To solve this problem it is necessary to apply the law of Malus which describes the change in the Intensity of Light when it crosses a polarized surface.

Mathematically the expression is given as

$I = I_0 cos^2\theta$

Where,

$I_0$ = Initial Intensity

I = Final Intensity after pass through the polarizer

$\theta$ = Angle between the polarizer and the light

Since it is sought to reduce the intensity by half the relationship between the two intensities will be given as

$\frac{I}{I_0} = \frac{1}{2}$

Using the Malus Law we have,

$I = I_0 cos^2\theta$

$cos^2\theta = \frac{I}{I_0}$

$cos^2\theta = \frac{1}{2}$

$\theta = cos^{-1}(\frac{1}{2})^2$

$\theta = 75.52\°$

Angle with respect to maximum is $90-75.52 = 14.48\°$

8 0

3 years ago

4. A car accelerating at 60 meters/second is hit in an accident by a bus. The net force exerted on

Licemer1 [7]

Answer:

500kg

Explanation:

mass = newtons/force divided by the acceleration rate

m = 30,000/60

m = 500

3 0

3 years ago

Select the arrangement of electromagnetic radiation which starts with the lowest wavelength and increases greatest wavelength. M

Scorpion4ik [409]

Answer:

gamma rays, ultraviolet, infrared, radio

Explanation:

Because on the Electromagnetic spectrum wavelength increases from the gamma end to radio end and frequency decrease in that order

8 0

3 years ago

Read 2 more answers

A 8.4-mH inductor carries a current I = Imaxsin ωt, with Imax = 4.00 A and f = ω/2π = 60.0 Hz. What is the self-induced emf as a

Aleks04 [339]

Answer:

E= -3.166 cosωt V

Explanation:

Given that

I = Imax sinωt

L= 8.4 m H

Imax= 4 A

f = ω/2π = 60.0 Hz

ω = 120π rad/s

We know that self induce E given as

$E=-L\dfrac{dI}{dt}$

$\dfrac{dI}{dt}= Imax \ \omega\ cos\omega t$

$E=-L\times Imax \ \omega\ cos\omega t$

$E=-8.4\times 120\times \pi \ cos\omega t$

E= -3166.72 cosωt m V

E= -3.166 cosωt V

This is the induce emf.

3 0

3 years ago