2007-05-13 07:09:04 · 7 answers · asked by Anonymous in Science & Mathematics ➔ Physics
KE = (1/2)mv^2
KE = (1/2)(10 kg)(3 m/s)^2
KE = 50 J (1 significant digit)
2007-05-13 07:24:30 · answer #1 · answered by Lkk814 3 · 0⤊ 1⤋
Might be more interesting to think about this, I assume you are standing still, but if some one was on the cart( still moving at 3m/s, he weighs 0 lol) and looked over to you now you are moving at 3m/s relative to him, you probably weight 70kg? So your KE is 0.5*70*9=315J If this was a closed system then either it's total internal energy is 315J or 45J(from you looking at him)...
How is this possible? (I'll leave you to think about it)
2007-05-13 08:37:12 · answer #2 · answered by Anonymous · 0⤊ 0⤋
Ek= 1/2 mv²
m=10 kg
v= 3m/s
Ek=?
Ek=1/2 mv²
Ek=1/2(10kg)(3m/s)²
Ek=45 J
2007-05-13 07:25:43 · answer #3 · answered by WELDER® 5 · 0⤊ 0⤋
(10*(3)^2)/2=45
2007-05-13 08:16:45 · answer #4 · answered by homosapien15 2 · 0⤊ 0⤋
Listen in class and apply the formulas you are given.
2007-05-13 07:49:12 · answer #5 · answered by msi_cord 7 · 0⤊ 0⤋
k.e = 0.5 * m * v^2 = 0.5 * 10 * 9 = 45 j
2007-05-13 07:13:04 · answer #6 · answered by the_warper 2 · 0⤊ 0⤋
simple.........apply the frmula ...1/2m*v*V
=>5*9=45
2007-05-13 07:15:33 · answer #7 · answered by kabir c 2 · 0⤊ 0⤋