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VikaD [51]
3 years ago
6

Find the angle for the third-order maximum for 591 nm wavelength light falling on a diffraction grating having 1460 lines per ce

ntimeter.
Physics
1 answer:
Marat540 [252]3 years ago
6 0

Answer:

15.32°

Explanation:

We have given the wavelength \lambda =591nm=591\times 10^{-9}m

Diffraction grating is 1460 lines per cm

So  d=\frac{10^{-2}}{1460}=6.71\times 10^{-6}m (as 1 m=100 cm )

For maximum diffraction

dsin\Theta =m\lambda here m is order of diffraction

So 6.71\times 10^{-6}sin\Theta =3\times 591\times 10^{-9}

sin\Theta =0.264

\Theta =15.32^{\circ}

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ankoles [38]
Given required solution

M=10kg W=? W=Fd
v=5.0m/s F=mg
t=2.40s =10*10=100N
S=VT
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so W=12*100
W=1200J
4 0
3 years ago
A stone is thrown in the upwards direction at the velocity of 4 m/s. It attains a certain height and then it falls back. During
Sati [7]

Answer: 0.8 m

Explanation:

Velocity of throw = 4m/s

Maximum Height attained(h) =?

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Using the relation:

v^2 = u^2 + 2aS

v = final Velocity = 0 (at maximum height)

u = Initial Velocity = 4

a = g downward acceleration = - 10

0 = 4^2 + 2(-10)(S)

0 = 16 - 20S

20S = 16

S = 16 / 20

S = 0.8m

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3 years ago
Study of the ___________
Bumek [7]

Answer: Photoelectric effect

Explanation:

3 0
3 years ago
A race car drives one lap around a race track that is 500 meters in length.
DIA [1.3K]

Answer:

a) 0

Explanation:

A) since he is going around a race track, his initial position is 0m, and his ending position is also 0m, so his displacement is 0. Yes, his distance is 500m, but his displacement is 0, and he did not end up going anywhere for his final position.

B) Distance is total amount traveled, which in this case is 500m, but displacement is 0 as he did not change his position.

4 0
3 years ago
The club was in contact with a ball, initially at rest, for about 0.0052 s. If the ball has a mass of 55 g and leaves the head o
taurus [48]

Answer:

2538.46N

Explanation:

Parameters given:

Initial speed, u = 0 m/s

Final velocity, v = 240 m/s

Time, t = 0.0052 s

Mass of club, m = 55g = 0.055 kg

First we find the acceleration:

a = (v - u)/t

a = (240 - 0)/0.0052

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F = m*a

F = 0.055 * 46153.85

F = 2538.46 N

3 0
3 years ago
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