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malfutka [58]
3 years ago
7

580 nm light shines on a double slit

Physics
1 answer:
Helga [31]3 years ago
5 0

Answer:

0.532

Explanation:

Your equation to find the second bright interference maximum is gonna be this: d sin (Θ) = m λ

First, find your variables.

λ = 580 · 10^-9

d = 0.000125

m = 2

Next, fill in the equation.

d sin (θ) = m λ

(0.000125) sin (θ) = (2) (580·10^-9)

Then isolate your variable.

θ = arcsin ( (2)(580·10^-9) / (0.000125) )

Run your equation and you will end up with 0.53171246 , which rounds to 0.532.

The main thing you have to watch out for is make sure you are calculating for the bright interference and not the dark interference, as well as checking you're calculating for the maximum, not the minimum.

I hope this helps :D

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So you would do 75 N x 10m x Cos (0 degrees)= 750 J
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Answer:

a. Quadruped arm and opposite leg raise

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Quadruped arm and opposite leg lift

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1. For a reaction to occur, energy must be
amm1812

For a reaction to occur, energy must be absorbed to break chemical bonds

<u>Explanation:</u>

Reactions can be classified as chemical reaction, nuclear reaction, thermal reaction. So in these three reaction types, the nature of energy will only be varying.

But in order to execute a reaction, there should be breaking of existing bonds and then formation of new bonds. So for breaking of the bonds of reactants, energy should be absorbed from the surrounding.

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HELPPP
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Answer:

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Explanation:

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4 0
2 years ago
A powerful searchlight shines on a man. The man's cross-sectional area is 0.500m2 perpendicular to the light beam, and the inten
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Answer:

The magnitude of the force the light beam exerts on the man is 5.9 x 10⁻⁵N

(b) the force the light beam exerts is much too small to be felt by the man.

Explanation:

Given;

cross-sectional area of the man, A = 0.500m²

intensity of light, I = 35.5kW/m²

If all the incident light were absorbed, the pressure of the incident light on the man can be calculated as follows;

P = I/c

where;

P is the pressure of the incident light

I is the intensity of the incident light

c is the speed of light

P = \frac{I}{c} =\frac{35500}{3*10^8} = 1.18*10^{-4} \ N/m^2

F = PA

where;

F is the force of the incident light on the man

P is the pressure of the incident light on the man

A is the cross-sectional area of the man

F = 1.18 x 10⁻⁴ x 0.5 = 5.9 x 10⁻⁵ N

The magnitude of the force the light beam exerts on the man is 5.9 x 10⁻⁵ N

Therefore, the force the light beam exerts is much too small to be felt by the man.

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