ok, here's another shot.
If (2/5xmxr^2)(2pi/t) = (2/5xmxr^2)(2pi/t) I can just solve for the Tfinal.
I can cancel some stuff out and get R^2/T = R^2/T and cross multiply.
(1.5x10^11)^2 x 25 days = 5.625x10^23
5.625x10^23 / (7x10^8)^2 = 1147959.18 daysHow am I looking there?
Homework Statement
The Sun rotates on its axis every 25 days. The sun currently has a radius of 7X10^8 m. When it expands into a Red Giant (in about 4 billion years) it will have a radius of 1.5X10^11 m. What will its rotational period be assuming the same mass for both, and they are both...
Ok, another shot.
F = Ke^2/r^2 = 8.99x10^9 x (1.602x10^-19)^2 / (4.399x10^26)^2 = 3.27 x 10^-81
C = PI X 2R = 3.14 x (2X (4.399 x 10^26)) = 2.76 X 10^ 27M
v^2 = (square root of) ((3.27 X 10^-81 x 4.399x10^26) / 9.11x10^-31) = 1.256 X 10^-43
P = V/C = 1.256 X 10^27 / 2.76 X 10^27 = 4.55 X...
Ok, here is a shot.
93bill/2= 46.5 billion x 3x10^8 (speed of light) x 365 days x 24 hours x 3600 seconds = r = 4.399x10^26
F = Ke^2/r^2 = 8.99x10^9 x (1.602x10^-19)^2 / 4.399x10^26 = .00327N
F-MxV^2/r = .00327 = 9.11x10^-31 x V^2 / r
v^2 = (square root of) ((.00327 x 4.399x10^26) /...
In the far future(10^85 years) an “element” called positronium will develop with a diameter of
the current observable universe of 93 billion light years. (Remember that light travels at 3 × 10^8 m/s). This element consists of an electron and a positron, both of which have a mass 9.11 × 10^−31...