Respuesta :
The
energy the element Iron draws is I=6 Ampers. The charge flows through the Iron
for t=1 min, which is equal to 60 seconds. The amount of charge which can flow
through the iron is denoted with q and can be calculated as: q=I*t=60*6=360 C
(coulombs).
The answer is "the same than the mercury in the bigger tube".
If one barometer tube has twice the cross-sectional area of another, mercury in the smaller tube will rise the same than the mercury in the bigger tube.
The mercury will rise to the point where the column of mercury has the same weight as the force exerted by the atmosphere.
The force exerted by the atmosphere is pressure * cross-sectional area
Anf the weight of the column of mercury, W, will be:
W = m* g
where m = density * volume, and volume = cross-sectional area * height
=> W = density * cross-sectional area * height
Then, you make W = F and get:
density * cross-sectional area * height = P * cross-sectional area
The term cress-sectional area appears on both sides so it gets cancelled, and the height of the column of mercury does not depend on the cross-sectional area of the barometer.
If one barometer tube has twice the cross-sectional area of another, mercury in the smaller tube will rise the same than the mercury in the bigger tube.
The mercury will rise to the point where the column of mercury has the same weight as the force exerted by the atmosphere.
The force exerted by the atmosphere is pressure * cross-sectional area
Anf the weight of the column of mercury, W, will be:
W = m* g
where m = density * volume, and volume = cross-sectional area * height
=> W = density * cross-sectional area * height
Then, you make W = F and get:
density * cross-sectional area * height = P * cross-sectional area
The term cress-sectional area appears on both sides so it gets cancelled, and the height of the column of mercury does not depend on the cross-sectional area of the barometer.