YumYum247
  • YumYum247
Yelp Me please!!!
Physics
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YumYum247
  • YumYum247
Yelp Me please!!!
Physics
jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
\[F_{net}=ma\] \[F_{applied}-F_{f}=m\frac{\Delta v}{t}\]
YumYum247
  • YumYum247
Normally the push force given to me in N but for some odd reason, it's given in velocity(m/sec). How do i change that m/s into N?!?!?
YumYum247
  • YumYum247
o it gunna look like this....hold up let me draw!!!

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anonymous
  • anonymous
They gave you a CHANGE in velocity (from rest to 1.6 m/s) and a time. Use that to find the acceleration and then it's straight Newton's 2nd
YumYum247
  • YumYum247
|dw:1442450685587:dw|
YumYum247
  • YumYum247
is it gunna be sec^2???????????????????????
anonymous
  • anonymous
yeah, they don't cancel it's kg-m/s², which is a Newton
YumYum247
  • YumYum247
ooooooooooooooooooooooooooooooooooooooooooooooooo!! i c
YumYum247
  • YumYum247
so 0.72Kg X 0.8m/SEc^2 right
YumYum247
  • YumYum247
so from here what do i need to do????????please help
anonymous
  • anonymous
yes. multiply that to get the net force
YumYum247
  • YumYum247
multiply that by wat tho :"(
YumYum247
  • YumYum247
i've already used up all my variables....
YumYum247
  • YumYum247
what's left to multiply tho????????/
anonymous
  • anonymous
|dw:1442451064756:dw|
anonymous
  • anonymous
0.72*0.8 is the right side of the equation
anonymous
  • anonymous
\(F_{applied} - F_f\) is the left side
anonymous
  • anonymous
You're solving for \(F_{applied}\)
YumYum247
  • YumYum247
yes i have the mass (0.72kg) and the 0.8m
anonymous
  • anonymous
|dw:1442451193773:dw|
YumYum247
  • YumYum247
|dw:1442451186851:dw|
anonymous
  • anonymous
|dw:1442451340214:dw|
YumYum247
  • YumYum247
|dw:1442451278883:dw|
anonymous
  • anonymous
sort of, you should end up adding 4.51 N
YumYum247
  • YumYum247
oh yah i see ....sorry....yah it should have been Ff and not FN
YumYum247
  • YumYum247
Alright Thanks.....
anonymous
  • anonymous
you're welcome

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