Explanation:
Given that,
Capacitor = 30μC
Resistor = 49.0Ω
Voltage = 30.0 V
Frequency = 60.0 Hz
We need to calculate the impedance
Using formula of impedance
.....(I)
We need to calculate the value of 
Using formula of
Put the value of into the formula of impedance
(a). We need to calculate the rms current in the circuit
Using formula of rms current
The rms current in the circuit is 0.30 A.
(b). We need to calculate the rms voltage drop across the resistor
Using formula of rms voltage
Put the value into the formula
The rms voltage drop across the resistor is 14.7 V
(c). We need to calculate the rms voltage drop across the capacitor
Using formula of rms voltage
The rms voltage drop across the capacitor is 26.53 V.
Hence, This is the required solution.
By someone wearing them doing some exhilarating thing
The at the top, the higher something is the more potential energy it has because if both fall the one at the top will have more kinetic energy
b.the covalent circuit is held throw magnetic waves
Answer:
1.74 m/s
Explanation:
From the question, we are given that the mass of the an object, m1= 2.7 kilogram(kg) and the mass of the can,m(can) is 0.72 Kilogram (kg). The velocity of the mass of an object(m1) , V1 is 1.1 metre per seconds(m/s) and the velocity of the mass of can[m(can)], V(can) is unknown- this is what we are to find.
Therefore, using the formula below, we can calculate the speed of the can, V(can);
===> Mass of object,m1 × velocity of object, V1 = mass of the can[m(can)] × velocity is of the can[V(can)].----------------------------------------------------(1).
Since the question says the collision was elastic, we use the formula below
Slotting in the given values into the equation (1) above, we have;
1/2×M1×V^2(initial velocity of the first object) + 1/2 ×M(can)×V^2(final velocy of the first object)= 1/2 × M1 × V^2 m( initial velocity of the first object).
Therefore, final velocity of the can= 2M1V1/M1+M2.
==> 2×2.7×1.1/ 2.7 + 0.72.
The velocity of the can after collision = 1.74 m/s
