Answer:
F = 2.40 × 
N
Explanation:
given data
charge q1 = 3.95 nC
x= 0.198 m
charge q2 = 4.96 nC
x= -0.297 m
solution
force on a point charge kept in electric field F = E × q ................1
here E is the magnitude of electric field and q is the magnitude of charge
and
first we will get here electric field at origin
So net field at origin is
E = (Kq2÷r2²) - (kq1÷r1²) ...............2
put here value
E = 9[(4.96÷0.297²)-(3.95÷0.198²)]
E = 400.72 N/C ( negative x direction )
so that force will be
F = 6 × 
× 400.72
F = 2.40 × N
C. a cello playing music at the concert
Answer:
First let's find the work.
Work = Force*Displacement
= 20*5
= 100 Joules
Power = Work/Time
= 100/4
= 25 Watts
<h2>
Answer:</h2>
4.2 C
<h2>
Explanation:</h2>
The charge (Q) moving is the product of the current(I) flowing through the torso of the person and the time taken (t) for the flow.
i.e
Q = I x t
Where;
I = current = 14.0A
t = time taken = 0.0300s
Substituting the values of I and t into the equation above gives
Q = 14.0 x 0.0300
Q = 4.2 C
Therefore quantity of charge moving is 4.2 C
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