Physics, 19.03.2020 21:32 emilyphillips1681

A resistor with resistance R is connected to a battery that has emf 13.0 V and internal resistance r = 0.350 Ω . For what two values of R will the power dissipated in the resistor be 81.0 W ?

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Answer from: adot1

The two values of R when the power dissipated is 81 W are 1.74 ohms or -0.35 ohms.

The given parameters;

emf of the battery, E = 13 Vinternal resistance of the battery, r = 0.35

The current flowing in the circuit is calculated as follows;

E = IR + Ir

E = I(R + r)

$I = \frac{E}{R+r}$ -----(1)

The power dissipated in the circuit is calculated as follows;

P = I²(R + r)

$I^2 = \frac{P}{R+ r} \\\\I = \sqrt{\frac{P}{R+ r} } \ \ ---(2)$

solve (1) and (2) together;

$\sqrt{\frac{P}{R+ r} } = \frac{E}{R+r} \\\\\frac{P}{R+ r} = \frac{E^2}{(R+r)^2} \\\\\frac{P}{R+ r} = \frac{E^2}{R^2 + 2Rr + r^2} \\\\E^2(R+r) = P(R^2 + 2Rr + r^2)\\\\13^2(R+ 0.35) = 81(R^2 + 2(0.35)R + (0.35)^2)\\\\169R + 59.15 = 81R^2 + 56.7R + 9.923\\\\81R^2 -112.3R -49.23 = 0\\\\Solve \ the \ quadratic \ equation\ using formula \ method;\\\\a = 81, \ b = -112.3, \ c = -49.23\\\\ R = \frac{-b \ \ +/- \ \ \sqrt{b^2 -4ac} }{2a} \\\\$

$R = \frac{-(-112.3)\ \ +/- \ \ \sqrt{(-112.3)^2 -4(81\times -49.23)} }{2(81)}\\\\R = 1.74 \ ohms \ \ or \ \ -0.35 \ ohms$

Thus, the two values of R when the power dissipated is 81 W are 1.74 ohms or -0.35 ohms.

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Answer from: noaheault

R = 1,746 Ω

Explanation:

The power dissipated in the circuit is

P = V I = V² / R

Let's find the current

R = V² / P

Let's calculate

R = 13²/81

R = 2,096 Ω

This is total resistance

R_total = R + r

R = R_total - r

R = 2,096 -0,350

R = 1,746 Ω

Answer from: Quest

scattering of light by tiny particles in air to produce the colors we see on sky, is dependent on the following factors:

(i) wavelength of light

(ii) size of the scattering particle

(iii) distance traveled by light through the atmosphere.

according to rayleigh's law of scattering, the amount of scattering is inversely proportional to the fourth power of its wavelength. shorter the wavelength, greater is its scattering. hence violet/blue light is scattered more than red.

sun's light is a composite light ( visible light) with wavelengths in the range 400 nm (violet) to 700 nm (red). as sunlight passes through our atmosphere, the molecules of air scatter light according to their wavelengths. during sunrise and sunset, the sun's position is below the horizon. hence light from the sun travels a greater distance in the atmosphere when compared to its position a noon.

as sunlight enters the earth's atmosphere, blue light is scattered away and at the end of its journey in the sky, only red and orange wavelengths remain in the beam.

hence at sunrise and sunset, the sky appears reddish orange.

Answer from: Quest

wavelength divides in half and speed stays the same.

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