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givi [52]
4 years ago
7

Find the angle of prism if the ray just fail to emerge from 2nd face When ray of light falls normallly on the face of prism of R

.I. 1.5​
Physics
1 answer:
Usimov [2.4K]4 years ago
3 0

The angle of prism is 41.81 degrees.

<u>Explanation:</u>

For no emergence to be taken place, inside a prism, Total Internal Reflection (TIR) should take place at the second surface.  For TIR, at second surface, angle of refraction must be greater than critical angle. Angle of prism is related to refraction as,

                            A>r_{1}+C

Since, r_{1} = C and A \geq 2 C

This implies A \geq C

                         \sin A \geq \sin C

                         \sin A \geq \frac{1}{\mu}

                         \sin A \geq \frac{2}{3}

when sin goes to other side become as sin inverse of value, and obtain the result as below,

                       A=41.81^{\circ}

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Mass and Weight: What is the mass of an object that experiences a gravitational force of 685 N near Earth's surface where g
11Alexandr11 [23.1K]

Answer:

m = 69.9 kg

Explanation:

The mass and the weight of an object are two different quantities. Mass is basically the amount of matter that is present in a body. It remains same everywhere in the universe and measured in kilograms.

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Therefore, the relation between mass and weight of an object is given by the following formula:

W = mg

m = W/g

where,

m = mass = ?

W = Weight = 685 N

g = 9.8 m/s²

Therefore,

m = (685 N)/(9.8 m/s²)

<u>m = 69.9 kg</u>

4 0
3 years ago
The maximum speed of a mass m on an oscillating spring is vmax . what is the speed of the mass at the instant when the kinetic a
umka21 [38]
Let
A =  the amplitude of vibration
k =  the spring constant
m =  the mass of the object

The displacement at time, t, is of the form
x(t) = A cos(ωt)
where
ω =  the circular frequency.

The velocity is
v(t) = -ωA sin(ωt)

The maximum velocity occurs when the sin function is either 1 or -1.
Therefore
v_{max} = \omega A
Therefore
v(t) = -V_{max} sin(\omega t)

The KE (kinetic energy) is given by
KE = \frac{m}{2}v^{2} = \frac{m}{2} V_{max}^{2} sin^{2} (\omega t)

The PE (potential energy) is given by
PE = \frac{k}{2} x^{2} = \frac{k}{2} A^{2} cos^{2} (\omega t)

When the KE and PE are equal, then
v^{2} = \frac{k}{m} A^{2} cos^{2} (\omega t)

For the oscillating spring,
\omega ^{2} =  \frac{k}{m} \\ V_{max} = \omega A =  \sqrt{ \frac{k}{m} } A
Therefore
v^{2} =  \frac{k}{m}  \frac{m}{k} V_{max}^{2} cos^{2} ( \sqrt{ \frac{k}{m} t} ) \\ v = V_{max}  \,cos( \sqrt{ \frac{k}{m} t} )

Answer: v(t) = V_{max} cos( \sqrt{ \frac{k}{m} t} )

3 0
4 years ago
Is work being done when you carry something?
wlad13 [49]

Answer:

Yes you are using energy to carry that thing.(Force is applied)

Explanation:

8 0
3 years ago
Will mark brainliest!
MaRussiya [10]

Answer:

Explanation:

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8 0
3 years ago
A weight lifter picks up a barbell and
kodGreya [7K]

Answer:

From highest to lowest: W_1>W_2>W_3

Explanation:

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This means that 0, so the work done is positive: W_1>0

In case 2), the barbell is held stationary: this means that the displacement is zero,

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W_2=0

In case 3), the barbell is put down slowly, without dropping it. This means that:

- The force applied is still upward (in fact, the force applied must be upward in order to overcome the force of gravity downward, and avoid the barbell to fall down)

- The displacement of the barbell is downward

This means that 90^{\circ}, so cos \theta, and therefore the work done is negative:

W_3

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W_1>W_2>W_3

8 0
3 years ago
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