but wasn’t the work on one calculated using 1/2d and the work on the other using 3/2d? because all the other factors are the same wouldn’t that make it 3x? what’ve I done wrong?
can someone help me with this?
r= 25.6m and m=885kg
the kinetic friction between tyres and road is 9420N
the velocity you calculate in part a) is 16.5m/s on a non-banked road. thing is I keep getting a lower velocity for the banked road.
ok i did some research apparently the increase in binding energy per nucleon in fusion/fission reactions corresponds to a release of energy (as binding energy is actually considered a negative but that’s beyond the hsc syllabus) and that release of energy must manifest as kinetic energy.
I have a question. How does binding energy fit into nuclear fission reactions? Does the lost mass become binding energy AND kinetic energy?
If this is the case, then why in the last question from the 2019 HSC do they not account for binding energy and assume that all the lost mass becomes...