An object of mass 20 kg is raised vertically through a distance of 8.0 m above ground level. Using g=10 m/s^2 what is the potent
1 answer:
You might be interested in
Answer:
1. A <em>series circuit </em>is a closed circuit which has all loads connected in a row and there is only one path for the current to flow.
2. The <em>Ohm's Law </em>state that a current flow through a resistor is directly proportional to the voltage across it
3. A <em>parallel circuit </em>is a closed circuit divided into branches that it has two o more paths for the current to flow and the loads are parallel to each other which mean the voltage across them is the same for all loads.
First we have to calculate the time taken to travel the distance 30 m, is
.
Now from equation of motion,
Given, .
As object starts from rest, so u = 0.
Substituting these values in above equation, we get
.
Thus, the acceleration is
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