Recent content by thyrgle

Ok thank you! I can take the integral of (KMm/r^3) and get: \frac{KMm}{2r^2} then: \frac{1}{2}mv^2 - \frac{KMm}{2r^2} = \frac{-KMm}{r_{max}} v^2 = 2KMm\times (\frac{1}{2Rr^2}) v = \sqrt{\frac{KM}{R^2}} Thanks again!

Homework Statement Suppose the gravitational force of the Earth on a body was F = \frac{KMm}{r^3}. What escape velocity would a body need to escape the gravitational field of the Earth? Homework Equations v_e = \sqrt{\frac{2GM}{R}} F_g = G*\frac{m_1*m_2}{r^2} The Attempt at a...

Oh, my bad. Haha the given answer was 4.1, for some reason I didn't even realize using h = 0.5 gives the answer approx. 4.1. My bad thanks again!

Actually never mind I should use h = 2 not 1.5 for the first part of my original attempt and then everything else should be the same. 1.5*9.8 = 14.7 14.7 = (1/2)(0.5833333)(w^2) 29.4 = 0.5833(w^2) 50.4 = w^2 w = 8.199 0.5*9.8*1 (mgh) 4.9 = (1/2)(0.5833)(w^2) 9.8 = (0.5833)w^2 16.8 = w^2 w =...

I was found using doing the following: I = I_cm + m*d^2 I = (1/12)m*L^2 + m*d^2 d = 0.5 since there is a 0.5 shift from the center. I = (1/12)*1*4 + 1*(0.5^2) I = (0.5833333)

I believe I have found I for the entire rod. The issue is with the height I think. I want to say the height is 1.5 but what about the 0.5 end? I try using m = 1, g = 9.8 and h = 1.5. 1.5*9.8 = 14.7 14.7 = (1/2)(0.5833333)(w^2) 29.4 = 0.5833(w^2) 50.4 = w^2 w = 7.099 But the answer is...

I first started thinking I could take the potential energy the system has before release as (mgh) and then at the bottom i could take mgh = (1/2)I*w^2 and find omega... But I couldn't figure out how to use it right...

Homework Statement A uniform rod (length = 2.0m) is mounted to rotate freely about a horizontal axis that is perpendicular to the rod and that passes through the rod at a point of 0.5m from one end of the rod. If the rod is released from rest in a horizontal position, what is the angular...

Using the Ion Electron Method (acidic conditions) I need to solve: MnO4- + C2O42- -> Mn2+ + CO2 So I tried doing: MnO4- -> Mn2+ And: C2O42- -> CO2 + 2H2O Then I balanced the oxygen atoms with H2O MnO4- -> Mn2+ + 4H2O And: C2O42- -> CO2 + 2H2O Then I added the...