KE = 1/2 mass X velocity squared
SO, what would KE be if mass = 100 kg and velocity = 5 m/s?
1/2 (100) x ???? = KE
2007-03-24 11:03:41 · 10 answers · asked by Anonymous in Science & Mathematics ➔ Physics
Hey Vito, your mother sleeps with nerds.
2007-03-24 11:09:52 · update #1
I'm thrown by the velocity SQUARED.....I'm trying to teach myself physics here folks.
2007-03-24 11:12:00 · update #2
This is extremely easy. All you do is plug in the different values for the different parts of the equation. KE=1/2mv(squared) so you do it like this:
KE = [1/2 x (100 kg)] x (5m/s)squared
KE= (50kg) x (25 m/s)
KE= 1250
2007-03-24 11:15:12 · answer #1 · answered by Pepper 2 · 1⤊ 0⤋
KE = 1/2 mass X velocity squared
KE=(1/2)x100xsquare of 5=50x25=1250 J
2007-03-29 08:11:17 · answer #2 · answered by ukmudgal 6 · 0⤊ 0⤋
If the standard units in calculating the kinetic energy for mass are the kg and for velocity m/s and KE (you say) =1/2 mass X velocity squared, then,
KE = (100 kg)/2 X(5m/s) squared
= 1250 kg m2/s2
is (kg m2/s2 ) the name of the unit for KE? Numerically like I said if the units are correct for this calculation, then I think that 's you answer, I m curious too.
2007-03-31 09:44:39 · answer #3 · answered by irrelevant to you 1 · 0⤊ 0⤋
Ke =1/2m x v2 mass m = 100kg
Velocity V2 = 5squared
Ke = 1/2 x 100 x 25
Ke = 100 x 25
Ke = 50 x 25
Ke = 1250 divide by 2
Ke = 625J
2007-03-27 17:23:01 · answer #4 · answered by mercytostody 1 · 0⤊ 0⤋
KE = 1/2mv^2
KE = 1/2 (100) (5)^2
KE = 1250 J
hope this helps
2007-03-24 11:13:09 · answer #5 · answered by 7 · 1⤊ 0⤋
1250 = KE
2007-03-24 11:09:53 · answer #6 · answered by Dave C 2 · 0⤊ 0⤋
um... it would be 1/2(100)x25=50*25=1250. it's too terribly difficult
2007-03-24 11:12:23 · answer #7 · answered by frickinaj 1 · 0⤊ 0⤋
750 kg m^2/s^2
2007-03-30 19:29:37 · answer #8 · answered by Anonymous · 0⤊ 0⤋
Your a Nerd.
2007-03-24 11:06:33 · answer #9 · answered by vito d 1 · 0⤊ 0⤋
seriously? you're asking this question? just plug in that 5 m/s in there!
2007-03-24 11:09:55 · answer #10 · answered by g. 4 · 0⤊ 1⤋