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NemiM [27]
3 years ago
9

At what angle should the axes of two polaroids be placed so as to reduce the intensity of the incident unpolarized light to 13.

Physics
1 answer:
prohojiy [21]3 years ago
5 0

Answer:

θ = 66.90°

Explanation:

we know that

I= \frac{I_0}{2}cos^2\theta

I= intensity of polarized light =1

I_o= intensity of unpolarized light = 13

putting vales we get

1= \frac{13}{2}cos^2\theta

⇒\theta=cos^{-1} \sqrt{\frac{1}{6.5} }

therefore θ = 66.90°

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This would be an example of deposition (Phase transition). 
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How did Newton use creativity and logic in his approach to investigating light?
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The Bohr model of the hydrogen atom pictures the electron as a tiny particle moving in a circular orbit about a stationary proto
igor_vitrenko [27]

a) The angular velocity of the electron is 4.12\cdot 10^{16} rad/s

b) The number of revolutions per second is 6.54\cdot 10^{15} rev/s

c) The centripetal acceleration of the electron is 8.98\cdot 10^{22} m/s^2

Explanation:

a)

This is a problem of uniform circular motion: in fact, the electron orbits around the proton in a uniform circular motion.

The angular velocity of an object in uniform circular motion is given by

\omega = \frac{v}{r}

where

v is the linear speed

r is the radius of the trajectory

For the electron orbiting around the proton, we have

v=2.18 \cdot 10^6 m/s

r=5.29\cdot 10^{-11} m

Therefore, the angular velocity is

\omega=\frac{2.18\cdot 10^6}{5.29\cdot 10^{-11}}=4.12\cdot 10^{16} rad/s

b)

The period of revolution of the electron is given by

T=\frac{2\pi}{\omega}

where

\omega = 4.12\cdot 10^{16}rad/s is the angular velocity

Substituting,

T=\frac{2\pi}{4.12\cdot 10^{16}}=1.53\cdot 10^{-16}s

The period is the time the electron takes to make one complete orbit around the proton; therefore, the number of revolutions of the electrons in one second is:

f=\frac{1}{T}=\frac{1}{1.53\cdot 10^{-16}}=6.54\cdot 10^{15} rev/s

c)

The centripetal acceleration of an object in circular motion is

a=\frac{v^2}{r}

where

v is the linear speed

r is the radius of the circle

For the electron, we have

v=2.18 \cdot 10^6 m/s

r=5.29\cdot 10^{-11} m

Therefore, the centripetal acceleration is

a=\frac{(2.18\cdot 10^6)^2}{5.29\cdot 10^{-11}}=8.98\cdot 10^{22} m/s^2

Learn more about circular motion:

brainly.com/question/2562955

brainly.com/question/6372960

#LearnwithBrainly

7 0
4 years ago
Approximately how many miles away from the earth is the moon?
Snezhnost [94]
It is about 238,900 miles between the Earth and the Moon.

Hope this helps!
5 0
3 years ago
A machine gun fires 60 bullets per minute, with a velocity of 700 m/s .If each bullet has a mass of 50 g .Find the power develop
MAXImum [283]

Answer:

2100 kg*m/s

Explanation:

The proper term for power in this case is to use the physics term "impuls" = p.

1. Find the "power" developed by the gun to fire just 1 bullet.

2. multiply by 60 to find the power produced by the gun per minute.

p = m * v

where

m = mass of the bullet = 50 g = 0.050 kg

v = velocity of the bullet = 700 m/s

p = 0.050 * 700

p = 35 kg*m/s

So the impuls for 1 bullet is 35 kg*m/s

The gun can fire 60 bullets per minute so it's fireing power must there fore be 60 * 35 = 2100 kg*m/s

4 0
3 years ago
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