In the following diagram, the voltage is 1.5 volts and the resistance is 3.0 ohms. Use Ohm's law to determine the current in the
2 answers:
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Part a)
As we know that energy stored inside the capacitor is given as
for a given capacitor we know
Now we can use it in above equation to find the energy
PART b)
If two negative charges are hold near to each other and then released
Then due to mutual repulsion they start moving away from each other
Due to mutual repulsion as the two charges moving away the electrostatic potential energy of two charges will convert into kinetic energy of the two charges.
So here as they move apart kinetic energy will increase while potential energy will decrease
Part c)
As we know that capacitance is given as
here we know that
First, you make a diagram of all the forces acting on the system. This is shown in the figure. We have to determine F1 and F4. Let's do a momentum balance. Momentum is conserved so the summation of all momentum is equal to zero. Momentum is force*distance.
To determine F1: (reference is F4, so F4=0)
∑Momentum = 0 = -F2 - F3 + F1
0 = (-4 kg)(9.81 m/s2)(0.25m)-(6kg)(9.81 m/s2)(0.5-0.3m)+F1(0.5-0.1m)
F1 = 53.96 N (left knife-edge)
To determine F4: (reference is F1, so F1=0)
∑Momentum = 0 = -F2 - F3 + F4
0 = (-4 kg)(9.81 m/s2)(0.25m)-(6kg)(9.81 m/s2)(0.5-0.2m)+F4(0.5-0.1m)
F4 = 68.67 N (right knife-edge)
To determine F1: (reference is F4, so F4=0)
∑Momentum = 0 = -F2 - F3 + F1
0 = (-4 kg)(9.81 m/s2)(0.25m)-(6kg)(9.81 m/s2)(0.5-0.3m)+F1(0.5-0.1m)
F1 = 53.96 N (left knife-edge)
To determine F4: (reference is F1, so F1=0)
∑Momentum = 0 = -F2 - F3 + F4
0 = (-4 kg)(9.81 m/s2)(0.25m)-(6kg)(9.81 m/s2)(0.5-0.2m)+F4(0.5-0.1m)
F4 = 68.67 N (right knife-edge)