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3 years ago

A light beam is incident on a piece of fluorite (n = 1.434) at the Brewster's angle.

1 answer:

5 0

The Brewster angle is the angle can be found through this equation
θ=arctan(n2/n1).
This came from Snell's Law:
n1sinθ=n2sin(90-θ)
=n2cosθ
n2/n1=tanθ --- θ=arctan(n2/n1)=arctan(1.434)=55.11 degrees

Next, the angle of refraction can be found by using Snell's Law
n1sinθ=n2sinθ'
sin(55.11)=1.434sinθ'
sinθ'=sin(55.11)/1.434=0.572 --- θ'=arcsin(0.572)=34.89 degrees

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3 years ago

Read 2 more answers

A 350-g mass is attached to a spring whose spring constant is 64 N/m. Its maximum acceleration is 5.3 m/s2. What is the frequenc

max2010maxim [7]

The frequency of oscillation is 2.153 Hz

What is the frequency of spring?

Spring Frequency is the natural frequency of spring with a weight at the lower end. Spring is fixed from the upper end and the lower end is free.

For the mass-spring system in this problem,

The Frequency of spring is calculated with the equation:

$f = \frac{1}{2\pi } \sqrt{\frac{k}{m} }$

Where,

f = frequency of spring

k = spring constant = 64 N/m

m = mass attached to spring = 350g = 0.350 kg

a = maximum acceleration = 5.3 m/s^2

Substituting the values in the equation,

$f = \frac{1}{2\pi } \sqrt{\frac{64}{0.350} }$

$f = \frac{1}{2\pi } ( 13.522)$

$f = 2.1535 Hz$

Hence,

The frequency of oscillation is 2.153 Hz

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1 year ago

5. What is the frequency of a sound wave when the speed of the sound is 340 m/s and has a wavelength of 1.21 m?

LekaFEV [45]

Explanation:

v = f × /\

340 = 1.21f

f = 280.99Hz

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3 years ago

During a baseball game, a batter hits a high pop-up. If the ball remains in the air for 6.22 s, how high does it rise? The accel

BigorU [14]

Answer:

47.4 m

Explanation:

When an object is thrown upward, it rises up, it reaches its maximum height, and then it goes down. The time at which it reaches its maximum height is half the total time of flight.

In this case, the time of flight is 6.22 s, so the time the ball takes to reach the maximum height is

$t=\frac{6.22}{2}=3.11 s$

Now we consider only the downward motion of the ball: it is a free fall motion, so we can find the vertical displacement by using the suvat equation

$s=ut+\frac{1}{2}gt^2$

where

s is the vertical displacement

u = 0 is the initial velocity

t = 3.11 s is the time

$g=9.8 m/s^2$ is the acceleration of gravity (taking downward as positive direction)

Solving the formula, we find

$s=\frac{1}{2}(9.8)(3.11)^2=47.4 m$

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3 years ago

a gym consists of a rectangular region with a semi-circle on each end. if the perimeter of the room is to be a 200 m running tra

Nikolay [14]

The dimensions of the rectangle are:

l = 50 m

b = $100/\pi$ m

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The perimeter is the length of the outline of a shape. To find the perimeter of a rectangle or square you have to add the lengths of all the four sides.

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The perimeter of a rectangle,denoted by P is given by the formula, P=2l+2b, where l is the length and b is the breadth of the rectangle.

<h3>Given:</h3>

As per the question:

Perimeter of the room is given as P = 200 m

The region is rectangular having a semicircle at each end.

Now,

Let 'l' be the length of the rectangle, 'b' be its breadth and 'r' be the radius of the semi-circle at each end.

Then, Area of the given rectangle, A = lb

Perimeter of the room, P is = $\pi r+l+\pi r+l=2\pi r+2l=\pi b+2l$