What is weightlessness ?

A ray is incident on a film of thickness t and the index of refraction n=1.6 at an angle θ = 35 degrees. Find the angle of refra

lidiya [134]

Answer

Given,

refractive index of film, n = 1.6

refractive index of air, n' = 1

angle of incidence, i = 35°

angle of refraction, r = ?

Using Snell's law

n' sin i = n sin r

1 x sin 35° = 1.6 x sin r

r = 21°

Angle of refraction is equal to 21°.

Now,

distance at which refractive angle comes out

d = 2.5 mm

α be the angle with horizontal surface and incident ray.

α = 90°-21° = 69°

t be the thickness of the film.

So,

$tan \alpha = \dfrac{t}{d/2}$

$tan 61^0 = \dfrac{t}{2.5/2}$

t = 2.26 mm

Hence, the thickness of the film is equal to 2.26 mm.

4 0

3 years ago

Define paralax error pls

stepladder [879]

Answer:

Parallax is a displacement or difference in the apparent position of an object viewed along two different lines of sight, and is measured by the angle or semi-angle of inclination between those two lines.

5 0

3 years ago

Monochromatic light with wavelength 590 nm passes through a single slit 2. 30 ?m wide and 1. 90 m from a screen. Find the distan

aivan3 [116]

Answer:

The fringes are 4.7*10^-7 m apart, such that they are adjacent.

Explanation:

Using the formula for adjacent fringes given a single slit:

Δ $x=\frac{(Wavelength)(Distance between slit and screen)}{Width}$

Δ $x=\frac{(590/10^{9})(1.90) }{(2.30)}$

Δ $x=0.000000487 m$

Hope this helps!

7 0

2 years ago

Read 2 more answers

A mover slides a refrigerator weighing 650 N at a constant velocity across the floor a distance of 8.1 m. The force of friction

boyakko [2]

Given :

A mover slides a refrigerator weighing 650 N at a constant velocity across the floor a distance of 8.1 m.

The force of friction between the refrigerator and the floor is 230 N.

To Find :

How much work has been performed by the mover on the refrigerator.

Solution :

Since, refrigerator is moving with constant velocity.

So, force applied by the mover is also 230 N ( equal to force of friction ).

Now, work done in order to move the refrigerator is :

$W = Force\times distance\\\\W = 230 \times 8.1\ N\ m\\\\W = 1863\ N\ m$

Hence, this is the required solution.

5 0

3 years ago

A police officer is parked by the side of the road, when a speeding car travelling at 50 mi/hrpasses. The police car immediately

Blababa [14]

Answer:

a) time taken to catch up with speeding car is 12.25 secs

b) the police car will travel 273.8 m to catch up with the speeding car

Explanation:

Given that;

speed of car $V_{c}$ = 50 mi/hr = 22.352 m/s

acceleration of police car = 10 mi/hr = 4.47 m/s²

$V_{f}$ = 70 mi/hr = 31.29 m/s

Now time taken to reach maximum speed is t₁

so

$V_{f}$ = $V_{i}$ + at₁

we substitute

31.29 = 0 + 4.47t₁

t₁ = 31.29 / 4.47

t₁ = 7 sec

now

d₁ = 0 + 1/2 × at₁²

d₁ = 0 + 1/2 × 0 + 4.47×(7)²

d₁ = 109.5 m

so distance travelled by the speeding car in time t₁ will be

$d_{c}$ = $V_{c}$ × t₁

we substitute

$d_{c}$ = 22.352 × 7

$d_{c}$ = 156.46 m

now distance between polive car and speeding car

Δd = $d_{c}$ - d₁

Δd = 156.46 - 109.5

Δd = 46.96 m

time taken to cover Δd will be

t₂ = Δd / ( $V_{f}$ - $V_{c}$ )

t₂ = 46.96 / ( 31.29 - 22.352 )

t₂ = 46.96 / 8.938

t₂ = 5.25 sec

distance travelled by the police in time t₂ will be

d₂ = $V_{f}$ × t₂

d₂ = 31.29 × 5.25

d₂ = 164.3 m

a) How long will it take before the officer catches up to the speeding car;

time taken to catch up with speeding car;

t = t₁ + t₂

t = 7 + 5.25

t = 12.25 secs

Therefore, time taken to catch up with speeding car is 12.25 secs

b) how far will it have travelled in order to do so;

distance = d₁ + d₂

distance = 109.5 + 164.3

distance = 273.8 m

Therefore, the police car will travel 273.8 m to catch up with the speeding car

6 0

3 years ago

What is weightlessness ?​

What is weightlessness ?