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mrs_skeptik [129]
3 years ago
11

I need help with...

Physics
1 answer:
Rom4ik [11]3 years ago
6 0
The car will gain new momentum if it's velocity is doubled or tripled.
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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
Usimov [2.4K]

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}

3 0
3 years ago
Explanation's of E=MC²
natita [175]

Answer:

E means energy

M= Mass

C=speed of light squared (the exponent means squared)

4 0
3 years ago
Read 2 more answers
A point P1 is located by the vector A = (3.74)î + (1.64)ĵ and a point P2 is located by the vector B = (1.60)î + (3.66)ĵ. The vec
seropon [69]

Answer:

\vec{C} = ( \ - 2.14 \ , \ 2.02 \ )

Explanation:

The vector that point from point P1 to point P2 its found simply by taking the vector at which point P2 its located and subtracting the vector at which point P1 its located:

\vec{C} = \vec{B} - \vec{A}

So:

\vec{C} = ( \ 1.60 \ , \ 3.66 \ ) - ( \ 3.74 \ , \ 1.64 \ )

\vec{C} = ( \ 1.60 \ - \ 3.74 \ , \ 3.66 \ - \ 1.64 \ )

\vec{C} = ( \ - 2.14 \ , \ 2.02 \ )

4 0
3 years ago
You stand on a straight desert road at night and observe a vehicle approaching. This vehicle is equipped with two small headligh
spayn [35]

To solve this problem we will apply the concepts related to Reyleigh's criteria. Here the resolution of the eye is defined as 1.22 times the wavelength over the diameter of the eye. Mathematically this is,

\theta = \frac{1.22 \lambda }{D}

Here,

D is diameter of the eye

D = \frac{1.22 (539nm)}{5.11 mm}

D= 1.287*10^{-4}m

The angle that relates the distance between the lights and the distance to the lamp is given by,

Sin\theta = \frac{d}{L}

For small angle, sin\theta = \theta

sin \theta = \frac{d}{L}

Here,

d = Distance between lights

L = Distance from eye to lamp

For small angle sin \theta = \theta

Therefore,

L = \frac{d}{sin\theta}

L = \frac{0.691m}{1.287*10^{-4}}

L = 5367m

Therefore the distance is 5.367km.

4 0
4 years ago
What is the magnitude of the force that lifts a 3-kilogram object straight upwards?
frez [133]

Any force of 29.4 Newtons or greater can do it.

4 0
3 years ago
Read 2 more answers
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