Where did the gold of the Inca civilization come from?

Light of wavelength 648.0 nm is incident on a narrow slit. The diffraction pattern is viewed on a screen 57.5 cm from the slit.

boyakko [2]

Answer:

$71.0 \mu m$

Explanation:

The formula for the single-slit diffraction is

$y=\frac{n\lambda D}{d}$

where

y is the distance of the n-minimum from the centre of the diffraction pattern

D is the distance of the screen from the slit

d is the width of the slit

$\lambda$ is the wavelength of the light

In this problem,

$\lambda=648.0 nm=6.48\cdot 10^{-7}m$

$D=57.5 cm=0.575 m$

$y=1.05 cm=0.0105 m$ , with n=2 (this is the distance of the 2nd-order minimum from the central maximum)

Solving the formula for d, we find:

$d=\frac{n\lambda D}{y}=\frac{2(6.48\cdot 10^{-7} m)(0.575 m)}{0.0105 m}=7.10\cdot 10^{-5} m= 71.0 \mu m$

6 0

3 years ago

The relationship between total energy and kinetic energy and potential energy is

horsena [70]

Answer: Potential energy is energy stored in an object due to its position or arrangement. Kinetic energy is the energy of an object due to its movement its motion. Potential energy can be converted into kinetic energy, and kinetic energy can be converted into potential energy.

7 0

4 years ago

In an LC circuit at one time the charge stored by the capacitor is 10 mC and the current is 3.0 A. If the frequency of the circu

Ronch [10]

Answer:

$i_2=3.61\ A$

Explanation:

<u>LC Circuit</u>

It's a special circuit made of three basic elements: The AC source, a capacitor, and an inductor. The charge, current, and voltage are oscillating when there is an interaction between the electric and magnetic fields of the elements. The following variables will be used for the formulas:

$q, q_1, q_2$ = charge of the capacitor in any time $t, t_1, t_2$

$q_o$ = initial charge of the capacitor

$\omega$ =angular frequency of the circuit

$i, i_1, i_2$ = current through the circuit in any time $t, t_1, t_2$

The charge in an LC circuit is given by

$q(t) = q_0 \, cos (\omega t )$

The current is the derivative of the charge

$\displaystyle i(t) = \frac{dq(t)}{dt} = - \omega q_0 \, sin(\omega t).$

We are given

$q_1=10\ mc=0.01\ c, i_1=3\ A,\ q_2=6\ mc=0.006\ c\ ,\ f=\frac{1000}{4\pi}$

It means that

$q(t_1) = q_0 \, cos (\omega t_1 )=q_1\ .......[eq 1]$

$i(t_1) = - \omega q_0 \, sin(\omega t_1)=i_1.........[eq 2]$

From eq 1:

$\displaystyle cos (\omega t_1 )=\frac{q_1}{q_0}$

From eq 2:

$\displaystyle sin(\omega t_1)=-\frac{i_1}{\omega q_0}$

Squaring and adding the last two equations, and knowing that

$sin^2x+cos^2x=1$

$\displaystyle \left ( \frac{q_1}{q_0} \right )^2+\left ( \frac{i_1}{\omega q_0} \right )^2=1$

Operating

$\displaystyle \omega^2q_1^2+i_1^2=\omega^2q_o^2$

Solving for $q_o$

$\displaystyle q_o=\frac{\sqrt{\omega^2q_1^2+i_1^2}}{\omega}$

Now we know the value of $q_0$ , we repeat the procedure of eq 1 and eq 2, but now at the second time $t_2$ , and solve for $i_2$

$\displaystyle \omega^2q_2^2+i_2^2=\omega^2q_o^2$

Solving for $i_2$

$\displaystyle i_2=w\sqrt{q_o^2-q_2^2}$

Now we replace the given values. We'll assume that the placeholder is a pi for the frequency, i.e.

$\displaystyle f=\frac{1}{4\pi}\ KHz$

$w=2\pi f=500\ rad/s$

$\displaystyle q_o=\frac{\sqrt{(500)^2(0.01)^2+3^2}}{500}$

$q_0=0.01166\ c$

Finally

$\displaystyle i_2=500\sqrt{0.01166^2-.006^2}$

$i_2=5\ A$

3 0

3 years ago

What role does gravity play in the universe

Mars2501 [29]

Answer:

Explanation:Gravity is what holds the planets in orbit around the sun and what keeps the moon in orbit around Earth. The gravitational pull of the moon pulls the seas towards it, causing the ocean tides. Gravity creates stars and planets by pulling together the material from which they are made.

6 0

3 years ago

Ben and Dan both weigh 600 N. They are doing pull-ups together. Each pull-up is 0.5 meters of distance. How much work do they ea

AlekseyPX

300J

Explanation:

Given parameters:

Weight of Dan = 600N

Weight of Ben = 600N

Pull-up Distance = 0.5m

Unknown:

Work done by Dan =?

Work done by Ben = ?

Solution:

Work done is the force applied to move a body through a distance in the direction of the force.

Work done = force x distance

Work done by Ben =600 x 0.5 = 300J

Work done by Dan = 600x 0.5 = 300J

learn more:

Work done brainly.com/question/9100769

#learnwithBrainly

6 0

3 years ago