Which is an example of a passive solar energy system?

monochromatic light from a distant source is incident on a slit 0.75 mm wide. on screen 2 m away, the distance from the central

hjlf

Displacement from the center line for minimum intensity is 1.35 mm , width of the slit is 0.75 so Wavelength of the light is 506.25.

<h3>How to find Wavelength of the light?</h3>

When a wave is bent by an obstruction whose dimensions are similar to the wavelength, diffraction is observed. We can disregard the effects of extremes because the Fraunhofer diffraction is the most straightforward scenario and the obstacle is a long, narrow slit.

This is a straightforward situation in which we can apply the

Fraunhofer single slit diffraction equation:

y = mλD/a

Where:

y = Displacement from the center line for minimum intensity = 1.35 mm

λ = wavelength of the light.

D = distance

a = width of the slit = 0.75

m = order number = 1

Solving for λ

λ = y + a/ mD

Changing the information that the issue has provided:

λ = 1.35 * 10^-3 + 0.75 * 10^-3 / 1*2

=5.0625 *10^-7 = 506.25

so

Wavelength of the light 506.25.

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5 0

1 year ago

****PLEASE HELP**** THERE ARE TWO QUESTIONS (ITS EASY)

notka56 [123]

but I think it's a and f

and

b and e

3 0

3 years ago

In thermodynamics, a closed system is a system where _____

Nady [450]

In a closed system, energy in form of heat (work) can be exchanged but not matter.

The answer to your question is C.

Hope it helped!

3 0

3 years ago

In young’s double slit experiment, the measured fringe width is 0.5 mm for a Sodium light of 589 nm at a distance of 1.5 m. A br

Levart [38]

Answer:

(A). The order of the bright fringe is 6.

(B). The width of the bright fringe is 3.33 μm.

Explanation:

Given that,

Fringe width d = 0.5 mm

Wavelength = 589 nm

Distance of screen and slit D = 1.5 m

Distance of bright fringe y = 1 cm

(A) We need to calculate the order of the bright fringe

Using formula of wavelength

$\lambda=\dfrac{dy}{mD}$

$m=\dfrac{d y}{\lambda D}$

Put the value into the formula

$m=\dfrac{1\times10^{-2}\times0.5\times10^{-3}}{589\times10^{-9}\times1.5}$

$m=5.65 = 6$

(B). We need to calculate the width of the bright fringe

Using formula of width of fringe

$\beta=\dfrac{yd}{D}$

Put the value in to the formula

$\beta=\dfrac{1\times10^{-2}\times0.5\times10^{-3}}{1.5}$

$\beta=3.33\times10^{-6}\ m$

$\beta=3.33\ \mu m$

Hence, (A). The order of the bright fringe is 6.

(B). The width of the bright fringe is 3.33 μm.

3 0

3 years ago

A tug boat pulls a ship with a constant net horizontal force of 5.00•10*3 N and causes the ship to move through a harbor. How mu

Murljashka [212]

The work done on the ship is $1.5\cdot 10^7 J$

Explanation:

The work done by a force on an object is given by:

$W=Fd cos \theta$

where

F is the magnitude of the force

d is the displacement

$\theta$ is the angle between the direction of the force and of the displacement

In this problem, we have:

$F=5.00\cdot 10^3 N$ (force acting on the ship)

d = 3.00 km = 3000 m (displacement of the ship)

$\theta=0^{\circ}$ (because the force is horizontal, and the displacement is horizontal as well)

Therefore, the work done on the ship is

$W=(5.00\cdot 10^3)(3000)(cos 0^{\circ})=1.5\cdot 10^7 J$

Learn more about work:

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8 0

3 years ago