Answer:
Work done on an object is equal to
FDcos(angle).
So, naturally, if you lift a book from the floor on top of the table you do work on it since you are applying a force through a distance.
However, I often see the example of carrying a book through a horizontal distance is not work. The reasoning given is this: The force you apply is in the vertical distance, countering gravity and thus not in the direction of motion.
But surely you must be applying a force (and thus work) in the horizontal direction as the book would stop due to air friction if not for your fingers?
Is applying a force through a distance only work if causes an acceleration? That wouldn't make sense in my mind. If you are dragging a sled through snow, you are still doing work on it, since the force is in the direction of motion. This goes even if velocity is constant due to friction.
Explanation:
Answer:
Explanation:
a) Power consumption is 4100 J/min / 60 s/min = 68.3 W(atts)
work done raised the potential energy
b) 75(9.8)(1000) / (3(3600)) = 68.055555... 68.1 W
c) efficiency is 68.1 / 68.3 = 0.99593... or nearly 100%
Not a very likely scenario.
From what I know; When a sample of liquid water vaporizes into water vapor, the electrons in the water sped up due to heat.
The EMF of the battery includes the force to to drive across its internal resistance. the total resistance:
R = internal resistance r + resistance connected rv
R = r + rv
Now find the current:
V 1= IR
I = R / V1
find the voltage at the battery terminal (which is net of internal resistance) using
V 2= IR
So the voltage at the terminal is:
V = V2 - V1
This is the potential difference vmeter measured by the voltmeter.
