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3 years ago

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The intensity at a distance of 6 m from a source that is radiating equally in all directions is 6.0\times 10^{-10} W/m^2. What i

s the power emitted by the source

1 answer:

Sophie [7]3 years ago

3 0

Answer:

P = 271 nW

Explanation:

If the source is radiating equally in all directions, it can be treated as a point source, so all points located at the same distance of the source, have the same intensity I, which is related to the power by the following expression:

$I = \frac{P}{A} =\frac{P}{4*\pi *r^{2} } (1)$

Solving for P, we get:

$P = I*4*\pi *r^{2} = 6.0e-10 W/m2 * 4 * \pi *(6m)^{2} =271 nW (2)$

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3 years ago

A system of mass 13 kg undergoes a process during which there is no work, the elevation decreases by 50 m, and the velocity incr

marin [14]

Answer:

The change in kinetic energy is 4.3875 kJ

The amount of energy transferred by heat for the process is -66.98 kJ

Explanation:

Given;

mass of the system, m = 13 kg

change in height, Δh = -50 m

initial velocity, u = 15 m/s

final velocity, v = 30 m/s

change in internal energy per mass, ΔU = -5 kJ/kg

The change in kinetic energy is given by;

ΔK.E = K.E₂ - K.E₁

ΔK.E = ¹/₂mv² - ¹/₂mu²

ΔK.E = ¹/₂m(v² - u²)

ΔK.E = ¹/₂ ₓ 13 (30² - 15²)

ΔK.E = 4387.5 J

ΔK.E = 4.3875 kJ

The amount of energy transferred by heat for the process;

Q = W + ΔP.E + ΔK.E + ΔU

Where;

ΔP.E = mgΔh

ΔP.E = 13 x 9.8 x (-50)

ΔP.E = -6370 J = -6.37 kJ

W = 0

ΔU = -5kJ/kg x 13kg

ΔU = -65 kJ

Q = W + ΔP.E + ΔK.E + ΔU

Q = 0 + (-6.37) + (4.3875) + (-65)

Q = -66.98 kJ

7 0

4 years ago

An oil bath maintained at 50.5°C loses heat to its surroundings at the rate of 4.68 kJ/min. Its temperature is maintained by an

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Answer:

The fraction of the time is 38.67%.

Explanation:

Given that,

Energy = 4.68 KJ

Resistance = 60

Voltage =110 V

If the rate of heat energy supplied by the coil to the oil bath = Q

$Q=4.68\ kJ/min$

We need to calculate the power released by the resistor at voltage

$P=\dfrac{V^2}{R}$

Put the value into the formula

$P=\dfrac{110^2}{60}$

$P=201.7\ W$

$P=12.102\ KJ/min$

We need to calculate the fraction of the time

$T=\dfrac{Q}{P}$

Put the value into the formula

$T=\dfrac{4.68}{12.102}$

$T=0.3867$

The percentage of time is

$T = 38.67\%$

Hence, The fraction of the time is 38.67%.

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3 years ago

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Answer:

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3 years ago

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Millikan is doing his oil drop experiment. He has a droplet with radius 1.6 µm suspended motionless in a uniform electric field

swat32

Answer:

The charge on the droplet is $3.106\times10^{-16}\ C$ .

Yes, quantization of charge is obeyed within experimental error.

Explanation:

Given that,

Radius = 1.6μm

Electric field = 46 N/C

Density of oil = 0.085 g/cm³

We need to calculate the charge on the droplet

Using formula of force

$F= qE$

$mg=qE$

$V\times\rho\times g=qE$

$q=\dfrac{V\times\rho}{E}$

$q=\dfrac{\dfrac{4}{3}\pi\times r^3\times\rho\times g}{E}$

Put the value into the formula

$q=\dfrac{\dfrac{4}{3}\times\pi\times(1.6\times10^{-6})^3\times85\times9.8}{46}$

$q=3.106\times10^{-16}\ C$

We need to calculate the quantization of charge

Using formula of quantization

$n = \dfrac{q}{e}$

Put the value into the formula

$n=\dfrac{3.106\times10^{-16}}{1.6\times10^{-19}}$

$n=1941.25$

Yes, quantization of charge is obeyed within experimental error.

Hence, The charge on the droplet is $3.106\times10^{-16}\ C$ .

Yes, quantization of charge is obeyed within experimental error.

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3 years ago

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