Which of the following is a technique for the effective communication? <br>

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.

To learn more about Wavelength of the light refer to:

brainly.com/question/15413360

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5 0

1 year ago

Derive the equation of motion of the block of mass m1 in terms of its displacement x. The friction between the block and the sur

Alenkasestr [34]

Answer:

the equivalent mass : $m_e = m_1+m_2+\frac{I}{R^2}$

the equation of the motion of the block of mass $m_1$ in terms of its displacement is = $(m_1+m_2+\frac{I}{R^2} )(\bar x) = (m_2gsin \phi) -(m_1gsin \beta)$

Explanation:

Let use m₁ to represent the mass of the block and m₂ to represent the mass of the cylinder

The radius of the cylinder be = R

The distance between the center of the pulley to center of the block to be = x

Also, the angles of inclinations of the cylinder and the block with respect to the ground to be $\phi$ and $\beta$ respectively.

The velocity of the block to be = v

The equivalent mass of the system = $m_e$

In the terms of the equivalent mass, the kinetic energy of the system can be written as:

$K.E = \frac{1}{2} m_ev^2$ --------------- equation (1)

The angular velocity of the cylinder = $\omega$ : &

The inertia of the cylinder about its center to be = I

The angular velocity of the cylinder can be written as:

$v = \omega R$

$\omega =\frac{v}{R}$

The kinetic energy of the system in terms of individual mass can be written as:

$K.E = \frac{1}{2}m_1v^2+\frac{1}{2} m_2v^2+\frac{1}{2}I\omega^2$

By replacing $\omega$ with $\frac{v}{R}$ ; we have:

$K.E = \frac{1}{2}m_1v^2+\frac{1}{2} m_2v^2+\frac{1}{2}I(\frac{v}{R})^2$

$K.E = \frac{1}{2}(m_1+ m_2+ \frac{I}{R} )v^2$ ------------------ equation (2)

Equating both equation (1) and (2); we have:

$m_e = m_1+m_2+\frac{I}{R^2}$

Therefore, the equivalent mass : $m_e = m_1+m_2+\frac{I}{R^2}$ which is read as;

The equivalent mass is equal to the mass of the block plus the mass of the cylinder plus the inertia by the square of the radius.

The expression for the force acting on equivalent mass due to the block is as follows:

$f_{block }=m_1gsin \beta$

Also; The expression for the force acting on equivalent mass due to the cylinder is as follows:

$f_{cylinder} = m_2gsin \phi$

Equating the above both equations; we have the equation of motion of the equivalent system to be

$m_e \bar x = f_{cylinder}-f_{block}$

which can be written as follows from the previous derivations

$(m_1+m_2+\frac{I}{R^2} )(\bar x) = (m_2gsin \phi) -(m_1gsin \beta)$

Finally; the equation of the motion of the block of mass $m_1$ in terms of its displacement is = $(m_1+m_2+\frac{I}{R^2} )(\bar x) = (m_2gsin \phi) -(m_1gsin \beta)$

8 0

2 years ago

I need help plzzzz!!!!!!!!!!!!!!!!

elena-s [515]

Answer:

You were a freeloader of my questions, so I'll be one too.

4 0

2 years ago

What is the net force acting upon this object? *

AnnZ [28]

2 N Right

.. .. .. .. .. .. .. ..

4 0

3 years ago

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Two planets A and B, where B has twice the mass of A, orbit the Sun in elliptical orbits. The semi-major axis of the elliptical

lozanna [386]

Answer:

2.83

Explanation:

Kepler's discovered that the square of the orbital period of a planet is proportional to the cube of the semi-major axis of its orbit, that is called Kepler's third law of planet motion and can be expressed as:

$T=\frac{2\pi a^{\frac{3}{2}}}{\sqrt{GM}}$ (1)

with T the orbital period, M the mass of the sun, G the Cavendish constant and a the semi major axis of the elliptical orbit of the planet. By (1) we can see that orbital period is independent of the mass of the planet and depends of the semi major axis, rearranging (1):

$\frac{T}{a^{\frac{3}{2}}}=\frac{2\pi}{\sqrt{GM}}$

$\frac{T^{2}}{a^{3}}=(\frac{2\pi }{\sqrt{GM}})^2$ (2)

Because in the right side of the equation (2) we have only constant quantities, that implies the ratio $\frac{T^{2}}{a^{3}}$ is constant for all the planets orbiting the same sun, so we can said that:

$\frac{T_{A}^{2}}{a_{A}^{3}}=\frac{T_{B}^{2}}{a_{B}^{3}}$

$\frac{T_{B}^{2}}{T_{A}^{2}}=\frac{a_{B}^{3}}{a_{A}^{3}}$

$\frac{T_{B}}{T_{A}}=\sqrt{\frac{a_{B}^{3}}{a_{A}^{3}}}=\sqrt{\frac{(2a_{A})^{3}}{a_{A}^{3}}}$

$\frac{T_{B}}{T_{A}}=\sqrt{\frac{2^3}{1}}=2.83$

6 0

3 years ago

Read 2 more answers

Which of the following is a technique for the effective communication? ​

Which of the following is a technique for the effective communication?