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Oliga [24]
3 years ago
5

A chamber 5.0 cm long with flat, parallel windows at the ends is placed in one arm of a Michelson interferometer (see below). Th

e light used has a wavelength of 500 nm in a vacuum. While all the air is being pumped out of the chamber, 29 fringes pass by a point on the observation screen. What is the refractive index of the air?
Physics
1 answer:
Ivahew [28]3 years ago
6 0

Answer:

1.00029

Explanation:

A wave is a disturbance which travels through a medium and transfers energy without displacing the medium itself.

interference is one of the characteristics of a wave motion

The Michelson interferometer produces interference fringes by splitting a beam of monochromatic light so that one beam strikes a fixed mirror and the other a movable mirror. When the reflected beams are reflected, an interference pattern results.

∝Δm=Δud

∝=wavelength  500nm

Δm=number of fringes

d=length 9f the  Michelson interferometer chamber

Δu= change in refractive index

500*10^-9*29=0.05*Δu

Δu=0.00029

REFRACTIVE INDEX=1+0.00029

refractive index=

1.00029

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A child is riding in a child-restraint chair, securely fastened to the seat of a car. Assume the car has speed 47 km/h when it h
Elina [12.6K]

Answer: F = 1235 N

Explanation: Newton's Second Law of Motion describes the effect of mass and net force upon acceleration: F_{net}=m.a

Acceleration is the change of velocity in a period of time: a=\frac{\Delta v}{\Delta t}

Velocity of the car is in km/h. Transforming it in m/s:

v=\frac{47.10^{3}}{36.10^{2}}

v = 13 m/s

At the moment the car decelerates, acceleration is

a=\frac{13}{0.2}

a = 65 m/s²

Then, force will be

F_{net}=19(65)

F_{net} = 1235 N

The horizontal net force the straps of the restraint chair exerted on the child to hold her is 1235 newtons.

5 0
2 years ago
At t = 0 a grinding wheel has an angular velocity of 24.0 rad/s. It has a constant angular acceleration of 30.0rad/s2 until a ci
kozerog [31]

Answer:

θ=108rad

t =10.29seconds

α=-8.17rad/s²

Explanation:

Given that

At t=0, Wo=24rad/sec

Constant angular acceleration =30rad/s²

At t=2, θ=432rad as it try to stop because the circuit break

Angular motion

W=Wo+αt

θ=Wot+1/2αt²

W²=Wo²+2αθ

We need to find θ between 0sec to 2sec when the wheel stop

a. θ=Wot+1/2αt²

θ=24×2+1/2×30×2²

θ=48+60

θ=108rad.

b. W=Wo+αt

W=24+30×2

W=84rad/s

This is the final angular velocity which is the initial angular velocity when the wheel starts to decelerate.

Wo=84rad/sec

W=0rad/s, because the wheel stop at θ=432rad

Using W²=Wo²+2αθ

0²=84²+2×α×432

-84²=864α

α=-8.17rad/s²

It is negative because it is decelerating

Now, time taken for the wheel to stop

W=Wo+αt

0=84-8.17t

-84=-8.17t

Then t =10.29seconds.

a. θ=108rad

b. t =10.29seconds

c. α=-8.17rad/s²

3 0
3 years ago
An argon ion laser puts out 5.0 W of continuous power at a wavelength of 532 nm. The diameter of the laser beam is 5.5 mm. If th
stepan [7]

Answer:

Number of photons travel through pin hole=6.4*10^{17}

Explanation:

First we will calculate the energy of single photon using below formula:

E=\frac{h*c}{λ}

Where :

h is plank's constant with value 6.626*10^{-34} J.s

c is the speed of light whch is3*10^{8}

λ is the wave length = 532nm

E=\frac{6.6268*10^{-34}* 3*10^{8}}{532nm}

E=3.73*10^{-19}J

Number of photons emitted per second:

\frac{5J/s}{1 photon/3.73*10^{-19} }

Number of photons emitted per second=1.34*10^{19} photons/s

\frac{A-hole}{A-beam}=\frac{\frac{pie*d-hole^2}{4}}{\frac{pie*d-beam^2}{4} }

Where:

A-hole is area of hole

A-beam is area of beam

d-hole is diameter of hole

d-beam is diameter if beam

\frac{A-hole}{A-beam}=\frac{d-hole^2}{d-beam^2}

\frac{Ahole}{A-beam}=\frac{1.22^2}{5.5^2}

\frac{A-hole}{A-beam}=\frac{144}{3025}

Number of photons travel through pin hole=1.34*10^{19} *\frac{144}{3025}

Number of photons travel through pin hole=6.4*10^{17}

7 0
2 years ago
A 40-cm-diameter, 300 g beach ball is dropped with a 4.0 mg ant riding on the top. The ball experiences air resistance, but the
mars1129 [50]

Answer:

The normal force exerted on the ant is 0.75 N.

Explanation:

Given;

diameter of the ball, D = 40 cm = 0.4m

radius of the ball, r = 0.2m

mass of the beach ball, m₁ = 300 g = 0.3 kg

mass of the ant, m₂ = 4 x 10⁻⁶ kg

speed of the ball, v = 4 m/s

The area of the ball, assuming spherical ball is given by;

A = 4πr²

A = 4π(0.2)² = 0.5027 m²

The drag force (resistance) experienced by the spherical ball is given as;

F_D = \frac{1}{2}C\rho Av^2

where;

C is the drag coefficient of the spherical ball = 0.45

ρ is density of air = 1.21 kg/m³

F_D = \frac{1}{2}C\rho Av^2\\\\F_D = \frac{1}{2}(0.45)(1.21) (0.5027)(4)^2\\\\F_D = 2.19 \ N

The downward force of the ball due to its weight and that of the ant is given by;

F_g = mg\\\\F_g =g(m_{ant} + m_{ball})\\\\F_g = g(4*10^{-6} \ kg\ + \ 0.3\ kg)\\\\F_g = g(0.300004 \ kg) \ \ \ (mass \ of \ the \ ant \ is \ insignificant)\\\\F_g = 9.8(0.3)\\\\F_g = 2.94 \ N

The net downward force experienced by the ball is given by;

F_{net} = F_g - F_D\\\\F_{net} = 2.94 \ N - 2.19 \ N\\\\F_{net} = 0.75 \ N

This downward force experienced by the ball is equal to the normal reaction it exerts on the ant.

Thus, the normal force exerted on the ant is 0.75 N.

5 0
2 years ago
Which of the following is an example of acceleration? Select one: a. A boy walks two miles in one hour. b. A girl walks two mile
defon

The answer is option C .Because as the direction of the boy changes ;velocity changes .

8 0
2 years ago
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