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Korvikt [17]
3 years ago
15

The intensity at distance from a spherically symmetric sound source is 100 W/m2. What is the intensity at five times this distan

ce from the source?
Physics
1 answer:
ss7ja [257]3 years ago
3 0

To solve this problem it is necessary to apply the concepts related to intensity as a function of power and area.

Intensity is defined to be the power per unit area carried by a wave. Power is the rate at which energy is transferred by the wave. In equation form, intensity I is

I = \frac{P}{A}

The area of a sphere is given by

A = 4\pi r^2

So replacing we have to

I = \frac{P}{4\pi r^2}

Since the question tells us to find the proportion when

r_1 = 5r_2 \rightarrow \frac{r_2}{r_1} = \frac{1}{5}

So considering the two intensities we have to

I_1 = \frac{P_1}{4\pi r_1^2}

I_2 = \frac{P_2}{4\pi r_2^2}

The ratio between the two intensities would be

\frac{I_1}{I_2} = \frac{ \frac{P_1}{4\pi r_1^2}}{\frac{P_2}{4\pi r_2^2}}

The power does not change therefore it remains constant, which allows summarizing the expression to

\frac{I_1}{I_2}=(\frac{r_2}{r_1})^2

Re-arrange to find I_2

I_2 = I_1 (\frac{r_1}{r_2})^2

I_2 = 100*(\frac{1}{5})^2

I_2 = 4W/m^2

Therefore the intensity at five times this distance from the source is 4W/m^2

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3 years ago
Name two objects that both use electric motors and are commonly found in houses
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Hope this helps!
3 0
2 years ago
Read 2 more answers
g Two radiation modes (one at the center frequency lIo and the other at lIO+?lI) are excited with 1000 photons each. Determine t
Schach [20]

Answer:

a) P=0.25x10^-7

b) R=B*N2*E

c) N=1.33x10^9 photons

Explanation:

a) the spontaneous emission rate is equal to:

1/tsp=1/3 ms

the stimulated emission rate is equal to:

pst=(N*C*o(v))/V

where

o(v)=((λ^2*A)/(8*π*u^2))g(v)

g(v)=2/(π*deltav)

o(v)=(λ^2)/(4*π*tp*deltav)

Replacing values:

o(v)=0.7^2/(4*π*3*50)=8.3x10^-19 cm^2

the probability is equal to:

P=(1000*3x10^10*8.3x10^-19)/(100)=0.25x10^-7

b) the rate of decay is equal to:

R=B*N2*E, where B is the Einstein´s coefficient and E is the energy system

c) the number of photons is equal to:

N=(1/tsp)*(V/C*o)

Replacing:

N=100/(3*3x10^10*8.3x10^-19)

N=1.33x10^9 photons

7 0
3 years ago
The rate (in mg carbon/m3/h) at which photosynthesis takes place for a species of phytoplankton is modeled by the function P = 9
Ivanshal [37]

Answer:

At light intensity I = 3, is P a maximum

Explanation:

Given:

P=\frac{90I}{I^2+I+9}

now differentiating the above equation with respect to Intensity 'I' we get

\frac{dp}{dI}=\frac{(I^2+I+9).\frac{d(90I)}{dI}-90I.\frac{d((I^2+I+9)}{dI}}{(I^2+I+9)^2}

or

\frac{dp}{dI}=\frac{(I^2+I+9).90-90I.(2I+1)}{(I^2+I+9)^2}

or

\frac{dp}{dI}=\frac{90I^2+90I+810)-(180I^2+90I)}{(I^2+I+9)^2}

or

\frac{dp}{dI}=\frac{-90I^2+810)}{(I^2+I+9)^2}

Now for the maxima \frac{dP}{dI}=0

thus,

0=\frac{-90I^2+810)}{(I^2+I+9)^2}

or

-90I^2+810=0

or

I^2=\frac{810}{90}

or

I^2=9

or

I = 3

thus, <u>for the value of intensity I = 3, the P is maximum</u>

at I = 3

P=\frac{90\times3}{3^2+3+9}

or

P=\frac{270}{21}

or

P=12.85

5 0
3 years ago
The power of the kettle was 2.6 kW
faltersainse [42]

Answer:

Cp = 4756 [J/kg*°C]

Explanation:

In order to calculate the specific heat of water, we must use the equation of energy for heat or heat transfer equation.

Q = m*Cp*(T_f - T_i)/t

where:

Q = heat transfer = 2.6 [kW] = 2600[W]

m = mass of the water = 0.8 [kg]

Cp = specific heat of water [J/kg*°C]

T_f  = final temperature of the water = 100 [°C]

T_i = initial temperature of the water = 18 [°C]

t = time = 120 [s]

Now clearing the Cp, we have:

Cp = Q*t/(m*(T_f - T_i))

Now replacing

Cp = (2600*120)/(0.8*(100-18))

Cp = 4756 [J/kg*°C]

7 0
2 years ago
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