Thanks for the hint.For Ice -> water:
m = 0.1kg c=2100 J kg^-1 K^-1 dt = 10 degrees (-10 -> 0)
therefore E= mc dt = 0.1(2100)(10) = 2100 J
Energy required to melt the ice:
E= specific latent heat of fusion of ice * mass
= 325000(0.1)
= 32500 J
and then for water to steam:
m = 0.1 kg...
1. Homework Statement
Calculate the heat energy required to convert 0.10 kg of ice at -10 degrees Celcius into steam at 100 degrees Celcius.
2. Homework Equations
E = m c dt (1)
where
E = heat energy (kJ)
m = unit mass (kg)
c = specific...
Oh cool, I did get ~1523.41 N and from the 1223.41 worked out for friction it would support F=ma.
Thanks a lot, you were a real help. If there was a reputation system I would show my appreciation :)
Not mentioned on the Question but the only forces I can think of is Weight of the car (perpendicular to the track, therefore 600(9.81)cos15 ), the frictional force and there should be a Normal reaction too.
How can I get to the Friction force from working out the resultant force?
Oh yh, and about the energy lost to friction bit:
Do I just work out the Net energy from before motion and after the car has traveled 100m?
i.e.:
Initial KE: 0 J
Initial GPE: mgh = 152340.8899...J
Total Energy...
First, a big hello to the Physics Forums community :)
Ok, now to business:
A car with mass 600 kg accelerates uniformly from rest down a steady hill inclined at 15 degrees to the horizontal[...]
- Work out the energy lost to friction after the car has traveled 100m
- The average...