Answer:
a. +10.9μC
b. 0.600N and downward
Explanation:
To determine the magnitude of the charge, we use the force rule that exist between two charges which us expressed as
F=(kq₁q₂)/r²
since q₁=-0.55μC and the force it applied on the charge above it is upward,we can conclude that the second charge is +ve, hence we calculate its magnitude as
q₂=Fr²/kq₁
q₂=(0.6N*0.3²)/(9*10⁹*0.55*10⁻⁶)
q₂=0.054/4950
q₂=1.09*10⁻⁵c
q₂=10.9μC.
Hence the second charge is +10.9μC
b. From the rule of charges which state that like charges repel and unlike charges attract, we can conclude that the two above charges will attract since they are unlike charges. Hence the direction of the force will be downward into the second charge and the magnitude of the force will remain the same as 0.600N
Answer:
The torque about his shoulder is 34.3Nm.
The solution approach assumes that the weight of the boy's arm acts at the center of the boy's arm length 35cm from the shoulder.
Explanation:
The solution to the problem can be found in the attachment below.
Answer:
Vdc=10V
Explanation:
in a closed loop consisting of a super charged capacitor and an inductor, the super charge capacitor acts as a supply when the loop is closed, at t=0, the emf stored in the capacitor is 10V (q/c); and at that same time Vl= voltage across the inductor or loop too would be 10V,
if the loop remains closed for a longer period, the inductor would absorb energy from the capacitor till it dissipates all charges with itself.
