*t₁

3 = 0 +(a₁)*(2)

a₁= (3)/(2)

a₁= 1.5 m/s²

(b) Determine the acceleration (magnitude only) of the car.

We apply the formula (1) for calculate acceleration :

vf₂= v₀₂+a₂*t₂

41 = 38 +(a₂)*(2)

a₂= (41 - 38)/(2)

a₂= 3 /2

a₂= 1.5 m/s²

(c) Does the car travel farther than the jogger during the 2.0 s? If so, how much farther?

We apply the formula (1) for calculate distance :

d₁= v₀₁*t₁+ (1/2)*a₁*t₁²= 0+ (1/2)*(1.5) *(2)² = 3 m

d₂= v₀₂*t₂+ (1/2)*a₂*t₂² =38*(2)+ (1/2)*(1.5) *(2)²= 79 m

d= 79 m-3 m

d= 76m : the car travels 76 meters longer than the jogger during the 2 seconds

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4 0
3 years ago
Read 2 more answers
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