Answer:
2.2 meters
Explanation:
Potential energy, PE created by a charge, q at a radius r from the charge source, Q, is expressed as:
is Coulomb's constant.
#The electric field,
at radius r is expressed as:
From i and ii, we have:
#Substitute actual values in our equation:
Hence, the distance between the charge and the source of the electric field is 2.2 meters
Answer:50ms-1
Explanation:use the formula v=d/t
in order to find the velocity,devide the distance with time taken.
since distance is 400 meters devide it with seconds whiuch gives us 50.
Hi there!
We can use the kinematic equation:
vf = Final velocity (? m/s)
vi = initial velocity (0 m/s, dropped from rest)
a = acceleration (due to gravity, 9.8 m/s²)
d = distance (9.8 m)
Simplify the equation to solve for vf:
Substitute in the given values:
0.4823 m/s
The initial velocity u1 of the ball=0
From the law of conservation of linear momentum.
m1u1+m2u2=m1v1+m2v2
(160×0)+(170×u1)=(160×0.3)+(170×0.2)
u1=0.4823m/s
