What can you say about the impedance of a series RLC circuit at the resonant frequency? The impedance of a series RLC circuit is
2 answers:
Answer:
At resonance impedance of series RLC circuit is minimum and equal to resistance of the circuit
Explanation:
Let the resistance of the series RLC circuit is R
Capacitive reactance of the circuit is and inductive reactance of the circuit is
We know that impedance of the series RLC circuit is given by , here R is the resistance
is inductive reactance and
is capacitive reactance
At resonance capacitive reactance will be equal to inductive reactance
So
So impedance will become
So from above calculation we can say that impedance of series RLC circuit is minimum and equal to resistance of the circuit
Answer:
Explanation:
Let R is the resistance of the circuit, L is the inductance of the circuit and c is the capacitance.
Inductive reactance= XL
capacitive reactance = Xc
The impedance of the RLC circuit is
As the circuit is in resonance, So, XL = Xc
In this condition, Z = R
So, the impedance is equal to the resistance of the circuit.
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Newton's third law of motion sates that force is directly proportional to the rate of change of momentum produced
(a) The final speeds of the ice sleds is approximately 0.49 m/s each
(b) The impulse on the cat is 11.0715 kg·m/s
(c) The average force on the right sled is 922.625 N
The reason for arriving at the above values is as follows:
The given parameters are;
The masses of the two ice sleds, m₁ = m₂ = 22.7 kg
The initial speed of the ice, v₁ = v₂ = 0
The mass of the cat, m₃ = 3.63 kg
The initial speed of the cat, v₃ = 0
The horizontal speed of the cat, v₃ = 3.05 m/s
(a) The required parameter:
The final speed of the two sleds
For the first jump to the right, we have;
By the law of conservation of momentum
Initial momentum = Final momentum
∴ m₁ × v₁ + m₃ × v₃ = m₁ × v₁' + m₃ × v₃'
Where;
v₁' = The final velocity of the ice sled on the left
v₃' = The final velocity of the cat
Plugging in the values gives;
22.7 kg × 0 + 3.63 × 0 = 22.7 × v₁' + 3.63 × 3.05
∴ 22.7 × v₁' = -3.63 × 3.05
v₁' = -3.63 × 3.05/22.7 ≈ -0.49
The final velocity of the ice sled on the left, v₁' ≈ -0.49 m/s (opposite to the direction to the motion of the cat)
The final speed ≈ 0.49 m/s
For the second jump to the left, we have;
By conservation of momentum law, m₂ × v₂ + m₃ × v₃ = m₂ × v₂' + m₃ × v₃'
Where;
v₂' = The final velocity of the ice sled on the right
v₃' = The final velocity of the cat
Plugging in the values gives;
22.7 kg × 0 + 3.63 × 0 = 22.7 × v₂' + 3.63 × 3.05
∴ 22.7 × v₂' = -3.63 × 3.05
v₂' = -3.63 × 3.05/22.7 ≈ -0.49
The final velocity of the ice sled on the right = -0.49 m/s (opposite to the direction to the motion of the cat)
The final speed ≈ 0.49 m/s
(b) The required parameter;
The impulse of the force
The impulse on the cat = Mass of the cat × Change in velocity
The change in velocity, Δv = Initial velocity - Final velocity
Where;
The initial velocity = The velocity of the cat before it lands = 3.05 m/s
The final velocity = The velocity of the cat after coming to rest =
∴ Δv = 3.05 m/s - 0 = 3.05 m/s
The impulse on the cat = 3.63 kg × 3.05 m/s = 11.0715 kg·m/s
(c) The required information
The average velocity
Impulse = × Δt
Where;
Δt = The time of collision = The time it takes the cat to finish landing = 12 ms
12 ms = 12/1000 s = 0.012 s
We get;
∴
The average force on the right sled applied by the cat while landing, = 922.625 N
Learn more about conservation of momentum here:
Answer:
(a). The change in the protons electric potential is 0.639 kV.
(b). The change in the potential energy of the proton is
(c). The work done on the proton is .
Explanation:
Given that,
Speed
Initial potential V=100 V
Final potential = 150 V
(a). We need to calculate the change in the protons electric potential
Potential energy of the proton is
Using conservation of energy
Put the value into the formula
(b). We need to calculate the change in the potential energy of the proton
Using formula of potential energy
Put the value into the formula
(c). We need to calculate the work done on the proton
Using formula of work done
Hence, (a). The change in the protons electric potential is 0.639 kV.
(b). The change in the potential energy of the proton is
(c). The work done on the proton is .