In short, when light illuminates a piece of metal, the light kicks off electrons from the metal’s surface and these electrons can be detected as a change in the electric charge of the metal or as an electric current. Hence the name: photo for light and electric for the current. The explanation behind this simple phenomenon opened the door to revolutionary modern physics concepts regarding the composition of light, quantum mechanics, and what is now referred to as the “wave-particle duality” of nature. The wave-particle duality of nature is perhaps one of the greatest mysteries of our universe and a very interesting philosophical subject! Your goal in this lab is to reproduce the photoelectric effect for yourselves and to understand how it demonstrates the particle behavior of light.
<span>Answer:
For a disc, the moment of inertia about the perpendicular axis through the center is given by 0.5MR^2.
where M is the mass of the disc and R is the radius of the disc.
For the axis through the edge, use parallel axis theorem.
I = I(axis through center of mass) + M(distance between the axes)^2
= 0.5MR^2 + MR^2 (since the axis through center of mass is the axis through the center)
= 1.5 MR^2</span>
Answer:
the magnitude of the magnetic force on the wire is 0.2298 N
Explanation:
Given the data in the question;
we know that, the magnitude of magnetic force is given as;
|F
| = I(
× 
)
given that
I = 2.6 A
= 0.17
= 0.52
so we substitute
|F | = 2.6( 0.17i" × 0.52j" )
|F | = 0.2298 N
Therefore, the magnitude of the magnetic force on the wire is 0.2298 N
Answer:
-48 N
Explanation:
mass of door (m) = 4 kg
acceleration of the door = 12 m/s^{2}
force exerted by the person = 48 N
From Newton's third law of motion, action and reaction are equal but opposite. Therefore the force exerted on the door by the person which is 48 N will be the same as the force exerted on the person by the door but opposite in its direction, and this would be - 48 N
The capacitive reactance is reduced by a factor of 2.
<h3>Calculation:</h3>
We know the capacitive reactance is given as,
where,
= capacitive reactance
f = frequency
C = capacitance
It is given that frequency is doubled, i.e.,
f' = 2f
To find,
=?
Therefore, the capacitive reactance is reduced by a factor of 2.
I understand the question you are looking for is this:
A capacitor is connected across an AC source. Suppose the frequency of the source is doubled. What happens to the capacitive reactant of the inductor?
- The capacitive reactance is doubled.
- The capacitive reactance is traduced by a factor of 4.
- The capacitive reactance remains constant.
- The capacitive reactance is quadrupled.
- The capacitive reactance is reduced by a factor of 2.
Learn more about capacitive reactance here:
brainly.com/question/23427243
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