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:
0 m/s²
Explanation:
The velocity is constant, so there is no acceleration.
Answer:
work = 1275.3 J
Explanation:
work = (force)(distance)cosø ------- force = ma
=(mass*acceleration)(distance)cosø
=(20*9.81)(6.5)cos0
=1275.3J
nite that the angle of cosine is the difference between the angle of force and the distance. in this case, the force and the distance are in the same direction. :)
1.2 x (2.2 x 10⁵) = 264,000 Ω
0.8 x (2.2 x 10⁵) = 176,000 Ω
With a 'nominal' value of 220,000 Ω, it could actually be anywhere <em>between 176,000Ω and 264,000Ω</em> .