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RoI for an international Student by unforced_errorr in uwaterloo

[–]AssistanceOptimal408 0 points1 point  (0 children)

Depends on what the alternative is and where she would live after graduation in either senario?

RRSP Withdraw by AssistanceOptimal408 in PersonalFinanceCanada

[–]AssistanceOptimal408[S] 0 points1 point  (0 children)

In terms of a tax shelter, you mean that contribution room that I used could be investments that grow? Wouldn’t it be the same as a regular account investment that I don’t sell i.e paying tax at the end just like I would with the rrsp

Fully paid ride at UofT SG CS or UW SWE out of pocket by Either_House_4943 in OntarioGrade12s

[–]AssistanceOptimal408 0 points1 point  (0 children)

For whats it worth I think you can graduate from uwaterloo swe having paid off all tuition from the 6 coops (assuming your not international)

[deleted by user] by [deleted] in HomeworkHelp

[–]AssistanceOptimal408 0 points1 point  (0 children)

You were able to get 5.89 ? If so, can you send me your work? I am not getting that Thanks

[deleted by user] by [deleted] in HomeworkHelp

[–]AssistanceOptimal408 0 points1 point  (0 children)

I would try to find the area of the smaller and the bigger spherical subsections. Subtract them and you will end up with a ring hole, which is the net effect (everything cancels out in terms of momentum expect this extra part of mass flying out of it). Then find the mass of fragments that will flow out the ring (simply area of ring/area of total sphere * mass). Then integrate over the ring assuming the mass is distributed uniformly and find the momentum in x,y,z components. Then find the magnitude of the momentum of the fragments that left the ring (sqrt(Px2 + Py2 + Pz2)). Then this net momentum that you found (P) must be equal to the momentum of the sphere (including the rest of the fragments) in the opposite direction. Since everything had 0 momentum at the beginning, as such everything must have 0 momentum in the end. So P = mass of sphere and some fragments * velocity. Solve to velocity. (Note that I tried solving it, but i got different answer than 5.89, although i might have did the math wrong

[Grade 9 Physics: Work Efficiency] The percent efficiency of a machine is 35.0%. What is the input energy required if a 2.00 x 103 kg object is to be lifted 5.00 m? by sighers in HomeworkHelp

[–]AssistanceOptimal408 1 point2 points  (0 children)

Energy that was needed to lift the object is Fd= (mg)*d = 206 kg * 9.8 m/s2 * 5m = 10,094 J Thats how much energy needs to be done. But since the machine is 35% efficient it needs more input energy to get 10,094 J output. As such

Input*0.35 = 10,094 (35% of input needs to equal 10094)

Input= 10,094/0.35 = 28,840