In that case, we would still need Merlin's magic wand and I hope he had a spare one for you and one for me.Originally Posted by mkirkpatrick
In that case, we would still need Merlin's magic wand and I hope he had a spare one for you and one for me.Originally Posted by mkirkpatrick
Time independence: [∂E(g)]²=[∂F(a)×∂r(a)]·[∂F(b)×∂r(b)] and Mass independence: ¶a(t)·¶r(t)=c²
Please remember to ask for an extra magic stick for me? Specially the kind that don't need any battery or recharging.Originally Posted by mkirkpatrick
Time independence: [∂E(g)]²=[∂F(a)×∂r(a)]·[∂F(b)×∂r(b)] and Mass independence: ¶a(t)·¶r(t)=c²
If it doesn't work can I return it for a cash refund? Just wondering how many electrons it have to work continuously and perpetually forever without decreasing effectiveness? I'm afraid since it's solar activated it would not work at night? Is there one activated by the moon or the stars?Originally Posted by mkirkpatrick
Time independence: [∂E(g)]²=[∂F(a)×∂r(a)]·[∂F(b)×∂r(b)] and Mass independence: ¶a(t)·¶r(t)=c²
If I'm truly in orbit within the current of space and time then I won't need it. Just like a planet, it orbits the sun forever and there is no resistance to slow it down.Originally Posted by mkirkpatrick
Time independence: [∂E(g)]²=[∂F(a)×∂r(a)]·[∂F(b)×∂r(b)] and Mass independence: ¶a(t)·¶r(t)=c²
In that case, we letgo of our inertial mass and exchange it with activia, a pure state of energy squared.Originally Posted by mkirkpatrick
Time independence: [∂E(g)]²=[∂F(a)×∂r(a)]·[∂F(b)×∂r(b)] and Mass independence: ¶a(t)·¶r(t)=c²
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