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Answer:
The force of car 3 on car 2 ≈ 1810.82 N
Explanation:
The equation for the change in momentum of the two cars are;
Conservation of linear momentum
150( 2.2 - v) = 265(1.5-v)
150 × 2.2 - 265×1.5 = (150+265)v
150 × 2.2 - 265×1.5 = -67.5 = 415×v
∴ v = -67.5/415 = -0.1627 m/s West = 0.1627 m/s East
The impulse of the net force is the amount of momentum change experienced given by the equation;
Impulse force = -
Where;
= The final velocity
= The initial velocity
For the the 265 kg mass, we have;
= 0.1627 m/s
= 1.5 m/s
Which gives the impulse a s F×Δt = 265×0.1627 - 265×1.5 = -354.38 kg·m/s
The change in kinetic energy of the collision = 1/2×265×(0.1627^2 - 1.5^2) =-294.62 J
Whereby the distance moved in one second is 0.1627 m, we have;
Work done = Force × Distance = Force × 0.1627 = 294.62
Force = 294.62/0.1627 = 1810.82 N.
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.
