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Levart [38]
3 years ago
15

Consider a single-slit diffraction pattern for λ=589nm , projected on a screen that is 1.00 m from a slit of width 0.25 mm. How

far from the center of the pattern are the centers of the first and second dark fringes?
Physics
1 answer:
const2013 [10]3 years ago
8 0

Answer: i) 2.356 × 10^-3 m = 2.356mm, ii) 4.712 × 10^-3 m = 4.712mm

Explanation: The formulae that relates the position of a fringe from the center to the wavelength, distance between slits and distance between slits and screen is given below as

y = R×(mλ/d)

Where y = distance between nth fringes and the center fringe.

m = order of fringe

λ = wavelength of light = 589nm = 589×10^-9m

R = distance between slits and screen = 1.0m

d = distance between slits = 0.25mm = 0.00025m

For distance between the first dark fringe and the center fringe.

This implies that m = 1

y = 1 × 589×10^-9 × 1/0.00025

y = 589×10^-9/0.00025

y = 2,356,000 × 10^-9

y = 2.356 × 10^-3 m = 2.356mm

For the second dark fringe, this implies that m = 2

y = 1 × 2 × 589×10^-9/0.00025

y = 1178 × 10^-9 /0.00025

y = 4,712,000 × 10^-9

y = 4.712 × 10^-3 m = 4.712mm

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A football is kicked with a velocity of 31 meters per second at an angle of 41 degrees what is the balls acceleration in the ver
IgorC [24]

Answer:

The vertical component of the acceleration is equal to the acceleration due to gravity g.

Explanation:

Given data,

The initial velocity of the football, u = 31 m/s

The angle of direction of the football with the horizontal, Ф = 41°

In free fall, the vertical component of acceleration always remains the same.

Since the ball is clicked in a particular direction that has a vertical component,  then the vertical component of the velocity of the ball is always acted upon by the gravitational force of the earth.

The horizontal component of the velocity of the projectile remains the same because the gravitational force doesn't act in the horizontal direction.

Since the only force is the gravitational force acts on the projectile, the vertical component of the acceleration is equal to the acceleration due to gravity g.

8 0
2 years ago
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An ac generator provides emf to a resistive load in a remote factory over a two-cable transmission line. At the factory a step-
Nataliya [291]

Answer:

a) 1.875 volts

b) 5.86 W

Explanation:

Given data:

transmission line resistance = 0.30/cable

power of the generator = 250 kW

if Vt = 80 kV

<u>a) Determine the decrease V along the transmission line </u>

Rms current in cable = P / Vt = 250 / 80 = 3.125 amps

hence the rms voltage drop ( Δrmsvoltage )

Δrmsvoltage = Irms* R  = 3.125 * 2 * 0.30 =<em> 1.875 volts </em>

<u>b) Determine the rate Pd at which energy is dissipated in line as thermal energy </u>

Pd = I^2rms*R

    =  3.125^2 * 2* 0.30  = <em>5.86 W</em>

3 0
2 years ago
A thief is trying to escape from a parking garage after completing a robbery, and the thief's car is speeding (v = 13 m/s) towar
rodikova [14]

<u>Answer</u>

Yes, the car reaches the door before the gate closes.

<u>Explanation</u>

The time taken by the car to reach at the door.

Time = distance / time

        = 22/13

          = 1.6923 seconds

Time taken by the door to close up to the height of the car.

Distance the door has to move to prevent the car from escaping = 9.1.4 = 7.6 m

From newton's 2nd law of motion;

s = ut + 1/2 gt²

7.6 = 0.6t + 1/2 × 10t²

7.6 = 0.6t + 5t²

50t² + 6t - 76 = 0

Solving this quadrilatic equation,

t = 28.537 seconds

Answer: Yes, the car reaches the door before the gate closes.


3 0
3 years ago
What will the reading of the voltmeter be at the instant the switch returns to position a if the inertia of the d'Arsonval movem
Step2247 [10]

Answer:

hello your question is incomplete attached below is the complete question

answer :

20.16 v

Explanation:

The reading of the voltmeter at the instant the switch returns  to position a

L = 5H

i ( current through inductor ) = 1/L ∫ V(t) d(t) + Vo

                                               = 1/5 ∫ 3*10^-3  d(t)  + 0 = 0.6 * 10^-3 t

iL ( 1.6 s ) = 0.6 * 10^-3 * 1.6 = 0.96 mA

Rm ( resistance ) = 21 * 1000 = 21 kΩ

 The reading of the voltmeter ( V )

V = IR

   = 0.96 mA * 21 k Ω  = 20.16 v

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