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3 years ago

((PLEASE HELP))

Physics

2 answers:

Degger [83]3 years ago

5 0

Answer:

Angle of refraction

Explanation:

The incident ray is the ray before it reaches the surface.

The refracted ray is the ray after it reaches the surface.

n₁ is called the index of incidence.

n₂ is called the index of refraction.

θ₁ is called the angle of incidence.

θ₂ is called the angle of refraction.

They are related by Snell's law:

n₁ sin θ₁ = n₂ sin θ₂

il63 [147K]3 years ago

5 0

Answer:

Angle of refraction

Explanation:

The angle θ2 represents the angle of refraction.

Refraction is the change in direction of light rays as it passes from one medium to another. As light rays passes from one medium to another, it changes direction.

If the incident light is passing from the less dense medium to denser medium, the refracted ray will bend towards the normal. In this case the angle of refraction 'r' is always less than the angle if incidence. (According to the diagram)

Similarly, If the incident light is passing from the denser medium to less dense medium, the refracted ray will bend away from the normal. In this case the angle of refraction 'r' is always greater than the angle of incidence.

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The strength of the electric field at a certain distance from a point charge is represented by E. What is the strength of the el

LUCKY_DIMON [66]

Answer:

e.)At twice the distance, the strength of the field is E/4.

Explanation:

The strength of the electric field at a certain distance from a point charge is given by:

$E=k\frac{Q}{r^2}$

where

k is the Coulomb's constant

Q is the charge

r is the distance from the point charge

In this problem, the distance from the point charge is doubled:

r' = 2r

So the new electric field strength is

$E'=k\frac{Q}{(2r)^2}=k \frac{Q}{4 r^2}=\frac{1}{4} (k\frac{Q}{r^2})=\frac{E}{4}$

so, at twice the distance the strength of the field is E/4.

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2 years ago

Starting at 1.0 m/s, a cheetah runs with a constant acceleration for 4.8 s reaching a speed of 28 m/s. What is the acceleration

LenaWriter [7]

Answer:

c. $5.6m/s^2$

Explanation:

Initial velocity of cheetah,u=1 m/s

Time taken by cheetah =4.8 s

Final velocity of cheetah,v=28 m/s

We have to find the acceleration of this cheetah.

We know that

Acceleration, $a=\frac{v-u}{t}$

Where v=Final velocity of object

u=Initial velocity of object

t=Time taken by object

Using the formula

Then, we get

Acceleration, a= $\frac{28-1}{4.8}=\frac{27}{4.8} m/s^2$

Acceleration= $a=5.6 m/s^2$

Hence, the acceleration of cheetah= $5.6m/s^2$

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2 years ago

Using monochromatic light of 410 nm wavelength, a single thin slit forms a diffraction pattern, with the first minimum at an ang

VLD [36.1K]

Answer:

c. 307 nm

Explanation:

angular position of first dark fringe = λ / d , λ is wavelength and d is width of slit .

(40 x π ) / 180 = 410 / d

angular position of second dark fringe = 2 x λ / d , λ is wavelength and d is width of slit .

(60 x π ) / 180 = 2 x λ / d

Dividing these equations

60 / 40 = 2 x λ / 410

λ = 307.5 nm.

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3 years ago

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kirill [66]

Answer 1) : 62.5 km/hour is the average velocity of the train.

2) The final velocity of the car at the end of 75 m is 14.69 m/s

Explanation:

1) Displacement of the train = 100 km + 150 km = 250 km

Total time train took =1 hour 15 min+ 45 min + 2 hours = 240 min = 4 hours

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62.5 km/hour is the average velocity of the train.

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Initial velocity of the car , $v_i$ = 6 m/s

Final velocity of thre car , $v_f$ =?

Using third equation of motion:

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The final velocity of the car at the end of 75 m is 14.69 m/s

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2 years ago

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Ganezh [65]

The answer is D that is the right question

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3 years ago