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
Answer:
V = a * t = 9.8 m/s^2 * 2.3 s = 22.5 m/s velocity after 2.3 s
S = 1/2 g t^2 since initial speed is zero
S = 1/2 * 9.8 m/s^2 * 5.29 s^2 = 25.9 m
Explanation:
It is given that,
Velocity in East, 
Velocity in North, 
(a) The resultant velocity is given by :
(b) The width of the river is, d = 80 m
Let t is the time taken by the boat to travel shore to shore. So,
t = 16 seconds
(c) Let x is the distance covered by the boat to reach the opposite shore. So,
x = 48 meters
Hence, this is the required solution.
Answer:
20.85 years
Explanation:
2.61 km = 2610 m
2.07 kW = 2070 W
First we need to calculate the potential energy required to take m = 
kg of rain cloud to an altitude of 2610 m is
With a P = 2070 W power pump, this can be done within a time frame of
or 658037739/(60*60) = 182788 hours or 182788 / 24 = 7616 days or 7616 / 365.25 = 20.85 years
