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3 years ago

If the light leaves a glass block ( n = 1.5 )with an angle of refraction of 70°. what is the angle of incidence?

Physics

1 answer:

julsineya [31]3 years ago

5 0

Answer:

The angle of incidence is 38.8°.

Explanation:

let n1 = 1.5 be the refractive index of glass and ∅1 be the angle of incidence, let n2 be the refractive index of air and ∅2 be the angle of refraction.

by Snell's law:

n1×sin(∅1) = n2×sin(∅2)

sin(∅1) = n2×sin(∅2)/n1

= (1)×sin(70)/1.5

∅1 = 38.8°

Therefore, the angle of incidence is 38.8°.

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What is the resultant force of 500g on abject accelerating at 5m/s2

max2010maxim [7]

Answer: 2.5 N

Explanation:

m = 500g = 0.5Kg

a = 5m/s2

F = ma = 0.5 x 5 = 2.5 N

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3 years ago

A radar antenna is tracking a satellite orbiting the earth. At a certain time, the radar screen shows the satellite to be 118 km

ruslelena [56]

Answer:

x component 60.85 m

y component 101.031 m

Explanation:

We have given distance r = 118 km

Angle which makes from ground = 58.9°

(a) X component of distance is given by $r_x=rcos\Theta =118\times cos58.9=118\times 0.5165=118=60.85m$

(b) Y component of distance is given by $r_Y=rcos\Theta =118\times sin58.9=118\times 0.8562=101.0316m$

These are the x and y component of position vector

6 0

3 years ago

Use the fact that 1inch=2.54cm and convert 52 inches into the equivalent length in centimeters.

Lelu [443]

Answer:

To convert inches to centimeters, use an easy formula and multiply the length by the conversion ratio.

Since one inch is equal to 2.54 centimeters, this is the inches to cm formula to conver

Explanation:

4 0

2 years ago

An object of mass, m1 with a velocity, v1 collides with another object at rest (v2 = 0) with a mass, m2. After the collision, m1

goblinko [34]

Answer:

$v"_{1} = v_{1} tan$ Θ

$v^{"} _{2} = \frac{m_{1}v_{1}}{m_{2}cos}$ Θ

Θ = $tan^{-1}(\frac{v^{"} _{1} }{v_{1} } )$

Explanation:

Applying the law of conservation of momentum, we have:

Δ $p_{x = 0}$

$p_{x} = p"_{x}$

$m_{1}v_{1} = m_{2}v"_{2} cos$ Θ (Equation 1)

Δ $p_{y} = 0$

$p_{y} = p"_{y}$

$0 = m_{1} v"_{1} - m_{2} v"_{2} sin$ Θ (Equation 2)

From Equation 1:

$v"_{2} = \frac{m_{1}v_{1}}{m_{2}cos}$ Θ

From Equation 2:

$m_{2} v"_{2}$ sinΘ = $m_{1} v_{1}$

$v"_{1} = \frac{m_{2} v"_{2}sinΘ}{m_{1} }$

Replacing Equation 3 in Equation 4:

$v"_{1}=\frac{m_{2}\frac{m_{1}v_{1}}{m_{2}cosΘ}sinΘ}{m_{1}}$

$v"_{1}=v_{1}\frac{sinΘ}{cosΘ}$

$v"_{1}=v_{1}tan$ Θ (Equation 5)

And we found Θ from the Equation 5:

tanΘ= $\frac{v"_{1}}{v_{1}}$

Θ= $tan^{-1}(\frac{v"_{1}}{v_{1}})$

7 0

3 years ago

A drone is flying horizontally when it runs out of power and begins to free fall from 16 m. No drag. If it lands 40 m away (in t

marysya [2.9K]

Answer:

the horizontal velocity while it was falling is 22.1 m/s.

Explanation:

Given;

height of fall, h = 16 m

horizontal distance, x = 40 m

The time to travel 16 m is calculated as;

$t = \sqrt{\frac{2h}{g} } \\\\t = \sqrt{\frac{2\times 16}{9.8} } \\\\t = 1.81 \ s$

The horizontal velocity is calculated as;

$v_x = \frac{X}{t} \\\\v_x = \frac{40}{1.81} \\\\v_x = 22.1 \ m/s$

Therefore, the horizontal velocity while it was falling is 22.1 m/s.

7 0

3 years ago