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fiasKO [112]
3 years ago
9

Light from a helium-neon laser (λ=633nm) passes through a circular aperture and is observed on a screen 5.80 mm behind the apert

ure. The width of the central maximum is 1.70 cm.What is the diameter (in mm) of the hole?
Physics
1 answer:
Helga [31]3 years ago
7 0

Explanation:

Whenever the light passes through hole or slit then it tends to bend that is actually a diffraction. It will then made the interference pattern of light and dark bands that due to constructive and destructive interference.

So by using the equation of diffraction,

dsinA = nL

SinA is a geometric component it can be written as,

d\frac{y}{x} = nL

∵ x is the distance from screen.

∵ y is the half of the width of central maximum.

now by putting the values in mm,

d\frac{17}{5.80} = 0.000633

d = 2.15×10^{-4} mm

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valentinak56 [21]

Answer:

A dump truck going 70 mph

Explanation:

hope it helps you

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Suppose a thin conducting wire connects two conducting spheres. A negatively charged rod is brought near one of the spheres, the
dangina [55]

Answer:

a. The spheres will attract each other.

Explanation:

When two conducting spheres are connected by a conducting wire and a negatively charged rod is brought near it then this will induce opposite (positive) charge at the nearest point on the sphere and by the conservation of charges there will also be equal amount of negative charge on the farthest end of this conducting system this is called induced polarization.

  • When the conducting wire which joins them is cut while the charged rod is still in proximity to of one of the metallic sphere then there will be physical separation of the two equal and unlike charges on the spheres which will not get any path to flow back and neutralize.
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4 0
3 years ago
What is the amplitude of a wave related to
Vanyuwa [196]

Answer:

intensity because square of the amplitude is proportional to the intensity of the wave

Explanation:

3 0
3 years ago
A 0.300 kg ball, moving with a speed of 2.5 m/s, has a head-on collision with at 0.600 kg ball initially at rest. Assuming a per
FrozenT [24]

Answer:

1.25 m/s

Explanation:

Given,

Mass of first ball=0.3 kg

Its speed before collision=2.5 m/s

Its speed after collision=2 m/s

Mass of second ball=0.6 kg

Momentum of 1st ball=mass of the ball*velocity

=0.3kg*2.5m/s

=0.75 kg m/s

Momentum of 2nd ball=mass of the ball*velocity

=0.6 kg*velocity of 2nd ball

Since the first ball undergoes head on collision with the second ball,

momentum of first ball=momentum of second ball

0.75 kg m/s=0.6 kg*velocity of 2nd ball

Velocity of 2nd ball=0.75 kg m/s ÷ 0.6 kg

=1.25 m/s

4 0
3 years ago
To test the performance of its tires, a car
velikii [3]

<u>Answer</u>:

The coefficient of  static friction between the tires and the road is 1.987

<u>Explanation</u>:

<u>Given</u>:

Radius of the track, r =  516 m

Tangential Acceleration a_r=  3.89 m/s^2

Speed,v =  32.8 m/s

<u>To Find:</u>

The coefficient of  static friction between the tires and the road = ?

<u>Solution</u>:

The radial Acceleration is given by,

a_{R = \frac{v^2}{r}

a_{R = \frac{(32.8)^2}{516}

a_{R = \frac{(1075.84)}{516}

a_{R = 2.085 m/s^2

Now the total acceleration is

\text{ total acceleration} = \sqrt{\text{(tangential acceleration)}^2 +{\text{(Radial acceleration)}^2

=>= \sqrt{ (a_r)^2+(a_R)^2}

=>\sqrt{ (3.89 )^2+( 2.085)^2}

=>\sqrt{ (15.1321)+(4.347)^2}

=>19.4791 m/s^2

The frictional force on the car will be f = ma------------(1)

And the force due to gravity is W = mg--------------------(2)

Now the coefficient of  static friction is

\mu =\frac{f}{W}

From (1) and (2)

\mu =\frac{ma}{mg}

\mu =\frac{a}{g}

Substituting the values, we get

\mu =\frac{19.4791}{9.8}

\mu =1.987

8 0
3 years ago
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