U unDRdog Full Member 10+ Year Member 15+ Year Member Joined Aug 21, 2008 Messages 413 Reaction score 1 Aug 9, 2009 #1 Members don't see this ad. v=1/2pi x sqrt (k/m) ..........if m doubles v doubles by a sqrt of 2 (holding everything else constant of course... Thanks
Members don't see this ad. v=1/2pi x sqrt (k/m) ..........if m doubles v doubles by a sqrt of 2 (holding everything else constant of course... Thanks
T thebillsfan Unseasoned Veteran 10+ Year Member Joined Dec 23, 2008 Messages 778 Reaction score 0 Aug 9, 2009 #2 unDRdog said: v=1/2pi x sqrt (k/m) ..........if m doubles v doubles by a sqrt of 2 (holding everything else constant of course... Thanks Click to expand... negative, chief. if m doubles then i believe v would decrease by a sq rt of two. also, v couldn't "double" by a sq rt of 2 anyway, hehe 🙂 Upvote 0 Downvote
unDRdog said: v=1/2pi x sqrt (k/m) ..........if m doubles v doubles by a sqrt of 2 (holding everything else constant of course... Thanks Click to expand... negative, chief. if m doubles then i believe v would decrease by a sq rt of two. also, v couldn't "double" by a sq rt of 2 anyway, hehe 🙂
Pulsar Full Member 10+ Year Member Joined Jun 25, 2009 Messages 54 Reaction score 0 Aug 9, 2009 #3 unDRdog said: v=1/2pi x sqrt (k/m) ..........if m doubles v doubles by a sqrt of 2 (holding everything else constant of course... Thanks Click to expand... thebillsfan is right. sqrt[a/(2*b)] = sqrt[1/2] * sqrt[a/b] = 1/sqrt[2] * sqrt[a/b] Upvote 0 Downvote
unDRdog said: v=1/2pi x sqrt (k/m) ..........if m doubles v doubles by a sqrt of 2 (holding everything else constant of course... Thanks Click to expand... thebillsfan is right. sqrt[a/(2*b)] = sqrt[1/2] * sqrt[a/b] = 1/sqrt[2] * sqrt[a/b]
U unDRdog Full Member 10+ Year Member 15+ Year Member Joined Aug 21, 2008 Messages 413 Reaction score 1 Aug 9, 2009 #4 Thats what i thought!!!! Gold Standard Test 2 number #44 says tough...i'm with you guys...sticking to the faithful EK trick... Upvote 0 Downvote
Thats what i thought!!!! Gold Standard Test 2 number #44 says tough...i'm with you guys...sticking to the faithful EK trick...