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Rasek [7]
3 years ago
13

In a double-slit experiment, light from two monochromatic light sources passes through the same double slit. The light from the

first light source has a wavelength of 645 nm. Two different interference patterns are observed. If the 8th order bright fringe from the first light source coincides with the 9th order bright fringe from the second light source, what is the wavelength of the light coming from the second monochromatic light source?
Physics
1 answer:
Dmitrij [34]3 years ago
4 0

Answer:

\lambda_2 = 573.3 nm

Explanation:

As we know that the position of maximum intensity on the screen is given as

y = \frac{N\lambda L}{d}

here we know that

\lambda = wavelength

L = distance of the screen

d = distance between two slits

now we know that the position of 8th maximum intensity is same as that of 9th maximum on the screen

so we have

\frac{N_1\lambda_1 L}{d} = \frac{N_2 \lambda_2 L}{d}

so here we have

8 (645 nm) = 9 \lambda_2

\lambda_2 = 573.3 nm

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slega [8]

Answer:

1/3 the distance from the fulcrum

Explanation:

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W_1 d_1 = W_2 d_2

where

W1 is the weight of the boy

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d2 is the distance of the partner from the fulcrum

In this problem, we know that the boy is three times as heavy as his partner, so

W_1 = 3 W_2

If we substitute this into the equation, we find:

(3 W_2) d_1 = W_2 d_2

and by simplifying:

3 d_1 = d_2\\d_1 = \frac{1}{3}d_2

which means that the boy sits at 1/3 the distance from the fulcrum.

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3 years ago
Which one of the following is a derived Si unit a. Newton, b. Meter, c. Mole, d. Kiogram<br>​
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4 0
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A block of mass m = 1.00 kg is attached to a spring of force constant k = 500 N/m. The block is pulled to a position xi= 5.00 cm
8_murik_8 [283]

Answer:

The speed of the block is 4.96 m/s.

Explanation:

Given that.

Mass of block = 1.00 kg

Spring constant = 500 N/m

Position x_{i}=5.00\ cm

Coefficient of friction = 0.350

(A). We need to calculate the speed the block has as it passes through equilibrium if the horizontal surface is friction less

Using formula of kinetic energy and potential energy

\dfrac{1}{2}mv^2=\dfrac{1}{2}kx^2-\mu mgx

Put the value into the formula

\dfrac{1}{2}\times1.00\times v^2=\dfrac{1}{2}\times500\times(5.00\times10^{-2})-0.350\times1.00\times9.8\times5.00\times10^{-2}

v^2=\dfrac{2\times12.3285}{1.00}

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v=4.96\ m/s

Hence, The speed of the block is 4.96 m/s.

6 0
2 years ago
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Answer:

C

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