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AlladinOne [14]
4 years ago
15

Monochromatic light from a point source is directed toward the sharp edge of a solid object. How does diffraction change the ima

ge that appears on a screen beyond the object?
Why are diffraction patterns not commonly observed in everyday situations?
Physics
2 answers:
Anna71 [15]4 years ago
6 0

Did you get the answers the quiz??? If so please write out the answers in the comment section :)

Otrada [13]4 years ago
5 0

Monochromatic light from a point source is directed toward the sharp edge of a solid object.

  • A spot will be seen on the screen after diffraction.
  • The size of the obstructing object is greater than the wavelength of the incident light in everyday situations, so we are not able to view diffraction patterns in everyday situations.

<u>Explanation:</u>

When a monochromatic light from a point source is directed toward the sharp edge of a solid object, the light beam will get diffracted from the edge as the body of the solid object will not be passing the light. So the incident light will occur as a spots on the screen beyond the object after diffraction.

Diffraction of light occurs when the incident light falls on a grater and the width of the grater or the obstructing object should be comparable or lesser to the size of the wavelength. So, in everyday situations we do not observe diffraction patterns as the obstructing objects in real world.

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If anyone can answer it'd be very appreciated, i need to pass this class
alexandr1967 [171]

Answer: 0.4

Explanation:

8 0
3 years ago
Read 2 more answers
A 0.31 kg cart on a horizontal frictionless track is attached to a string. The string passes over a disk-shaped pulley of mass 0
Stella [2.4K]

To solve this problem it is necessary to apply the concepts related to Newton's second law and its derived expressions for angular and linear movements.

Our values are given by,

M_{cart} = 0.31kg\\m_{pulley} = 0.08kg\\r_{pulley} = 0.012m\\F = 1.1N\\

If we carry out summation of Torques on the pulley we will have to,

F_2*d-F_1*d = I \alpha

Where,

I = Inertia moment

\alpha =Angular acceleration, which is equal in linear terms to a/r (acceleration and radius)

The moment of inertia for this object is given as

I = \frac{1}{2} mr^2

Replacing this equations we have know that

(F_2 - F_1)d = (\frac{1}{2}(m_{pulley})r^2) (\frac{a}{r})

F_2 - F_1 = \frac{1}{2}m_{pulley} \frac{F_1}{M_{cart}}

F_2 = (1+\frac{1}{2}(\frac{m_{pulley}}{M_{cart}}))F_1

Or

F_1 = \frac{F_2}{(1+\frac{1}{2}(\frac{m_{pulley}}{M_{cart}}))}

Replacing our values we have that

F_1 = \frac{1.1}{(1 + (0.5)(\frac{0.08}{0.31}))}

F_1 = 0.974 N

Therefore the tension in the string between the pulley and the cart is  0.974 N

6 0
4 years ago
all bearings are made by lebng spherical drops of molten metal fall inside a tall tower – called a shot tower – and solidify as
Wewaii [24]

Answer:

Part b)

h = 78.5 m

Part c)

v = 39.24 m/s

Explanation:

Part b)

If ball need t = 0 to t = 4 s then height of the tower is the total displacement of the ball in t = 4 s interval

here if ball start from rest

then its displacement is given as

\Delta y = \frac{1}{2}gt^2

\Delta y = \frac{1}{2}(9.81)(4^2)

\Delta y = 78.5 m

Part c)

Speed of the bearing at the end of the motion of the ball

v_f = v_i + at

v_f = 0 + (9.81)(4)

v_f = 39.24 m/s

7 0
3 years ago
The sun's radiation has different wavelengths. the shorter is the radiation's wavelength, the weaker is the radiation's energy.
NeX [460]
It is false. Because the amount of energy carried in the wave is inversely related to the length of the waves wavelength. To correct the statement it should be that the shorter the radiation's wavelength the stronger is the radiation's energy.
7 0
3 years ago
A 0.550 kg air-track glider is attached to each end of the track by two coil springs. It takes a horizontal force of 0.500 N to
Inessa [10]

(1) The effective spring constant of the system is 7.14 N/m.

(2) The maximum x-acceleration of the glider is 0.9 m/s².

(3) The x-coordinate of the glider at time t= 0.650T is 0.28 m.

(4) The kinetic energy of the glider at x=0.00 m is 0.0175 J.

<h3>The effective spring constant of the system</h3>

The effective spring constant of the system is calculated as follows;

F = kx

where;

  • k is spring constant

k = F/x

k = 0.5/0.07

k = 7.14 N/m

<h3>Maximum acceleration of the glider</h3>

a = ω²x

where;

  • ω is angular speed

ω = √k/m

ω = √(7.14/0.55)

ω = 3.6 rad/s

a =  (3.6)² x 0.07

a = 0.9 m/s²

<h3>Period of the oscillation</h3>

T = 2πx/v

T = 2πx/(ωx)

T = 2π/ω

T = 2π/(3.6)

T = 1.75 seconds

t = 0.65T

t = 0.65 x 1.75

t = 1.14 seconds

x = vt

x = (ωx)t

x = (3.6 x 0.07) x 1.14

x = 0.28 m

<h3>kinetic energy of the glider</h3>

apply the principle of conservation of energy

Kinetic energy = work done by the spring

Kinetic energy = average force x distance

Kinetic energy = ¹/₂(0.5 N) x 0.07 m

Kinetic energy = 0.0175 J

Learn more about kinetic energy here: brainly.com/question/25959744

#SPJ1

The complete question is below:

A 0.550 kg air-track glider is attached to each end of the track by two coil springs. It takes a horizontal force of 0.500 N to displace the glider to a new equilibrium position, x= 0.070 m.

1. Find the effective spring constant of the system.

2. The glider is now released from rest at x= 0.070 m. Find the maximum x-acceleration of the glider.

3. Find the x-coordinate of the glider at time t= 0.650T, where T is the period of the oscillation.

4. Find the kinetic energy of the glider at x=0.00 m.

3 0
1 year ago
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