The number of electrons emitted from the metal per second increases if the intensity of the incident light is increased.
Answer: Option B
<u>Explanation:</u>
As a result of photoelectric effect, electrons are emitted by the light incident on a metal surface. The emitted electrons count and its kinetic energy can measure as the function of light intensity and frequency. Like physicists, at the 20th century beginning, it should be expected that the light wave's energy (its intensity) will be transformed into the kinetic energy of emitted electrons.
In addition, the electrons count emitting from metal must vary with light wave frequency. This frequency relationship was expected because the electric field oscillates due to the light wave and the metal electrons react to different frequencies. In other words, the number of electrons emitted was expected to be frequency dependent and their kinetic energy should be dependent on the intensity (constant wavelength) of light.
Thus, the maximum in kinetic energy of electrons emitted increases with increase in light's frequency and is experimentally independent of light intensity. So, the number of emitted electrons is proportionate to the intensity of the incident light.
The answer is D. As the ambulance gets closer, the sound waves are compressed relative to the person; so the frequency increases.
1.igneous rocks are formed when magma or lava cools and hardens
2. the cooling rate of the rock
Answer:
The diameter of wire should be
m
Explanation:
Given:
Current density

Current
A
From the formula of current density,

Where
area of cylindrical wire = 



m
For finding the diameter of wire,

m
Therefore, the diameter of wire should be
m