Answer:
V_{average} = 
, V_{average} = 2 V
Explanation:
he average or effective voltage of a wave is the value of the wave in a period
V_average = ∫ V dt
in this case the given volage is a square wave that can be described by the function
V (t) = 
to substitute in the equation let us separate the into two pairs
V_average = 
V_average = 
V_{average} =
we evaluate V₀ = 4 V
V_{average} = 4 / 2)
V_{average} = 2 V
I think its 13...........
Answer:
Volt
Explanation:
Voltage is what makes electric charges move. ... Voltage is also called, in certain circumstances, electromotive force (EMF). Voltage is an electrical potential difference, the difference in electric potential between two places. The unit for electrical potential difference, or voltage, is the volt.
The ohm is defined as an electrical resistance between two points of a conductor when a constant potential difference of one volt, applied to these points, produces in the conductor a current of one ampere, the conductor not being the seat of any electromotive force.
The coulomb (symbolized C) is the standard unit of electric charge in the International System of Units (SI). ... In terms of SI base units, the coulomb is the equivalent of one ampere-second. Conversely, an electric current of A represents 1 C of unit electric charge carriers flowing past a specific point in 1 s.
An ampere is a unit of measure of the rate of electron flow or current in an electrical conductor. One ampere of current represents one coulomb of electrical charge (6.24 x 1018 charge carriers) moving past a specific point in one second.
Answer:
The photoelectric effect occurs only for frequencies above the cutoff frequency, regardless of the intensity.
Explanation:
The photoelectric effect occurs when light is shined on metals such as zinc beyond a certain frequency (the threshold frequency), which causes electrons to escape from the zinc. The electrons which are fleeing are called photo electrons.
Therefore photo electric effect is
The photoelectric effect occurs only for frequencies above the cutoff frequency, regardless of the intensity.
Answer:
202.8m
Explanation:
Given that A pirate fires his cannon parallel to the water but 3.5 m above the water. The cannonball leaves the cannon with a velocity of 120 m/s. He misses his target and the cannonball splashes into the briny deep.
First calculate the total time travelled by using the second equation of motion
h = Ut + 1/2gt^2
Let assume that u = 0
And h = 3.5
Substitute all the parameters into the formula
3.5 = 1/2 × 9.8 × t^2
3.5 = 4.9t^2
t^2 = 3.5/4.9
t^2 = 0.7
t = 0.845s
To know how far the cannonball travel, let's use the equation
S = UT + 1/2at^2
But acceleration a = 0
T = 2t
T = 1.69s
S = 120 × 1.69
S = 202.834 m
Therefore, the distance travelled by the cannon ball is approximately 202.8m.
