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stellarik [79]
3 years ago
6

The rhinestones in costume jewelry are glass with index of refraction 1.50. To make them more reflective, they are often coated

with a layer of silicon monoxide of index of refraction 2.00. What is the minimum coating thickness needed to ensure that light of wavelength 576 nm and of perpendicular incidence will be reflected from the two surfaces of the coating with fully constructive interference
Physics
1 answer:
anyanavicka [17]3 years ago
4 0

Answer:

T=62.9*10^{-9}

Explanation:

From the question we are told that:

Index of refraction of Rinestones \gamma_1 =1.5

Index of refraction of silicon \gamma_2 =2.0

Wavelength  \lambda=576nm=576*10^{-9}

Let each layer have thickness T

Therefore

Total Thickness =2T

Generally the equation for  Constructive interference is mathematically given by

2T=(m+0.5)\frac{l\lambda}{\gamma_2}

Where

 M=0

 2T=(0+0.5)\frac{576*10^{-9}}{2*2.0}

 T=62.9*10^{-9}

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Trucks can be run on energy stored in a rotating flywheel, with an electric motor getting the flywheel up to its top speed of 20
Genrish500 [490]

Answer:

The time it can operate between chargins in minutes is

t=102.8 minutes

Explanation:

Given: m=500kg, r=1.0m, w=200\pi rad/s

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K_R=\frac{1}{2}*I*w^2

I=\frac{1}{2}*m*r^2

I=\frac{1}{2}*500kg*(1.0m)^2

I=250 kg*m^2

K_R=\frac{1}{2}*250kg*m^2*(200\pi rad/s)^2

K_R=49.348x10^6J

b). The power average 0.8kW un range time can be find

P=\frac{K_R}{t'}

Solve to t'

t=\frac{K_R}{P}

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3 0
3 years ago
What current is needed in the solenoid's wires?
marta [7]
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B = uo N I / h
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5 0
3 years ago
A box of mass 60 kg is at rest on a horizontal floor that has a static coefficient of friction of 0.6 and a kinetic coefficient
gavmur [86]

Answer:

a) The minimum force required to start moving the box is 352.86 N

b) i) The friction force for the box in motion is 147.025 N

ii) The acceleration of box is 4.21625 m/s²

Explanation:

The parameters of the box at rest and the floor are;

The mass of the box = 60 kg

The static coefficient friction of the floor = 0.6

The kinetic coefficient friction of the floor = 0.25

Frictional force = Normal force × Friction coefficient

For an horizontal floor and the box laying on the floor, we have;

The normal force = The weight of the box = Mass of the box × Acceleration due to gravity, g

The acceleration due to gravity, g = 9.81 m/s²

The weight of the box  = 60 × 9.81 = 588.6 N

a) The static coefficient gives the frictional force observed by the box and which must be surpassed to bring about motion

Therefore;

The minimum force required to start moving the box = The static frictional force = Weight of the box × The static coefficient of friction

The minimum force required to start moving the box = 588.1 × 0.6 = 352.86 N

The minimum force required to start moving the box = 352.86 N

b) i) When an horizontal force of 400 N is applied, the applied force is larger than the static friction force, and the box will be in motion with the kinetic coefficient of friction being the source of friction

The friction force for the box in motion = 588.1 × 0.25 = 147.025 N

ii) The force, F with the box is in motion, is given as follows;

F = Mass of box × Acceleration of box, a = Applied force - Kinematic friction force

F = 60 × a = 400 - 147.025 = 252.975 N

60 × a = 252.975 N

a = 252.975 N/(60 kg) = 4.21625 m/s²

Acceleration of box, a = 4.21625 m/s².

6 0
3 years ago
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