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murzikaleks [220]
3 years ago
12

Monochromatic light with a wavelength of 384 nm passes through a single slit and falls on a screen 86 cm away. If the distance o

f the first-order dark band is 0.22 cm from the center of the pattern, what is the width of the slit?
Answer in units of cm.
Physics
1 answer:
valkas [14]3 years ago
3 0
This is a Fraunhofer single slit experiment, where the light passing through the slit produces an interference pattern on the screen, and where the dark bands (minima of diffraction) are located at a distance of
y= \frac{m\lambda D}{a}
from the center of the pattern. In the formula, m is the order of the minimum, \lambda the wavelenght, D the distance of the screen from the slit and a the width of the slit.

In our problem, the distance of the first-order band (m=1) is y=0.22 cm. The distance of the screen is D=86 cm while the wavelength is \lambda = 384 nm=384 \cdot 10^{-7}cm. Using these data and re-arranging the formula, we can find a, the width of the slit:
a= \frac{m \lambda D}{y}= \frac{1 \cdot 384 \cdot 10^{-7}cm \cdot 86 cm}{0.22 cm}=0.015 cm
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faust18 [17]

Complete Question:

In the same configuration of the previous problem 3, four long straight wires are perpendicular to the page, and their cross sections form a square of edge length a = 13.5 cm. Each wire carries 7.50 A, and the currents are out of the page in wires 1 and 4 and into the page in wires 2 and 3.

a) Draw a diagram in a (x,y) plane of the four wires with wire 4 perpendicular to the origin. Indicate the current's directions.

b) Draw a diagram of all magnetic fields produced at the position of wire 3 by the other three currents.

c) Draw a diagram of all magnetic forces produced at the position of wire 3 by the other three currents.

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Answer:

force, 1.318 ₓ 10⁻⁴

direction, 18.435°

Explanation:

The attached file gives a breakdown step by step solution to the questions

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6 0
3 years ago
Master of physics needed
Delicious77 [7]
Hey JayDilla, I get 1/3.  Here's how:
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E_{linear}= \frac{1}{2}mv^2
where
v=r \omega
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The rotational part requires the moment of inertia of a solid cylinder
I_{cyl} =  \frac{1}{2}mr^2
Then the rotational kinetic energy is
E_{rot}= \frac{1}{2}I \omega ^2= \frac{1}{4}mr^2 \omega ^2
Adding the two types of energy and factoring out common terms gives
\frac{1}{2}mr^2 \omega ^2(1+ \frac{1}{2})
Here the "1" in the parenthesis is due to linear motion and the "1/2" is due to the rotational part.  Since this gives a total of 3/2 altogether, and the rotational part is due to a third of this (1/2), I say it's 1/3.

8 0
3 years ago
What resistance must be connected in parallel with a 633-Ω resistor to produce an equivalent resistance of 205 Ω?
alukav5142 [94]

Answer:

303 Ω

Explanation:

Given

Represent the resistors with R1, R2 and RT

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RT = 205

Required

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Since it's a parallel connection, it can be solved using.

1/Rt = 1/R1 + 1/R2

Substitute values for R1 and RT

1/205 = 1/633 + 1/R2

Collect Like Terms

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Take LCM

1/R2 = (633 - 205)/(205 * 633)

1/R2 = 428/129765

Take reciprocal of both sides

R2 = 129765/428

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5 0
3 years ago
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Marrrta [24]

Answer:

ax = 6.43m/s²

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So the first step is to calculate the time at which the velocity =12.0m/s and with this time calculate the acceleration. Detailed solution can be found in the attachment below.

7 0
3 years ago
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