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From the graph, when amplitude (x) is .60 cm energy is 4mJ.
I thought that depended on the values given in the question. If they are all given in 2 s.f shouldn't t your final answer be in 2 s.f as well?
Both are acceptable.
So why don't we consider the Ke while calculating the total energy?
Yeah high frequency means high photon energy. But why does it mean that there will be less photons arriving per unit time?
I thought that depended on the values given in the question. If they are all given in 2 s.f shouldn't t your final answer be in 2 s.f as well?
Why does the ms say that there are less photons arriving per unit time?The no. of photons arriving per unit time depends on intensity. It doesn't depend on frequency. If less photons are arriving that means the intensity is less!
oh can u tell me which year is it from?
no no, I'm talking about ppr 2 and all. for eg: if you are required to find the force and you are given m=5.0kg and a=1.2 m s^-2 then wont force=ma=6.0 N and not 6.00N?Yeah that too! but more significant number means less uncertainty...so 3 sf is much reliable. im talking about paper 4 here.
For paper 5, you're right!
paper 3 practical? i prefer using a cross since u circle a point which is anomalous
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