Answer:
A. Voltage
Explanation:
I got a 100% on the quiz on edg so trust me
Answer:
ω = 1.83 rad/s clockwise
Explanation:
We are given:
I1 = 3.0kg.m2
ω1 = -5.4rad/s (clockwise being negative)
I2 = 1.3kg.m2
ω2 = 6.4rad/s (counterclockwise being positive)
By conservation of the momentum:
I1 * ω1 + I2 * ω2 = (I1 + I2) * ω
Solving for ω:
Since it is negative, the direction is clockwise.
Answer:
The tube should be held vertically, perpendicular to the ground.
Explanation:
As the power lines of ground are equal, so its electrical field is perpendicular to the ground and the equipotential surface is cylindrical. Therefore, if we put the position fluorescent tube parallel to the ground so the both ends of the tube lie on the same equipotential surface and the difference is zero when its potential.
And the ends of the tube must be on separate equipotential surfaces to optimize potential. The surface near the power line has a greater potential value and the surface farther from the line has a lower potential value, so the tube must be placed perpendicular to the floor to maximize the potential difference.
