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ehidna [41]
3 years ago
9

What principle is responsible for alternating light and dark bands when light passes through two or more narrow slits? What prin

ciple is responsible for alternating light and dark bands when light passes through two or more narrow slits? reflection refraction dispersion interference polarization
Physics
1 answer:
kipiarov [429]3 years ago
4 0

Answer:

Interference

Explanation:

When two waves of same frequency and constant phase difference super impose at a point on the screen then due to their superposition we will get different intensity of light at different positions of the screen

This phenomenon of redistribution of energy is known as interference of light.

So at the position of screen where the light intensity is maximum on the screen is known as constructive interference while the positions on the screen where it will get minimum intensity on the screen is known as destructive interference of the light

So correct answer would be

Interference

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g In a certain underdamped RLC circuit, the voltage across the capacitor decreases in one cycle from 5.0 V to 3.8 V. The period
NeTakaya

The question is incomplete. The complete question is :

In a certain underdamped RLC circuit, the voltage across the capacitor decreases in one cycle from 5.0 V to 3.8 V. The period of the oscillations is 1.2 microseconds (1.2*10^-6). What is Q?

Solution :

The underdamped RLC circuit

$v_{t} = ve^{-\frac{R}{2L}t} \cos \omega t$

$\omega = \sqrt{\frac{1}{LC}-\frac{R^2}{4L^2}}= \frac{2 \pi}{T}$

We know in one time period, v = 2v, at t = T, $v_t = 3.8 v$

so, $9.8 = 5 e^{-\frac{R}{2L}T} \cos \frac{2 \pi}{T}T$

   $e^{-\frac{R}{2L}T} = \frac{3.8}{5} \times 1$

   $\frac{R}{2L}T= \ln \frac{5}{3.8}$

  $\frac{R}{L}= \frac{2}{1.2 \times 10^{-6}} \ln \frac{5}{3.8}$

 $\frac{R}{L} = 457.3 \times 10^3$

Now, Q value $= \frac{1}{R}\sqrt{\frac{L}{C}}$

                     $=\sqrt{\frac{L}{R^2C}\times \frac{L}{L}}$

                     $=\sqrt{(\frac{L}{R})^2 \times \frac{1}{LC}}$

              $\frac{1}{LC}=27.43 \times 10^{12}$

∴ $Q=\sqrt{\left(\frac{1}{457.3 \times 10^3}\right)^2 \times 27.43 \times 10^{12}}$

  $Q=\sqrt{131.166}$

      = 11.45

4 0
2 years ago
The Kinetic energy, K, of an object with mass m moving with velocity v can be found using the formula - E_{\text{k}}={\tfrac {1}
tester [92]

Answer:

The ratio of kinetic energies of 5 kg object to 20 kg object is 1:1.

Explanation:

Kinetic energy is defined as energy possessed by an object due to its motion.It is calculated by:

K.E=\frac{1}{2}mv^2

Kinetic energy of the 5 kg object.

Mass of object,m = 5 kg

Velocity of an object = v

K.E=\frac{1}{2}mv^2=\frac{1}{2}\times 5kg\times v^2

Kinetic energy of the 20 kg object.

Mass of object,m' = 20 kg

Velocity of an object = v'

K.E=\frac{1}{2}mv^2=\frac{1}{2}\times 20kg\times v'^2

The ratio of the kinetic energy of the 5 kilogram object to the kinetic energy of the 20-kilogram object:

\frac{K.E}{K.E'}=\frac{\frac{1}{2}\times 5kg\times v^2}{\frac{1}{2}\times 20kg\times v'^2}

Given that, v = 2v'

\frac{K.E}{K.E'}=\frac{1}{1}

The ratio of kinetic energies of 5 kg object to 20 kg object is 1:1.

3 0
3 years ago
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