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OleMash [197]
3 years ago
7

A diffraction grating with 1000 lines per mm is used in a spectrometer to measure the wavelengths of light emitted from a gas di

scharge tube. You measure the diffraction angle to be at 20° for one emission line. What would happen to this emission line’s diffraction angle if you replaced the diffraction grating with one having 500 lines per mm instead
Physics
1 answer:
frez [133]3 years ago
4 0

Answer:

= 9.8°

Explanation:

Width of one slit (a₁ ) = 1 / 1000 mm=0.001 mm = 10⁻⁶ m.

width of one slit in case 2 (a₂ ) = 1/500 =2 x 10⁻⁶ m

angular position of fringe, Sinθ  = n λ /a

n is order of fringe , λ is wave length of light and a  is slit aperture

So Sinθ  ∝ 1 / a

Sin θ₁ /Sin θ₂ = a₂/a₁ ;

Sin20°/sinθ₂ = 2 / 1

sinθ₂ = Sin 20° / 2 = .342/2 = .171

θ₂ = 9.8 °

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Ray Of Light [21]
Go here to get some help now a=can you help me i posted it on your profile


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8 0
4 years ago
Calculate the focal length of a lens needed by a woman whose near point is 50cm from her eyes, assuming the least distance of di
son4ous [18]

The focal length of a lens needed by a woman whose near point is 50cm from her eyes is 50cm.

To find the answer, we have to know about the focal length of correcting lens.

<h3>How to find the focal length of correcting lens?</h3>
  • If x is the distance of nearest point of the defective eye and D is the least distance of distinct vision, then, the expression for focal length of the correcting lens will be,

                           f=\frac{XD}{X-D}

  • It is given that, the woman whose near point is 50cm from her eyes, assuming the least distance of distinct vision for a normal eye is 25cm. Thus, the focal length will be,

                       f=\frac{50*25}{50-25} =50cm.

Thus, we can conclude that, the focal length of a lens needed by a woman whose near point is 50cm from her eyes is 50cm.

Learn more about the focal length here:

brainly.com/question/27915592

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4 0
2 years ago
Bob is pulling a 30kg filing cabinet with a force of 200N , but the filing cabinet refuses to move. The coefficient of static fr
vredina [299]

Answer:

2.0\cdot 10^2 N

Explanation:

The cabinet does not move: this means that the net force acting on it is zero.

Along the horizontal direction, we have two forces:

- The push exerted by Bob, F = 200 N, forward

- The frictional force, F_f, which acts in the opposite direction (backward)

Since the net force must be zero, we have:

F-F_f = 0

So solving the equation we can find the magnitude of the friction force:

F_f = F = 200 N=2.0 \cdot 10^2 N

7 0
4 years ago
Two basketballs of equal mass are rolling toward each other at constant velocities. The first basketball (B1) has a velocity of
slamgirl [31]

v'_2 = \frac{2m_1}{m_1+m_2} (4.3) - \frac{m_1-m_2}{m_1+m_2} (4.3)\\\\v'_1 = \frac{m_1-m_2}{m_1+m_2} (4.3) + \frac{2m_2}{m_1+m_2} (4.3)

<u>Explanation:</u>

Velocity of B₁ = 4.3m/s

Velocity of B₂ = -4.3m/s

For perfectly elastic collision:, momentum is conserved

m_1v_1 + m_2v_2 = m_1v'_1 + m_2v'_2

where,

m₁ = mass of Ball 1

m₂ = mass of Ball 2

v₁ = initial velocity of Ball 1

v₂ = initial velocity of ball 2

v'₁ = final velocity of ball 1

v'₂ = final velocity of ball 2

The final velocity of the balls after head on elastic collision would be

v'_2 = \frac{2m_1}{m_1+m_2} v_1 - \frac{m_1-m_2}{m_1+m_2} v_2\\\\v'_1 = \frac{m_1-m_2}{m_1+m_2} v_1 + \frac{2m_2}{m_1+m_2} v_2

Substituting the velocities in the equation

v'_2 = \frac{2m_1}{m_1+m_2} (4.3) - \frac{m_1-m_2}{m_1+m_2} (4.3)\\\\v'_1 = \frac{m_1-m_2}{m_1+m_2} (4.3) + \frac{2m_2}{m_1+m_2} (4.3)

If the masses of the ball is known then substitute the value in the above equation to get the final velocity of the ball.

5 0
3 years ago
Inside a NASA test vehicle, a 3.50-kg ball is pulled along by a horizontal ideal spring fixed to a friction-free table. The forc
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F(of spring)=230x=ma=3.5(5)=17.5=230x; x=0.07m.
3 0
3 years ago
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