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Oksanka [162]
3 years ago
11

gAn optical engineer needs to ensure that the bright fringes from a double-slit are 15.7 mm apart on a detector that is from the

slits. If the slits are illuminated with coherent light of wavelength 633 nm, how far apart should the slits be
Physics
1 answer:
igomit [66]3 years ago
7 0

Answer:

d = 68.5 x 10⁻⁶ m = 68.5 μm

Explanation:

The complete question is as follows:

An optical engineer needs to ensure that the bright fringes from a double-slit are 15.7 mm apart on a detector that is  1.70m from the slits. If the slits are illuminated with coherent light of wavelength 633 nm, how far apart should the slits be?

The answer can be given by using the formula derived from Young's Double Slit Experiment:

y = \frac{\lambda L}{d}\\\\d  =\frac{\lambda L}{y}\\\\

where,

d = slit separation = ?

λ = wavelength = 633 nm = 6.33 x 10⁻⁷ m

L = distance from screen (detector) = 1.7 m

y = distance between bright fringes = 15.7 mm = 0.0157 m

Therefore,

d = \frac{(6.33\ x\ 10^{-7}\ m)(1.7\ m)}{0.0157\ m}\\\\

<u>d = 68.5 x 10⁻⁶ m = 68.5 μm</u>

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Answer:

35 N to the right.

Explanation:

450 is going to the right so you subtract what is going against it. Which gives you 35. And because 450 is bigger than 415, it'll be going to the right.

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Explanation:

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A driver with a 0.80-s reaction time applies the brakes, causing the car to have acceleration opposite the direction of motion.
jeka94

Answer:

a) During the reaction time, the car travels 21 m

b) After applying the brake, the car travels 48 m before coming to stop

Explanation:

The equation for the position of a straight movement with variable speed is as follows:

x = x0 + v0 t + 1/2 a t²

where

x: position at time t

v0: initial speed

a: acceleration

t: time

When the speed is constant (as before applying the brake), the equation would be:

x = x0 + v t

a)Before applying the brake, the car travels at constant speed. In 0.80 s the car will travel:

x = 0m + 26 m/s * 0.80 s = <u>21 m  </u>

b) After applying the brake, the car has an acceleration of -7.0 m/s². Using the equation for velocity, we can calculate how much time it takes the car to stop (v = 0):

v = v0 + a* t

0 = 26 m/s + (-7.0 m/s²) * t

-26 m/s / - 7.0 m/s² = t

t = 3.7 s

With this time, we can calculate how far the car traveled during the deacceleration.

x = x0 +v0 t + 1/2 a t²

x = 0m + 26 m/s * 3.7 s - 1/2 * 7.0m/s² * (3.7 s)² = <u>48 m</u>

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3 years ago
A girl, standing 150m in front of a tall building, fires a shot using a starting pistol. A boy, standing
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I believe 650 or C

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plz mark brainnlyist :)

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3 years ago
A roller coaster starts from rest at its highest point and then descends on its (frictionless) track. Its speed is 36 m/s when i
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Answer:

its speed when its height was half that of its starting point is 25.46 m/s

Explanation:

Given;

final speed of the roller coaster, v = 36 m/s

Applying general equation of motion;

V² = U² + 2gh

where;

V is the final speed of the roller coaster

U is the initial speed of the roller coaster = 0

h is the height attained at a given velocity

36² = 0 + (2 x 9.8)h

1296 = 19.6 h

h = 1296/19.6

h = 66.1224 m

when its height was half that of its starting point, h₂ = ¹/₂ h

h₂ = ¹/₂(66.1224 m) = 33.061 m

At h = 33.061 m, V = ?

V² = U² + 2gh

V² = 0 + 2 x 9.8 x 33.061

V² = 648

V = √648

V = 25.46 m/s

Therefore, its speed when its height was half that of its starting point is 25.46 m/s

8 0
3 years ago
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