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Roman55 [17]
3 years ago
10

If the distance from a light source triples, how does light intensity change? The intensity will be 3x greater. The intensity wi

ll be 1/3 as much. The intensity will be 1/9 as much. The intensity will not change.
Physics
1 answer:
Tcecarenko [31]3 years ago
3 0

Answer:

The intensity will be 1/9 as much.

Explanation:

The intensity of the light or any source is inversely related to the square of the distance.

I\alpha \frac{1}{r^{2} }

Now according to the question the distance is increased by three times than,

\frac{I_{2} }{I_{1} }=\frac{r_{1}^{2} }{r_{2}^{2} }

Therefore,

\frac{I_{2} }{I_{1} }=\frac{r_{1}^{2} }{(3r_{1})^{2} }\\\frac{I_{2} }{I_{1} }=\frac{1}{9} \\{I_{2}=\frac{1}{9}{I_{1} }

Therefore the intensity will become 1/9 times to the initial intensity.

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A magnetic field is entering into a coil of wire with radius of 2(mm) and 200 turns. The direction of magnetic field makes an an
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Answer:

a) <em>2.278 x 10^-5 volts</em>

b) <em>1.139 x 10^-6 Ampere</em>

c) <em>2.59 x 10^-11 W</em>

Explanation:

The radius of the wire r = 2 mm = 0.002 m

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direction of the magnetic field ∅ = 25°

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varying time = 2 sec

The cross sectional area of the wire = \pi r^{2}

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The induced EMF is given as

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where dΦ/dt = (2.278 x 10^-7)/2 = 1.139 x 10^-7

E = 200 x 1.139 x 10^-7 = <em>2.278 x 10^-5 volts</em>

<em></em>

<em></em>

b) If the two ends are connected to a resistor of 20 Ω, the current through the resistor is given as

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I = (2.278 x 10^-5)/20 = <em>1.139 x 10^-6 Ampere</em>

<em></em>

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<em> </em>c) power delivered to the resistor is given as

P = IE

P = (1.139 x 10^-6) x (2.278 x 10^-5) = <em>2.59 x 10^-11 W</em>

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