Recent content by JohnTheGreat101

as for syllabi of each class, there is no online copy. I can post the class descriptions so hopefully this helps EnM: When presented with a physical situation and asked to solve a particular problem in, for example, electricity and magnetism (i.e. the creation of an electric current by a...

I have not asked. Per my community college, I cannot talk to someone until the next semester starts as they are off for the summer. My Fall semester starts around Mid August.(This also applies to help with transferring since I will be applying this fall .-. ) This is what they told me in my last...

Yes, I am looking to transfer to a four year university from a cc. It follows a semester system.

Hello, I have just taken "Mechanics" or probably better written as Newtonian Mechanics and now I am moving on. I have the option of going into Electricity and Magnetism or Waves, Optics, and Thermodynamics as they both only have the requirement of calculus 2 and the Newtonian Mechanics class I...

Ok so I found the mass of the cable. mgr/gr = m 170/9.8x1.2 = 14.46 kg. COM = 5 kg x 1.5m + 14.46 kg x 1.2 / 14.46 + 5 = 1.28 m

I know the answer is 170 but I am not sure how to get there. I tried doing things backwards g=9.8 t = fr = mgr 0= 170 + tbl - tbr - 5x9.8x1.5 0= 170 + tbl - tbr - 73.5 -96.5 = tbl - tbr -96.5 = 18*9.8 * 0.2 - 18*9.8*1.4 -96.5 does not equal -211.68

I see so then I would just apply the equation: m1v1=m2v2 and solve for V1?

This is a practice problem so I know that the answer is 750 m/.s. Not totally sure what to do with the friction or if any of my listed equations are relevant here but here goes what I've tried: .45 kg = v(0.8 sec) + 1/2(-9.8 *0.1435) * (0.8)^2 .45 kg = v(0.8) + 1/2(-1.4063) * 0.64 .45=v(0.8) -...