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Alex17521 [72]
3 years ago
14

A parallel beam of light from a laser with a wavelength 450 nm, falls on a grating whose slits are 1.28 x 10^-4 cm apart. How fa

r (in cm) is the 1t order bright spot from the center of the pattern on a screen 3.4 m away?
Physics
1 answer:
Katena32 [7]3 years ago
7 0

Answer:

Y_1 = 1.195 m = 119.5 cm

Explanation:

For nth order bright fringe, the distance from the center of the pattern is given by :-

Y_n = \frac{n\lambda D}{d}.

Where,

n = order of nth  bright fringe. = 1

\lambda = wavelength of  given light  =  450 nm = 450*10-{9} m

D = distance between the screen and slits. =  3.4 m

and d = separation between the slits. = 1.28 *10^{-4} cm =1.28 *10^{-6} m

Y_1 = \frac{(1 *450 *10{-9} *3.4)}{(1.28*10{-6})}

Y_1 = 1.195 m = 119.5 cm

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What kind of thermal transfer 1.conduction 2.convective 3.radiation from the sun?
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Radiation

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Hope this helps :)
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3 years ago
A 160-m long ski lift carries skiers from a station at the foot of a slope to a second station 40 m above. What is the IMA of th
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How much power is needed to move a 2000 kg mass from the bottom of the 150 m talk great pyramid in Egypt up a ramp if the ramp h
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How Heavy? More than 2,300,000 limestone and granite blocks were pushed, pulled, and dragged into place on the Great Pyramid. The average weight of a block is about 2.3 metric tons (2.5 tons).

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2 years ago
An object in the shape of a thin ring has radius a and mass M. A uniform sphere with mass m and radius R is placed with its cent
madreJ [45]

Answer:

F = GMmx/[√(a² + x²)]³

Explanation:

The force dF on the mass element dm of the ring due to the sphere of mass, m at a distance L from the mass element is

dF = GmdM/L²

Since the ring is symmetrical, the vertical components of this force cancel out leaving the horizontal components to add.

So, the horizontal components add from two symmetrically opposite mass elements dM,

Thus, the horizontal component of the force is

dF' = dFcosФ where Ф is the angle between L and the x axis

dF' = GmdMcosФ/L²

L² = a² + x² where a = radius of ring and x = distance of axis of ring from sphere.

L = √(a² + x²)

cosФ = x/L

dF' = GmdMcosФ/L²

dF' = GmdMx/L³

dF' = GmdMx/[√(a² + x²)]³

Integrating both sides we have

∫dF' = ∫GmdMx/[√(a² + x²)]³

∫dF' = Gm∫dMx/[√(a² + x²)]³    ∫dM = M

F = GmMx/[√(a² + x²)]³  

F = GMmx/[√(a² + x²)]³

So, the force due to the sphere of mass m is

F = GMmx/[√(a² + x²)]³

3 0
2 years ago
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