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.
Take the stone's position at ground level to be the origin, and the downward direction to be negative. Then its position in the air 
at time 
is given by
Let 
be the depth of the well. The stone hits the bottom of the well after 5.00 s, so that
ANSWER
Velocity of the mass reaches zero
EXPLANATION
We want to identify what hapens to a mass attached toa a spring at maximum displacement.
When a mass attached to a spring is at its maximum position of displacement, the direction of the mass begins to change. This implies that the velocity of the mass will reach zero.
Hence, at maximum displacement, the velocity of the mass reaches zero.
A quadrilateral with only one pair of parallel sides.
<span>a small carpal bone in the base of the hand, articulating with the metacarpal of the index finger.
</span>
Answer:
E= 55.53 x 10³ V/m
Explanation:
Given that
a= 3.63 cm
Area ,A= a²
distance ,d= 0.473 mm
Stored energy ,U = 8.49 nJ
Value of capacitor given as
By putting the values
C=2.46 x 10⁻¹¹ F
V=Voltage difference
V=26.27 V
V= E d
E=Electric filed
26.27 = E x 0.473 x 10⁻³
E= 55.53 x 10³ V/m
