Recent content by lozzajp

I am hoping I have this right but want to make sure. I have 9x10^9 x N x m^2/C^2 and want to use units of cm and μC (micro) 9x10^9 x N x 1m^2/1C^2 x 10000cm^2/1m^2 x 1C^2/1x10^12μC^2 9x10^9 x N x 10000cm^2/1x10^12μC^2 90 x N x 1cm^2/1μC^2 Have I got that right? Havent fallen into any...

So it is impossible without quadratic formula.. guess I have some reading to do thanks for all your help!

thats a good way to do it! thanks for that. do post the intense algebra? :)

i understand that but i don't know how to solve for r (R = s in my part)

Homework Statement where does the gravity cancel out between the Earth and moon?Homework Equations s=3.84405x10^8m earth m = 5.98x10^24kg moon m =7.35x10^22kg G = 6.67x10^-11N m^2/kg^2 mass of object between = 1kg for simplicity F = G x m / r^2 r = ? F Earth = F moon G x Earth / r^2 = G x moon...

person 1, 100kg 10m/s. person 2 50kg (made these up on spot) KE = 50kg x 100m/s KE = 5000kJ (person 2 has 0) post collision, 150kg, 6.7m/s KE = 75kg x 44.89m/s KE = 3366.75kJ a person double the mass and half the velocity only has a KE of 2500kJ?

heat, sound and things. but a higher kinetic energy results in more "damage", where as the person will still be knocked back at the same velocity as momentum is the same. I can't quite get my head around it :/

If they rebound then they will encounter a second collision with the floor.

Homework Statement What is more likely to cause greater injury, a collision with a light person at fast speed or a person with twice the mass at half the velocity? Homework Equations momentum = mv KE = 1/2mv^2 The Attempt at a Solution I am having a little trouble at which is...

Do you mean m1+m2? I did the exercise again using my own way of things and ended up with the right answer. force 1 (gravity and m2) - force 2 (m1 down the incline) / m1+m2 (as in f=ma or a=f/m) =2.9m/s/s which is the answer the book gives, I think i have the idea of tension figured out now...

One thing I enjoy as a new physics student is using formulas. I have been playing with the fundamental formulas in terms of units rather than their names. That is instead of a=v/t m/s/s=(m/s)/s or v=at m/s=(m/s/s) x s m/s=m/s Simple maths yea, but when I am getting to the bigger...

ill look at above post over the next while and see what I can get from it thanks :). just figured out problem 2 on my own, i was rearranging it wrong. s= ut + (0.5) x a x t^2 s/a = ut + (0.5) x t^2 2s/a = ut + t^2 no initial velocity so ut = 0 t^2 = 2s/a

I can manage that for easier problems, but if you look at what I am having trouble with is what equation to use at the right time and how to get there/know that its that one to use. just to add, friction, pressure and electrostatic forces I have not encountered yet :( simply having trouble with...

Ok this will be a long post, sorry in advance. I am an online student of a university and the help I am getting from them is ridiculous, I am also using a book called quicksmart introductory physics, and for the most part it is quite rubbish. 1st Problem: (not homework questions, I am trying...