Answer:
Let I and j be the unit vector along x and y axis respectively.
Electric field at origin is given by
E= kq1/r1^2 i + kq2/r2^2j
= 9*10^9*1.6*10^-19*/10^-6*(2i+ j)
= (2.88i + 1.44j)*10^-3 N/C
Force on charge= qE= 3*10^-19*1.6*(2.88i +1. 44 j) *10^-3
F= (1.382 i + 0.691 j) *10^-21
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Explanation:
Answer:
below
Explanation:
Net accelerating force becomes 12-8 = 4 N
F = ma
4 = 2 * a
a = 2 m/s^2
By equation of motion we have v = u + at
Where u = Initial velocity, v = final velocity, t = time taken and a = acceleration
Here v = 141 m/s, u = 17.7 m/s and t = 6 s
On substitution we will get
141 = 17.7+ 6a
So, a = (141-17.7)/6 = 20. 55 m/
Aceeleration = 20. 55 m/
along north direction.
According to the statements the number of electrons is 150, then
e = 150
But there is a positive charge of +22e, then the number of protons would be
p = 150+172
If the mass of the electrons is

And the mass of the protong is

We have that the total mass of the system would be


