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2 years ago

Light is polarized by using:

Physics

1 answer:

maxonik [38]2 years ago

8 0

Answer:

Polaroid fliter

Explanation:

light can be polarized by using Polaroid filters

Polaroid fliter are made of special material that is capable of blocking one of the two planes of vibration of an electromagnetic wave

hope this is useful--(have a good day)

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Sallie's recipe calls for a total of cups of flour to make biscuits. It says to use cup of flour from the total for rolling out

zloy xaker [14]

Answer:

Answer is D

Explanation:

7 0

2 years ago

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A beam of light has a wavelength of 650 nm in vacuum. (a) What is the speed of this light in a liquid whose index of refraction

Lady_Fox [76]

Answer:

The speed of this light and wavelength in a liquid are $2.04\times10^{8}\ m/s$ and 442 nm.

Explanation:

Given that,

Wavelength = 650 nm

Index refraction = 1.47

(a). We need to calculate the speed

Using formula of speed

$n = \dfrac{c}{v}$

Where, n = refraction index

c = speed of light in vacuum

v = speed of light in medium

Put the value into the formula

$1.47=\dfrac{3\times10^{8}}{v}$

$v=\dfrac{3\times10^{8}}{1.47}$

$v= 2.04\times10^{8}\ m/s$

(b). We need to calculate the wavelength

Using formula of wavelength

$n=\dfrac{\lambda_{0}}{\lambda}$

$\lambda=\dfrac{\lambda_{0}}{n}$

Where, $\lambda_{0}$ = wavelength in vacuum

$\lambda$ = wavelength in medium

Put the value into the formula

$\lambda=\dfrac{650\times10^{-9}}{1.47}$

$\lambda=442\times10^{-9}\ m$

Hence, The speed of this light and wavelength in a liquid are $2.04\times10^{8}\ m/s$ and 442 nm.

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3 years ago

A 0.15 kg project hits the ground with a speed of 11 m/s. The project comes to rest in 0.015 seconds. What is the net force?

Blababa [14]

Answer:

Explanation:

ACCORDING TO NEWTONS SECOND LAW;

F = mass * acceleration

F = m(v-u/t)

m is the mass = 0.15kg

v is the final velocity = 11m/s

u is the initial velocity = 0m/s

t is the time = 0.015

Substitute;

F = 0.15(11-0)/0.015

F = 0.15(11)/0.015

F = 1.65/0.015

F = 110N

Hence the net force is 110N

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2 years ago

A charge of 25 nC is uniformly distributed along a straight rod of length 3.0 m that is bent into a circular arc with a radius o

Greeley [361]

Answer:

E = 31.329 N/C.

Explanation:

The differential electric field $dE$ at the center of curvature of the arc is

$dE = k\dfrac{dQ}{r^2}cos(\theta )$ <em>(we have a cosine because vertical components cancel, leaving only horizontal cosine components of E. )</em>

where $r$ is the radius of curvature.

Now

$dQ = \lambda rd\theta$ ,

where $\lambda$ is the charge per unit length, and it has the value

$\lambda = \dfrac{25*10^{-9}C}{3.0m} = 8.3*10^{-9}C/m.$

Thus, the electric field at the center of the curvature of the arc is:

$E = \int_{\theta_1}^{\theta_2} k\dfrac{\lambda rd\theta }{r^2} cos(\theta)$

$E = \dfrac{\lambda k}{r} \int_{\theta_1}^{\theta_2}cos(\theta) d\theta.$

Now, we find $\theta_1$ and $\theta_2$ . To do this we ask ourselves what fraction is the arc length 3.0 of the circumference of the circle:

$fraction = \dfrac{3.0m}{2\pi (2.3m)} = 0.2076$

and this is

$0.2076*2\pi =1.304$ radians.

Therefore,

$E = \dfrac{\lambda k}{r} \int_{\theta_1}^{\theta_2} cos(\theta)d\theta= \dfrac{\lambda k}{r} \int_{0}^{1.304}cos(\theta) d\theta.$

evaluating the integral, and putting in the numerical values we get:

$E = \dfrac{8.3*10^{-9} *9*10^9}{2.3} *(sin(1.304)-sin(0))\\$

$\boxed{ E = 31.329N/C.}$

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3 years ago

If are spaced closely together on the map,there is a drastic temperature change over the distance

AlexFokin [52]

If <em>the isotherms</em> are spaced closely together over some portion of the map, there is a drastic temperature change over that portion.

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3 years ago

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