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Anna71 [15]
2 years ago
10

Plane-polarized light is incident on a single polarizing disk, with the direction of E0 parallel to the direction of the transmi

ssion axis. Through what angle should the disk be rotated so that the intensity in the transmitted beam is reduced by a factor of 6.00?
Physics
1 answer:
Vladimir [108]2 years ago
7 0

Answer: 65.9^{\circ}

Explanation:

Given

Intensity must be reduced by a factor of 6

Intensity is given by I=I_o\cos ^2\theta\\

Substitute I by \frac{I_o}{6}

\Rightarrow \dfrac{I_o}{6}=I_o\cos ^2\theta\\\\\Rightarrow \cos^2\theta =\dfrac{1}{6}\\\\\Rightarrow \cos \theta=\dfrac{1}{\sqrt{6}}\\\\\Rightarrow \theta=65.9^{\circ}

So, the disk must be rotated by an angle of 65.9^{\circ} .

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Select the statements that are TRUE. Electrophilic aromatic substitution reactions have energies of activation that are very low
NikAS [45]

Answer:

Addition reactions with benzenes lead to the loss of aromaticity.

Benzene and its derivatives undergo a type of substitution reaction in which a hydrogen atom is replaced by a substituent, but the stable aromatic benzene ring is regenerated at the end of the mechanism.

Benzene and its derivatives tend to undergo electrophilic aromatic substitution reactions.

Explanation:

5 0
3 years ago
At a given instant an object has an angular velocity. It also has an angular acceleration due to torques that are present. There
katen-ka-za [31]

a) Constant

b) Constant

Explanation:

a)

We can answer this question by using the equivalent of Newton's second law of motion of rotational motion, which can be written as:

\tau_{net} = I \alpha (1)

where

\tau_{net} is the net torque acting on the object in rotation

I is the moment of inertia of the object

\alpha is the angular acceleration

The angular acceleration is the rate of change of the angular velocity, so it can be written as

\alpha = \frac{\Delta \omega}{\Delta t}

where

\Delta \omega is the change in angular velocity

\Delta t is the time interval

So we can rewrite eq.(1) as

\tau_{net}=I\frac{\Delta \omega}{\Delta t}

In this problem, we are told that at a given instant, the object has an angular acceleration due to the presence of torques, so there is a non-zero change in angular velocity.

Then, additional torques are applied, so that the net torque suddenly equal to zero, so:

\tau_{net}=0

From the previous equation, this implies that

\Delta \omega =0

Which means that the angular velocity at that instant does not change anymore.

b)

In this second case instead, all the torques are suddenly removed.

This also means that the net torque becomes zero as well:

\tau_{net}=0

Therefore, this means that

\Delta \omega =0

So also in this case, there is no change in angular velocity: this means that the angular velocity of the object will remain constant.

So cases (a) and (b) are basically the same situation, as the net torque is zero in both cases, so the object acts in the same way.

8 0
3 years ago
The caste system in India is illegal. True or False
anzhelika [568]
<span>Discrimination is illegal, but caste system is legal.
So answer: False</span>
4 0
3 years ago
a 100 kg person travels from sea level to an altitude of 5000 m. By how mans newtons does their weight change?
Bumek [7]

Answer:

Thus, the change in the weight of the person is 1.6N , option c is correct.

Explanation:

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2 years ago
A star that is moving toward an observer has its visible light shifted toward which end of the spectrum? A. blue B. red C. yello
xxTIMURxx [149]
La D mijo es blanco por que al pasar con rapides el color se torna blanco


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3 years ago
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