Recent content by Physicsnuubie

Hi! Thanks for replying. I couldn't establish the link between factorial and expontential. 2^n < kn! = kn(n-1)(n-2)...1 for n>n0 this would be the definition of big O. but how can i prove it? what values of k and n0 do i have to take?? thank you very much!

Hi experts, I came across a question to determine whether 2^n = O(n!) is true or false. If its true, i need to prove it. I am aware of the list of growing order functions. So i suppose the statement is true. However, I can't prove it. anyone can help me with this? thank you very much!

Hi, I came across this question which i couldn't figure out why the kinetic energy of the table-tennis ball is greater than the bowling ball after collision. Qn: A table-tennis ball is thrown at a stationary bowling ball. The table-tennis ball makes a one-dimensional elastic collision and...

Hi. I have to first say that I am not a teacher. I can't explain this in terms of energies. but, this problem can also be explained using kinematics. When the stone is being thrown 30 degrees upwards at initial velocity, u, at the roof of the building, the velocity deceases until 0 ms-1...

Oops. sorry. I suppose your mgh refers to work done by gravity and not by car. yea? In that sense, there will have no contradictions. hehh. :shy:

Exactly. This is why previously, I said: So, work done by car is +mgh. yea? This somehow contradicts to "Work Done (by all forces except gravity) + Work Done by gravity = ΔKE" you wrote. In conclusion, Net Work Done = ΔKE cannot be used directly if the movement involves moving...

What my solution to my question is: Resolving the horizontal component of velocity at point B, then using K.E. formula, 0.5(m)(v2) to find the K.E. gained. Next step, Finding GPE, which is, GPE = (m)(9.81)(10) Next, Sum up K.E. and GPE., and that will be the NET work done for the car...

alright, does your including gravity mean the weight of the car? or do u mean the sum of upward force & the weight of the car? because, now that the car is moving 45 degrees upwards at a constant acceleration. to find the total work done from point A(bottom of a slope) at rest, to point...

Hi, thank you very much for enlightening me. If I did not mistaken you, what you mean is, Work-Energy Theorem Net Work Done = Change in K.E., where change in K.E. = Change in G.P.E. + Change in K.E1. so the problem here lies in the term(change in K.E.) being used. correct? if the Net...

Hi, i am having a slight confusion with this theorem. I understand that if a car travels horizontally for s m at the uniform acceleration, the Net Work Done = Change in K.E. (by Work-Energy Theorem) The change in K.E. is the amount of joules required to exert the amount of Net Force on the...