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victus00 [196]
3 years ago
5

Light of wavelength 480 nm illuminates a double slit, and the interference pattern is observed on a screen. At the position of t

he m = 4 bright fringe, how much farther is it to the more distant slit than to the nearer slit?
Physics
1 answer:
vova2212 [387]3 years ago
4 0

Answer:

BC =980nm

Explanation:

Given wavelength =490nm

m=2

Required to find (OB - OC) = BC (more distant slit - nearest slit ) = dsin theta

We have dsin theta = m*wavelength

or BC = m* wavelength = 2*490nm

= 980nm = 98.000A

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disa [49]

The acceleration due to gravity is g/4

The acceleration above the earth surface is given by the relation

g^'=gr^2/〖(h+r)〗^2

Since the satellite orbits the earth in a orbit of radius equal to earth radius, therefore

g^'=(gr^2)/〖(r+r)〗^2 =g/4

Thus the acceleration due to gravity on the satellite is g/4.

6 0
3 years ago
The exit nozzle in a jet engine receives air at 1200 K, 150 kPa with negligible kinetic energy. The exit pressure is 80 kPa, and
nikitadnepr [17]

Complete question:

The exit nozzle in a jet engine receives air at 1200 K, 150 kPa with negligible kinetic energy. The exit pressure is 80 kPa, and the process is reversible and adiabatic. Use constant specific heat at 300 K to find the exit velocity.

Answer:

The exit velocity is 629.41 m/s

Explanation:

Given;

initial temperature, T₁ = 1200K

initial pressure, P₁ = 150 kPa

final pressure, P₂ = 80 kPa

specific heat at 300 K, Cp = 1004 J/kgK

k = 1.4

Calculate final temperature;

T_2 = T_1(\frac{P_2}{P_1})^{\frac{k-1 }{k}

k = 1.4

T_2 = T_1(\frac{P_2}{P_1})^{\frac{k-1 }{k}}\\\\T_2 = 1200(\frac{80}{150})^{\frac{1.4-1 }{1.4}}\\\\T_2 = 1002.714K

Work done is given as;

W = \frac{1}{2} *m*(v_i^2 - v_e^2)

inlet velocity is negligible;

v_e = \sqrt{\frac{2W}{m} } = \sqrt{2*C_p(T_1-T_2)} \\\\v_e = \sqrt{2*1004(1200-1002.714)}\\\\v_e = \sqrt{396150.288} \\\\v_e = 629.41  \ m/s

Therefore, the exit velocity is 629.41 m/s

6 0
3 years ago
Determine the magnitude of the electrostatic force on a 0.06000 C charged object when it is placed in an electric field of magni
joja [24]

Answer:

Explanation:

Use the following equation:

E=\frac{q}{F} and solve for F:

F=\frac{q}{E} and filling in:

F=\frac{.0600}{1500}

F = 4.0 × 10^{-4 N

3 0
2 years ago
What kind of object are the light rays interacting with in the model below?
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Answer:

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5 0
2 years ago
Please help :<
Artemon [7]

Answer: methyl

Explanation:

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