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V125BC [204]
3 years ago
5

When a student shines a 470 nm laser through this grating, how many bright spots could be seen on a screen behind the grating

Physics
1 answer:
scZoUnD [109]3 years ago
3 0

Answer:

15

Explanation:

Assuming that commercial diffraction grating has 300 lines per mm.

also, given that a 470 nm laser through this grating

d = width of the slit  must be = \frac{10^{-3}}{300}

λ = wavelength = 470 nm = 470 x 10^{-9} m

Using the equation

d Sinθ = nλ

n = order

N = number of spots

θ = angle between the diffracted ray and the normal ray= 90°

Assuming the incident ray to be normal to the surface

Plugging the values we get

3.33\times10^{-6}sin90 = n(470\times10^{-9})

solving we get

n = 7

Number of spots are given as

N = 2n + 1

N = 2(7) + 1

N = 15

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Convert 3 hours 21 minutes to decimal hours.
Oxana [17]

Total hours is 3.35 hours

<u>Explanation:</u>

Given:

convert 3 hours 21 minutes to decimal hours

We know:

1 hour = 60 minutes

and

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1 minute = 1 / 60 hours

So,

21 minutes = \frac{21}{60} hours

                  = 0.35 hours

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7 0
3 years ago
1. A particle is projected vertically upwards with a velocity of 30 ms from a point 0. Find (a) the maximum height reached(b) th
Bogdan [553]

Assuming the particle is in free fall once it is shot up, its vertical velocity <em>v</em> at time <em>t</em> is

<em>v</em> = 30 m/s - <em>g t</em>

where <em>g</em> = 9.8 m/s² is the magnitude of the acceleration due to gravity, and its height <em>y</em> is given by

<em>y</em> = (30 m/s) <em>t</em> - 1/2 <em>g t</em> ²

(a) At its maximum height, the particle has 0 velocity, which occurs for

0 = 30 m/s - <em>g t</em>

<em>t</em> = (30 m/s) / <em>g</em> ≈ 3.06 s

at which point the particle's maximum height would be

<em>y</em> = (30 m/s) (3.06 s) - 1/2 <em>g</em> (3.06 s)² ≈ 45.9184 m ≈ 46 m

(b) It takes twice the time found in part (a) to return to 0 height, <em>t</em> ≈ 6.1 s.

(c) The particle falls 35 m below its starting point when

-35 m = (30 m/s) <em>t</em> - 1/2 <em>g t</em> ²

Solve for <em>t</em> to get a time of about <em>t</em> ≈ 7.1 s

7 0
3 years ago
A car is traveling at a steady 73 km/h in a 50 km/h zone. A police motorcycle takes off at the instant the car passes it, accele
iVinArrow [24]

Answer:

29 m

Explanation:

given,

speed of the car = 73 Km/hr

                           = 73 x 0.278 m/s  = 20.3 m/s

Time elapses before the motorcycle is moving as fast as the car

using equation of motion

v = u + at

initial speed = 0 m/s

20.3= 0 + 7 x t

t = 2.9 s

b) distance traveled by the car in that time

  d = s x t

  d = 20.3 x 2.9 = 58.87 m

 distance traveled by the police

s = ut + \dfrac{1}{2} a t^2

s = 0 + \dfrac{1}{2}\times 7\times 2.9^2

s = 29.435 m

distance car is ahead = 58.57 - 29.44 = 29.13

                                    = 29 m

3 0
3 years ago
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