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Rashid
3 months ago
15

Early investigators (including Thomas Young) measured the thickness of wool fibers using diffraction. One early instrument used

a collimated beam of 560 nm light to produce a diffraction pattern on a screen placed 30 cm from a single wool fiber.
part a: If the fiber's diameter was 18 μm, what was the width of the central maximum on the screen?
Physics
1 answer:
ValentinkaMS [3.4K]3 months ago
8 0

Answer:

w= 1.867\times10^{-2}

Explanation:

Provided:

fiber diameter d= 18 μm

screen distance D= 30 cm

wavelength λ= 560 nm

from this, we can determine the fringe width

w=\frac{2\lambda D}{d}

substituting the values yield

w=\frac{2\times560\times10^{-9}\times0.3}{18\times10^{-6}

w= 1.867\times10^{-2}

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A resistor with resistance R and an air-gap capacitor of capacitance C are connected in series to a battery (whose strength is "
kicyunya [3294]

Answer:

a) Q = C*emf

b)  Decrease in electric field strength and electric potential

c) Initial current through the resistor = emf/R

d) The final charge = K*C*emf

Explanation:

a) The resistors and capacitors are linked in series with the battery

According to Kirchoff's voltage law, the total voltage in the circuit must equal zero

Let V_{R}represent the Voltage across the Resistor

V_{c}and

represent the Voltage across the capacitor

Implementing KVL;

emf - V_{R} - V_{c} = 0\\

.........................(1)

Since this is a series connection, the same current traverses through the circuit

V_{R} = IR\\Q = CV_{c} \\V_{c} = Q/C

Integrating V_{c}and V_{R}into equation (1)

emf - IR - Q/C = 0

Initially, as the capacitor reaches full charge, the current will drop to zero because of equilibrium

I = 0A\\emf = Q/C\\Q = C* emf

b) When the plastic sheet is inserted between the plates, current begins to flow because both the electric field intensity and electric potential decrease. As a result, charge diminishes, leading to current flow

c) The current through the resistor equates to the total current within the circuit (given the series connection)

I = I_{o} \exp(\frac{-t}{RC} )\\At time the initial time, t\\t = 0\\ I_{o} = \frac{emf}{R} \\

Substituting the values of t and I₀ into the aforementioned formula for I

I = \frac{emf}{R} \exp(0)\\I = \frac{emf}{R}

d) Note: The initial charge on the capacitor equals C * emf

Following the insertion of the plastic, the new charge will be:

Q = K* Q_{initial} \\Q_{initial} = C *emf\\Q_{final} = KCemf

4 0
2 months ago
A transverse wave on a string has an amplitude A. A tiny spot on the string is colored red. As one cycle of the wave passes by,
Sav [3153]
Since it's classified as a transverse wave, the particle on the string moves horizontally as the wave progresses, without actual forward or backward travel. Consequently, the red dot shifts 'A' to the left, returns 'A' to the center, moves 'A' to the right, and goes back 'A' to the center once again. Thus, the red dot collectively travels a distance totaling 4A.
6 0
2 months ago
A solid conducting sphere carrying charge q has radius a. It is inside a concentric hollow conducting sphere with inner radius b
Softa [3030]

Response:

Clarification:

Refer to the diagram indicating the charges on the specified sphere (see attachment).

The electric field at the stated positions is

E(r) = 0 for r≤a.  Equation 1

E(r) = kq/r² for a<r<b.   Equation 2

E(r) = 0 for b<r<c.      Equation 3

E(r) = kq/r² for r>c.    Equation 4.

We understand that electric potential correlates with the electric field through

V = Ed

A. To compute the potential at the outer surface of the hollow sphere (r=c), we determine that the electric field there is

E = kQ / r²

Then,

V = Ed,

At d = r = c

Thus,

Vc = (kQ / c²) × c

Vc = kQ / c

As a result, the total charge Q consists of +q, -q, and +q

Hence, Q = q - q + q = q

V = kq / c

B. To calculate the potential at the inner surface of the hollow sphere (r=b), we have

V = kQ/r

V = kQ / b,   noting that r = b

So, Q = q

V = kq / b

C. At r = a

Following from equation 1:

E(r) = 0 for r≤a.  Equation 1

The electric field at the surface of the solid sphere is 0, E = 0N/C

Thus,

V = Ed = 0 V

Consequently, the electric potential at the solid sphere's surface is 0.

D. At r = 0

The electric potential can be determined by

V = kq / r

As r approaches 0,

V = kq / 0

V approaches infinity.

8 0
3 months ago
Ronnie kicks a playground ball with an initial velocity of 16 m/s at an angle of 40° relative to the ground. What is the approxi
Softa [3030]
The calculation for the horizontal component is performed as follows:
Vhorizontal = V · cos(angle)

For your instance, Vhorizontal = 16 · cos(40) equates to 12.3 m/s

Conclusion: 12.3 m/s
7 0
3 months ago
Read 2 more answers
A boy is whirling a stone around his head by means of a string. The string makes one complete revolution every second; and the m
Maru [3345]

Answer:

(A) The tension's magnitude grows to four times the initial value, 4F.

Explanation:

When an object travels in a circular path, a centripetal force is exerted upon it. In this instance, the centripetal force acting on the stone can be represented by \frac { m{ v }^{ 2 } }{ r }.

                   Here, m denotes the mass of the object

                               v is the velocity or speed of the object

                               r signifies the radius of the circular path

Importantly, the tension corresponds to the centripetal force.

Initially, the string completes one revolution each second, and subsequently, it accelerates to perform two revolutions in the same time frame. This signifies that the speed has increased twofold.

Applying our formula:F =\frac { m{ v }^{ 2 } }{ r }

                               where F indicates the tension in the string

assuming the starting speed is v, after doubling it becomes 2v

Maintaining the circle's radius, we arrive at:

F=\frac { m{ (2v) }^{ 2 } }{ r } =\frac { 4m{ v }^{ 2 } }{ r }

From this equation, it's clear that the initial tension has quadrupled.

Consequently, the magnitude of the tension increases to four times its original value, 4F.

3 0
2 months ago
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