Given Information:
Resistance = R = 14 Ω
Inductance = L = 2.3 H
voltage = V = 100 V
time = t = 0.13 s
Required Information:
(a) energy is being stored in the magnetic field
(b) thermal energy is appearing in the resistance
(c) energy is being delivered by the battery?
Answer:
(a) energy is being stored in the magnetic field ≈ 219 watts
(b) thermal energy is appearing in the resistance ≈ 267 watts
(c) energy is being delivered by the battery ≈ 481 watts
Explanation:
The energy stored in the inductor is given by

The rate at which the energy is being stored in the inductor is given by

The current through the RL circuit is given by

Where τ is the the time constant and is given by


Therefore, eq. 1 becomes

At t = 0.13 seconds

(b) thermal energy is appearing in the resistance
The thermal energy is given by

(c) energy is being delivered by the battery?
The energy delivered by battery is

Your grade will probably go down to a D 68% or little higher than that
30km. 24 the first two hours and 6 the half hour
Ethylene glycol is termed as the primary ingredients in antifreeze.
The ethylene glycol molecular formula is C₂H₆O₂.
Molar mass of C₂H₆O₂ is = (2×12) +(6×1) + (216) = 62g/mol
Now that antifreeze by mass is 50%, then there is 1kg of ethylene glycol which is present in 1kg of water.
ΔTf = Kf×m
ΔTf = depression in the freezing point.
= freezing point of water freezing point of the solution
= O°c - Tf
= -Tf
Kf = depression in freezing constant of water = 1.86°C/m
M is the molarity of the solution.
=(mass/molar mass) mass of solvent in kg
=1000g/62 (g/mol) /1kg
=16.13m
If we plug the value we get
-Tf = 1.86 × 16.13 = 30
Tf = -30°c