' W ' is the symbol for 'Watt' ... the unit of power equal to 1 joule/second.
That's all the physics we need to know to answer this question.
The rest is just arithmetic.
(60 joules/sec) · (30 days) · (8 hours/day) · (3600 sec/hour)
= (60 · 30 · 8 · 3600) (joule · day · hour · sec) / (sec · day · hour)
= 51,840,000 joules
__________________________________
Wait a minute ! Hold up ! Hee haw ! Whoa !
Excuse me. That will never do.
I see they want the answer in units of kilowatt-hours (kWh).
In that case, it's
(60 watts) · (30 days) · (8 hours/day) · (1 kW/1,000 watts)
= (60 · 30 · 8 · 1 / 1,000) (watt · day · hour · kW / day · watt)
= 14.4 kW·hour
Rounded to the nearest whole number:
14 kWh
The solution for this problem would be:(10 - 500x) / (5 - x)
so start by doing:
x(5*50*2) - xV + 5V = 0.02(5*50*2)
500x - xV + 5V = 10
V(5 - x) = 10 - 500x
V = (10 - 500x) / (5 - x)
(V stands for the volume, but leaves us with the expression for x)
Amazingly awesome !
Go Luke !
See Luke investigate !
Water, water, water.
Heat, heat, heat.
Luke rocks !
He probably learned something.
Explanation and answer:
This type of question can be clarified and sometimes solved by drawing a proper diagram or sketch. (see below)
Solution:
Since we do not know the reaction of the support, we can take moments about the support (thereby eliminating its involvement).
CCW moment = 0.900(5.00/2 - x) kg-m
CW moment = 0.300*(5.00/2-2.00)) = 0.150 kg-m
At equilibrium, CCW moment = CW moment, so
0.900(2.50-x) = 0.150
Expand and solve
2.25 - 0.900x = 0.150
0.900x = 2.25-0.15 = 2.10
x = 2.10 / 0.900 = 2.33 m (to nearest cm)
Answer:
yes, the potential difference across the terminals of the battery can be equal to its emf.
Explanation:
when the current in the battery is zero, meaning the current though, and hence the potential drop across the internal resistance is zero. This only happens when there is no load placed on the battery.
