Answer:
Explanation:
Given that,
First Capacitor is 10 µF
C_1 = 10 µF
Potential difference is
V_1 = 10 V.
The charge on the plate is
q_1 = C_1 × V_1 = 10 × 10^-6 × 10 = 100µC
q_1 = 100 µC
A second capacitor is 5 µF
C_2 = 5 µF
Potential difference is
V_2 = 5V.
Then, the charge on the capacitor 2 is.
q_2 = C_2 × V_2
q_2 = 5µF × 5 = 25 µC
Then, the average capacitance is
q = (q_1 + q_2) / 2
q = (25 + 100) / 2
q = 62.5µC
B. The two capacitor are connected together, then the equivalent capacitance is
Ceq = C_1 + C_2.
Ceq = 10 µF + 5 µF.
Ceq = 15 µF.
The average voltage is
V = (V_1 + V_2) / 2
V = (10 + 5)/2
V = 15 / 2 = 7.5V
Energy dissipated is
U = ½Ceq•V²
U = ½ × 15 × 10^-6 × 7.5²
U = 4.22 × 10^-4 J
U = 422 × 10^-6
U = 422 µJ
Wave speed= Frequency x Wavelength
-> Wavelength= Wave speed divide Frequency
-> 50m/s divide 100Hz= 0.5m
Answer:
1. Yes, if the elephant is standing still.
2. impulse acting on it.
Explanation:
Momentum is the product of mass and velocity. A change in momentum occurs when any of the quantities change. A change in momentum is equal to impulse acting on it. A change in momentum per unite time generates force.
P = m v
ΔP = Δm v or mΔv
F = ΔP /Δt
Impulse, F Δt = ΔP.
An ant has negligible mass in comparison to an elephant. Thus, an ant can have more momentum only when the elephant is standing still. Then, the momentum of the elephant would be zero as it would have zero velocity. and a moving ant would have more Momentum.
The change in momentum of an object is impulse acting on it.
Impulse, F Δt = ΔP.
Answer:
the correct answer is
D.Remaining well-hydrated while exercising is best
