High pressure systems entail sinking air, while lows entail rising air. High pressure systems form where air converges in the higher levels of the atmosphere. The converging air has nowhere to go and is forced to sink toward the ground. This sinking effect inhibits cloud formation and therefore precipitation.
Answer: λ (wavelength) = 3.3 m → ΔS = 3.3 m
v (speed) = 5.6 m/s → ΔV = 5.6 m/s
T (period) → ΔT = ?
f (frequency) = ?
If:
Now: (if, ΔT = T), Using the formula of the period of a wave, find the frequency:
Explanation: Your Welcome u.u
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

Answer:
I think its A plz tell me if im right
Answer:
Explanation:
Given

Frictional Force is balanced by force due to car acceleration
Frictional force 



