Answer:
time will elapse before it return to its staring point is 23.6 ns
Explanation:
given data
speed u = 2.45 × 
m/s
uniform electric field E = 1.18 × 
N/C
to find out
How much time will elapse before it returns to its starting point
solution
we find acceleration first by electrostatic force that is
F = Eq
here
F = ma by newton law
so
ma = Eq
here m is mass , a is acceleration and E is uniform electric field and q is charge of electron
so
put here all value
9.11 × 
kg ×a = 1.18 × × 1.602 × 
a = 20.75 × 
m/s²
so acceleration is 20.75 × m/s²
and
time required by electron before come rest is
use equation of motion
v = u + at
here v is zero and u is speed given and t is time so put all value
2.45 × = 0 + 20.75 × (t)
t = 11.80 × 
s
so time will elapse before it return to its staring point is
time = 2t
time = 2 ×11.80 ×
time is 23.6 × s
time will elapse before it return to its staring point is 23.6 ns
The amount of power change if less work is done in more time"then the amount of power will decrease".
<u>Option: B</u>
<u>Explanation:</u>
The rate of performing any work or activity by transferring amount of energy per unit time is understood as power. The unit of power is watt
Here this equation showcase that power is directly proportional to the work but dependent upon time as time is inversely proportional to the power i.e as time increases power decreases and vice versa.
This can be understood from an instance, on moving a load up a flight of stairs, the similar amount of work is done, no matter how heavy but when the work is done in a shorter period of time more power is required.
For purposes of completing our calculations, we're going to assume that
the experiment takes place on or near the surface of the Earth.
The acceleration of gravity on Earth is about 9.8 m/s², directed toward the
center of the planet. That means that the downward speed of a falling object
increases by 9.8 m/s for every second that it falls.
3 seconds after being dropped, a stone is falling at (3 x 9.8) = 29.4 m/s.
That's the vertical component of its velocity. The horizontal component is
the same as it was at the instant of the drop, provided there is no horizontal
force on the stone during its fall.
