Answer:
A) B = 5.4 10⁻⁵ T, B) the positive side of the bar is to the West
Explanation:
A) For this exercise we must use the expression of Faraday's law for a moving body
fem = 
fem =
- d (B l y) / dt = - B lv
B = 
we calculate
B = - 7.9 10⁻⁴ /(0.73 20)
B = 5.4 10⁻⁵ T
B) to determine which side of the bar is positive, we must use the right hand rule
the thumb points in the direction of the rod movement to the south, the magnetic field points in the horizontal direction and the rod is in the east-west direction.
Therefore the force points in the direction perpendicular to the velocity and the magnetic field is in the east direction; therefore the positive side of the bar is to the West
Answer: 20
Explanation: Mass number is the number of neutrons plus the number of protons. 8 + 12 gives 20.
Both the birth and death rate are expressed per 1000 of the population.
Answer: 
Explanation:
Given
Capacitance 
Resistance 
Inductance 
In LCR circuit, current is maximum at resonance frequency i.e.

Insert the values

Also, frequency is given by


When the box IS on the shelf 2m above the ground,
its potential energy is
(weight) x (height) = (3 N) x (2 m) = 6 joules .
THAT's the work you have to do, to lift the box up to there.