the state of being thick, sticky, and semifluid in consistency, due to internal friction.
Our values can be defined like this,
The problem can be solved for part A, through the Work Theorem that says the following,
Where
KE = Kinetic energy,
Given things like that and replacing we have that the work is given by
W = Fd
and kinetic energy by
So,
Clearing F,
Replacing the values
B) The work done by the wall is zero since there was no displacement of the wall, that is d = 0.
Answer: 0.2 hours
Explanation: In order to solve this question we have to considerer that a recargeable battery can supply 1800 mA in one hour then we have to determine how long could this battery drive current through a long, thin wire of resistance 34 Ω .
Besides, this battery has a voltage of 12 V
so by using the Ohm law we also know that V=R*I,
Fron this we can obtain:
I= V/R= 12 V/ 34 Ω=0.35 A= 350 mA
then considering that this battery can supply 1800 mA in one hour we have this battery can supply 350 mA in x time in the form:
1hour------- 1800 mA
x hour--------350 mA
time= 350/1800= 0.2 hour
