Mechanical to electrical - generator
Electrical to mechanical - motor
Hope it helps
<span>It's another energy balance equation, though: energy to start with is the same as energy that you end with. Suppose that we start a distance r0 from the Earth and end a distance r1 from the Moon, then the energy balance gives:
1 v02 - G M / r0 - G m / (D - r0) = 1 v12 - G M / (D - r1) - G m / r1
...where m is the moon's mass.
One simple limit takes D ? ? and 1 v02 ? G M / r0 (the escape velocity equation), to yield:
1 v12 ? G M / r1
v1 ? ?( 2 G M / r1 ) = 2377 m/s.</span>
vf=vi-gt (at peak vf = 0)
vi=gt
44=9.8t
t=4.49 s
time return = time upward = 4.49 s
-- a party balloon that's full of helium and resting on the ceiling
-- a piece of styrofoam
-- a cloud of smoke
-- a bag of hot air
-- a cirrus cloud in the sky
Setting reference frame so that the x axis is along the incline and y is perpendicular to the incline
<span>X: mgsin65 - F = mAx </span>
<span>Y: N - mgcos65 = 0 (N is the normal force on the incline) N = mgcos65 (which we knew) </span>
<span>Moment about center of mass: </span>
<span>Fr = Iα </span>
<span>Now Ax = rα </span>
<span>and F = umgcos65 </span>
<span>mgsin65 - umgcos65 = mrα -------------> gsin65 - ugcos65 = rα (this is the X equation m's cancel) </span>
<span>umgcos65(r) = 0.4mr^2(α) -----------> ugcos65(r) = 0.4r(rα) (This is the moment equation m's cancel) </span>
<span>ugcos65(r) = 0.4r(gsin65 - ugcos65) ( moment equation subbing in X equation for rα) </span>
<span>ugcos65 = 0.4(gsin65 - ugcos65) </span>
<span>1.4ugcos65 = 0.4gsin65 </span>
<span>1.4ucos65 = 0.4sin65 </span>
<span>u = 0.4sin65/1.4cos65 </span>
<span>u = 0.613 </span>