A centrifuge rotor rotating at 10,000 rpm is shut off and is eventually brought to rest by a frictional force of 1.20m n. if the
1 answer:
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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:
Explanation:
Given that,
Initially, the spaceship was at rest, u = 0
Final velocity of the spaceship, v = 11 m/s
Distance accelerated by the spaceship, d = 213 m
We need to find the acceleration experienced by the occupants of the spaceship during the launch. It is a concept based on the equation of kinematics. Using the third equation of motion to find acceleration.
So, the acceleration experienced by the occupants of the spaceship is .