Please help me with this

A single-slit diffraction pattern is formed by monochromatic electromagnetic radiation from a distant source passing through a s

tatiyna

Answer:

a. λ = 647.2 nm

b. I₀ 9.36 x 10⁻⁵

Explanation:

Given:

β = 56.0 rad , θ = 3.09 ° , γ = 0.170 mm = 0.170 x 10⁻³ m

a.

The wavelength of the radiation can be find using

β = 2 π / γ * sin θ

λ = [ 2π * γ * sin θ ] / β

λ = [ 2π * 0.107 x 10⁻³m * sin (3.09°) ] / 56.0 rad

λ = 647.14 x 10⁻⁹ m ⇒ λ = 647.2 nm

b.

The intensity of the central maximum I₀

I = I₀ (4 / β² ) * sin ( β / 2)²

I = I₀ (4 / 56.0²) * [ sin (56.0 /2) ]²

I = I₀ 9.36 x 10⁻⁵

8 0

4 years ago

What is a simple machine? A. any object that makes work easier when a single force is applied by changing the strength or the di

Otrada [13]

The best answer is a

5 0

3 years ago

Read 2 more answers

A bicycle tire is spinning counterclockwise at 2.70 rad/s. During a time period Δt = 1.50 s, the tire is stopped and spun in the

Andru [333]

Answer:

Δω = -5.4 rad/s

αav = -3.6 rad/s²

Explanation:

Given:

Initial angular velocity = ωi = 2.70 rad/s

Final angular velocity = ωf = -2.70 rad/s (negative sign is

due to the movement in opposite direction)

Change in time period = Δt = 1.50 s

Required:

Change in angular velocity = Δω = ?

Average angular acceleration = αav = ?

Solution:

Angular velocity (Δω):

Δω = ωf - ωi

Δω = -2.70 - 2.70

Δω = -5.4 rad/s.

Average angular acceleration (αav):

αav = Δω/Δt

αav = -5.4/1.50

αav = -3.6 rad/s²

Since, the angular velocity is decreasing from 2.70 rad/s (in counter clockwise direction) to rest and then to -2.70 rad/s (in clockwise direction) so, the change in angular velocity is negative.

7 0

3 years ago

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Describe the ratio of kinetic energy and potential energy the skateboarder will have at each of the following points in her ride

amm1812

Answer:

no answer

Explanation:

7 0

3 years ago

Two particles with charges of 2nC and 5nC are separated by a distance of 3m. The charge 2nC is placed on the left. Find the forc

Vanyuwa [196]

Answer:

$1\cdot 10^{-8} N$ to the left

Explanation:

The magnitude of the electrostatic force between two charges is given by the following equation:

$F=k\frac{q_1 q_2}{r^2}$

where:

$k=9\cdot 10^9 Nm^{-2}C^{-2}$ is the Coulomb's constant

$q_1, q_2$ are the magnitude of the two charges

r is the distance between the two charges

Moreover, the force is:

- Attractive if the charges have opposite sign

- Repulsive if the charges have same sign

In this problem, we have:

$q_1=2nC=2\cdot 10^{-9}C$ is the magnitude of charge 1

$q_2=5nC =5\cdot 10^{-9}C$ is the magnitude of charge 2

r = 3 m is the distance between the two charges

Substituting, we find the force on both charges:

$F=\frac{(9\cdot 10^9)(5\cdot 10^{-9})(2\cdot 10^{-9}}{3^2}=1\cdot 10^{-8} N$

Here, the two charges are both positive, so the force is repulsive; since the 2 nC charge is on the left, this means that the force on this charge is to the left (away from the 5 nC charge).

5 0

3 years ago