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.
The answer to this question is False.
Since f=m(v^2/r),or fnet is equal to ma.
force = unknown
velocity=22m/s
radius=75m
f=m(v^2/r)
f=925(22^2/75)
f=5969.333N
