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skelet666 [1.2K]
3 years ago
11

*PLEASE HURRY ITS FOR A QUIZ*

Physics
1 answer:
kotegsom [21]3 years ago
7 0
The correct answers are :
<span>B. The wheels of a bike riding down the street
C. The blades of a fan that is turned on
D. A child sitting on a moving merry-go-round

Let's see why. Keep in mind that the angular momentum is defined as
</span>L=mvr
<span>where m is the mass of the rotating object, v its velocity and r the radius of the circular trajectory.

</span><span>A. stationary Ferris wheel --> this one has no angular momentum, because its velocity is zero (the wheel is stationary), so v=0 and the angular momentum is zero: L=0
</span>
<span>B. The wheels of a bike riding down the street --> this one has angular momentum, because the edge of the wheel is moving with speed v

</span><span>C. The blades of a fan that is turned on --> this one also has angular momentum, because the fan is moving as well

</span><span>D. A child sitting on a moving merry-go-round --> also this one has angular momentum, which is given by 
</span>L=mvr
<span>where m is the child mass, v its velocity and r the radius of the merry-go-round.</span>
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Answer:

1.696 nm

Explanation:

For a diffraction grating, dsinθ = mλ where d = number of lines per metre of grating = 5510 lines per cm = 551000 lines per metre and λ = wavelength of light = 467 nm = 467 × 10⁻⁹ m. For a principal maximum, m = 1. So,

dsinθ = mλ = (1)λ = λ

dsinθ = λ

sinθ = λ/d.

Also tanθ = w/D where w = distance of center of screen to principal maximum and D = distance of grating to screen = 1.03 m

From trig ratios 1 + cot²θ = cosec²θ

1 + (1/tan²θ) = 1/(sin²θ)

substituting the values of sinθ and tanθ we have

1 + (D/w)² = (d/λ)²

(D/w)² = (d/λ)² - 1

(w/D)² = 1/[(d/λ)² - 1]

(w/D) = 1/√[(d/λ)² - 1]

w = D/√[(d/λ)² - 1] = 1.03 m/√[(551000/467 × 10⁻⁹ )² - 1] = 1.03 m/√[(1179.87 × 10⁹ )² - 1] = 1.03 m/1179.87 × 10⁹  = 0.000848 × 10⁻⁹ = 0.848 × 10⁻¹² m = 0.848 nm.

w is also the distance from the center to the other principal maximum on the other side.

So for both principal maxima to be on the screen, its minimum width must be 2w = 2 × 0.848 nm = 1.696 nm

So, the minimum width of the screen must be 1.696 nm

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