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kakasveta [241]
3 years ago
12

unpolarized light of intensity Io is incident on an ideal linear polariser (no absorption) . what is the transmitted intensity?​

Physics
1 answer:
amid [387]3 years ago
3 0

Answer:

A Polarizing sheet transmits only the component of light polarized along a particular direction and absorbs the component perpendicular to that direction.

Consider a light beam in the z direction incident on a Polaroid which has its transmission axis in the y direction. On the average, half of the incident light has its polarization axis in the y direction and half in the x direction. Thus half the intensity is transmitted,and the transmitted light is linearly polarized in the y direction.

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A spring with a spring constant of 350 N/m pulls a door closed. How much work is done as the spring pulls the door at a constant
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The work done to stretch the spring will be 112 J.

<h3>What is spring force?</h3>

The force required to extend or compress a spring by some distance scales linearly with respect to that distance is known as the spring force. Its formula is

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K is the spring constant= 8.5 N/m

x is the length by which spring got stretched = 1.2m

The work is done to stretch the spring is;

\rm W= \frac{1}{2} kx^2 \\\\ W=\frac{1}{2} \times 350 \times (0.850-0.050)^2 \\\\ W=0.5 \times 350 \times (0.80)^2 \\\\W=112 \ J

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A 8.2-V battery is connected in series with a 38-mH inductor, a 150-Ω resistor, and an open switch.A 8.2-V battery is connected
tigry1 [53]

Answer:

(A). The current in the circuit is 19.25 mA.

(B). The store energy in the inductor is 7.04 μJ.

Explanation:

Given that,

Voltage = 8.2 V

Inductor = 38 mH

Resistance = 150 Ω

Time t = 0.110 ms

The battery has negligible internal resistance, so that the total resistance  in the circuit is 150 ohms. Then use this equation for current at time t in terms of inductance

We need to calculate the current

Using formula of current

I(t)=\dfrac{V}{R}\times(1-e^{-t\times\dfrac{R}{L}})

Put the value into the formula

I(t)=\dfrac{8.2}{150}\times(1-e^{-0.110\times10^{-3}\times\dfrac{150}{38\times10^{-3}}})

I(t)=0.01925\ A

I(t) = 19.25\ mA

(B). We need to calculate the store energy in the inductor

Using formula of energy

E=\dfrac{1}{2}LI^2

Put the value into the formula

E=\dfrac{1}{2}\times38\times10^{-3}\times(0.01925)^2

E=7.04\times10^{-6}\ J

{tex]E=7.04\ \mu J[/tex]

Hence, (A). The current in the circuit is 19.25 mA.

(B). The store energy in the inductor is 7.04 μJ.

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