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MA_775_DIABLO [31]
4 years ago
14

A diamond is cut such that the angle between its top surface and its bottom surface is α. For α=45∘, find the largest possible v

alue of the incident angle θa such that the blue light is totally internally reflected off the bottom surface. Express your answer in degrees to four significant figures.
Physics
1 answer:
Reil [10]4 years ago
4 0

Answer:

i = 61 degree

Explanation:

Given,

\alpha =45^{o}

Now, by the snell's law

sin\theta_c=\frac{1}{n}\\sin\theta_c=\frac{1}{2.450}\\\theta_c=24.09^o

Now,

Sin i / sin r = n 2 / n 1

sin i / sin r (45 - 24.09) = 2.45 / 1

i = 60.97 degree

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What happens to the magnitude of the gravitational force as the distance between two bodies increase?
Anna [14]

Answer:

The magnitude of the force will decrease

Explanation:

The gravitational force is one of the four fundamental forces of nature. It is an attractive force exerted between every object having mass.

Its magnitude is given by the equation:

F=\frac{Gm_1 m_2}{r^2}

where

G is the gravitational constant

m1 is the mass of the first object

m2 is the mass of the second object

r is the separation between the objects

As we see from the equation, the magnitude of the gravitational force is inversely proportional to the square of the distance between the objects:

F\propto \frac{1}{r^2}

Therefore, this means that as the distance between two bodies increases, the gravitational force will decrease.

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A 0.5 kg cheeseburger is lobbed at a particularly unhappy customer with a force of 10 N.
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Which word in the sentence is a gerund?
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Which statements describe the motion of the object represented by the graph? Check all that apply.
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Explanation:

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Charge Q is distributed uniformly throughout the volume of an insulating sphere of radius R = 4.00 cm. At a distance of r = 8.00
Elena L [17]

Answer:

2.62898\times 10^{-6}\ C/m^3

1979.99974\ N/C

Explanation:

k = Coulomb constant = 8.99\times 10^{9}\ Nm^2/C^2

Q = Charge

r = Distance = 8 cm

R = Radius = 4 cm

Electric field is given by

E=\dfrac{kQ}{r^2}\\\Rightarrow Q=\dfrac{Er^2}{k}\\\Rightarrow E=\dfrac{990\times 0.08^2}{8.99\times 10^{9}}\\\Rightarrow Q=7.04783\times 10^{-10}\ C

Volume charge density is given by

\sigma=\dfrac{Q}{\dfrac{4}{3}\pi R^3}\\\Rightarrow \sigma=\dfrac{7.04783\times 10^{-10}}{\dfrac{4}{3}\pi (0.04)^3}\\\Rightarrow \sigma=2.62898\times 10^{-6}\ C/m^3

The volume charge density for the sphere is 2.62898\times 10^{-6}\ C/m^3

E=\dfrac{kQr}{R^3}\\\Rightarrow E=\dfrac{8.99\times 10^9\times 7.04783\times 10^{-10}\times 0.02}{0.04^3}\\\Rightarrow E=1979.99974\ N/C

The magnitude of the electric field is 1979.99974\ N/C

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