Explanation:
The 11Ω, 22Ω, and 33Ω resistors are in parallel. That combination is in series with the 4Ω and 10Ω resistors.
The net resistance is:
R = 4Ω + 10Ω + 1/(1/11Ω + 1/22Ω + 1/33Ω)
R = 20Ω
Using Ohm's law, we can find the current going through the 4Ω and 10Ω resistors:
V = IR
120 V = I (20Ω)
I = 6 A
So the voltage drops are:
V = (4Ω) (6A) = 24 V
V = (10Ω) (6A) = 60 V
That means the voltage drop across the 11Ω, 22Ω, and 33Ω resistors is:
V = 120 V − 24 V − 60 V
V = 36 V
So the currents are:
I = 36 V / 11 Ω = 3.27 A
I = 36 V / 22 Ω = 1.64 A
I = 36 V / 33 Ω = 1.09 A
If we wanted to, we could also show this using Kirchhoff's laws.
Answer:
a = -04978 m / s²
Explanation:
For this exercise we can use the kinematics equations, as the initial speed, distance and time indicate, we can use
x = x₀ + v₀ t + 1 /2 a t²
½ a t² = x-x₀ - v₀ t
a = (x-x₀ - v₀ t) 2 / t²
let's reduce the magnitudes to the SI system
x₀ = 1000 mm = 1 m
x = 5000mm = 5m
let's calculate
a = (5 - 1 - 15 60) 2/60²
a = -04978 m / s²
negative sign indicates that braking is related
The phases of the Moon and Venus