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Andre45 [30]
2 years 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:
umka21 [38]2 years ago
8 0

Answer:

w= 1.867\times10^{-2}

Explanation:

Given:

fibers diameter d= 18 μm

distance of the screen D= 30 cm

wavelength λ= 560 nm

we know that fringe width

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

putting values we get

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

w= 1.867\times10^{-2}

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An electric ceiling fan is rotating about a fixed axis with an initial angular velocity magnitude of 0.240 rev/s. The magnitude
bazaltina [42]

Explanation:

Given that,

Angular velocity = 0.240 rev/s

Angular acceleration = 0.917 rev/s²

Diameter = 0.720 m

(a). We need to calculate the angular velocity after time 0.203 s

Using equation of angular motion

\omega_{f}=\omega_{i}+\alpha t

Put the value in the equation

\omega_{f}=0.240+0.917\times0.203

\omega_{f}=0.426\ rev/s

The angular velocity is 0.426 rev/s.

(b). We need to calculate the tangential speed of the blade

Using formula of  tangential speed

v= r\omega

Put the value into the formula

v = \dfrac{0.720 }{2}\times0.426\times2\pi

v=0.963\ m/s

The tangential speed of the blade is 0.963 m/s.

(c). We need to calculate the magnitude at of the tangential acceleration

Using formula of tangential acceleration

a_{t}=r\alpha

Put the value into the formula

a_{t}=0.36\times0.917\times2\pi

a_{c}=2.074\ m/s^2

The tangential acceleration is 2.074 m/s².

Hence, This is required solution.

4 0
2 years ago
3. What conclusion can you make about the electric field strength between two parallel plates? Explain your answer referencing P
KIM [24]

Answer:

From the relation above we can conclude that the  as the distance between the two plate increases the electric field strength decreases

Explanation:

I cannot  find any attached photo, but we can proceed anyways theoretically.

The electric field strength (E) at any point in an electric field is the force experienced by a unit positive charge (Q) at that point

i.e

E=\frac{F}{Q}

But the force F

F= \frac{kQ1Q2}{r^2}

But the electric field intensity due to a point charge Q at a distance r meters away is given by

E= \frac{\frac{kQ1Q2}{r^2}}{Q} \\\\\E= \frac{Q1}{4\pi er^2 }

<em>From the relation above we can conclude that the  as the distance between the two plate increases the electric field strength decreases</em>

6 0
2 years ago
A galloping pony speeds past you at 5 m/s. The frequency of the sound produced by the hooves on the dirt is 221 Hz. Assume the s
joja [24]
Given:
speed of passing pony 5 m/s
frequency of the sound produced: 221 Hz
speed of sound 342 m/s

Let us use the Doppler Shift Formula:
Where the <span>source is moving away from the observer at rest
</span>
f' = (v / v+vs) f
Where, vs<span> = Velocity of the Source,</span>
           v = Velocity of sound or light in medium,
           f = Real frequency,
           f' = Apparent frequency.

f'= [342 m/s / (342 m/s+5m/s)] * 221 Hz
f' = 0.9856 * 221Hz
f' = 217.8176 Hz or 218 Hz

The observed frequency <span>of the hooves after the pony has passed your position is 218 Hz.</span>
7 0
2 years ago
Read 2 more answers
A package is dropped from a helicopter moving upward at 1.5 m/s. If it takes 16.0 s before the package strikes the ground, how h
pashok25 [27]
Well we know acceleration from free fall due to gravity is 9.8m/s^2

Lay out

S = displacement is what we need

U

V = 1.5m/s

A = 9.8m/s2

T = 16.0s

Use the equation s=vt-1/2at^2

Where a = acceleration t= time and v= velocity

Sub in the values to get displacement or height from ground

= -1230.4 metres which would be positive as you’re measuring distance (scalar quantity) so it’s 1230.4 metres
8 0
2 years ago
A circular loop of wire with radius r=0.0250 m and resistance r=0.390 ohms is in a region of spatially uniform magnetic field. t
slavikrds [6]

Answer:

0.0133 A

Explanation:

The time at which B=1.33 T is given by  

1.33 = 0.38*t^3  

t = (1.33/0.38)^(1/3) = 1.52 s  

Using Faraday's Law, we have  

emf = - dΦ/dt = - A dB/dt = - A d/dt ( 0.380 t^3 )  

Area A = pi * r² = 3.141 *(0.025 *0.025) = 0.00196 m²

emf = - A*(3*0.38)*t^2  

thus, the emf at t=1.52 s is  

emf = - 0.00196*(3*0.38)*(1.52)^2 = -0.0052 V  

if the resistance is 0.390 ohms, then the current is given by  

I = V/R = 0.0052/0.390 = 0.0133 A

3 0
2 years ago
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