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bija089 [108]
3 years ago
15

You're using a monochromatic beam of light with wavelength 500 nm in an interferometer. What is the minimum distance you would n

eed to move one of the interferometer's mirrors to change a spot of constructive interference to a spot of destructive interference?
Physics
1 answer:
xxTIMURxx [149]3 years ago
4 0

Answer:

The minimum distance moved by mirror is 125 nm

Explanation:

Given:

Wavelength of light \lambda = 500 \times 10^{-9} m

Here given in question, spot of constructive interference change to a destructive interference.

Hence minimum distance moved by one of mirror is given by,

   t = \frac{1}{2}(\frac{\lambda}{2} )

Where t = distance moved by mirror

   t = \frac{\lambda }{4}

   t = \frac{500\times 10^{-9} }{4}

   t = 125 \times 10 ^{-9}

Therefore, the minimum distance moved by mirror is 125 nm

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ziro4ka [17]

Answer:

9. A 1500kg car traveling +6m/s with a 2000kg truck at rest. The vehicles collide, but do not stick together. The car has a velocity -3m/s after the collision. What is the velocity of the truck? a. What type of collision occurred above?

6 0
3 years ago
The electric field strength between the plates of a simple air capacitor is equal to the voltage across the plates divided by th
Dmitry [639]

Answer:

d = V/E

Explanation:

From the definition, we can say that the electric field strength between the plates of a parallel plate capacitor is

E = v/d

where

E = electric field strength

V = potential difference

d = distance between the plates

On rearranging the equation and making d subject of the formula, we have

d = V/E

From the question, we're given that

V = 112 V

E = 1.12 kV/cm converting to V/m, we have 110000 V/cm

d = 112 / 110000

d = 0.00102 m

d = 1.02*10^-3 m

5 0
3 years ago
The dragster has a mass of 1.3 Mg and a center of mass at G. A parachute is attached at C provides a horizontal braking force of
adell [148]

Answer:

The deceleration of the dragster upon releasing the parachute such that the wheels at B are on the verge of leaving the ground is  16.33 m/s²

Explanation:

The additional information to the question is embedded in the diagram attached below:

The height between the dragster and ground is considered to be 0.35 m since is not given ; thus in addition win 0.75 m between the dragster and the parachute; we have: (0.75 + 0.35) m = 1.1 m

Balancing the equilibrium about point A;

F(1.1) - mg (1.25) = ma_a (0.35)

1.8v^2(1.1) - 1200(9.8)(1.25) = 1200a(0.35)

1.8v^2(1.1) - 14700 = 420 a   ------- equation (1)

F_x = ma_x \\ \\ = 1.8v^2 = 1200 \ a             --------- equation (2)

Replacing equation 2 into equation 1 ; we have :

{1.1 * 1200 \ a} - 14700 = 420 a

1320 a - 14700 = 420 a

1320 a -  420 a =14700

900 a = 14700

a = 14700/900

a = 16.33 m/s²

The deceleration of the dragster upon releasing the parachute such that the wheels at B are on the verge of leaving the ground is  16.33 m/s²

5 0
3 years ago
Which part of an atom has most of its mass? A. electrons B. neutrons C. nucleus D. protons
vredina [299]

<em>The </em><em>nucleus</em><em> has most of the atomic mass in an atom. The </em><em>nucleus</em><em> is made up of protons and neutrons.</em>

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6 0
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