anonymous
  • anonymous
Please Help with Practice question!!! An electrically charged particle accelerates uniformly from rest to speed v while traveling a distance x. a) Show the acceleration of the particle is a= v^2/2x. b) If the particle starts from rest and reaches a speed of 1.8 x 10^7 m/s over a distance of 0.10m, show that its acceleration is 1.6 x 10^15 m/s^2.
Physics
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anonymous
  • anonymous
Please Help with Practice question!!! An electrically charged particle accelerates uniformly from rest to speed v while traveling a distance x. a) Show the acceleration of the particle is a= v^2/2x. b) If the particle starts from rest and reaches a speed of 1.8 x 10^7 m/s over a distance of 0.10m, show that its acceleration is 1.6 x 10^15 m/s^2.
Physics
chestercat
  • chestercat
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anonymous
  • anonymous
I know for B i just plug in the numbers...
shivaniits
  • shivaniits
you know \[v ^{2}-u ^{2}=2as\] and here s=x and u=0 since it started from rest so we have:- \[v ^{2}-0=2ax\] \[a=v ^{2}/2x\]
shivaniits
  • shivaniits
now for second part just put the values for the equation.. hope this helps..!!

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anonymous
  • anonymous
so the equation for part A is ...
anonymous
  • anonymous
v^2-0=2ax?
shivaniits
  • shivaniits
yup that'w what you get after putting in the values x=s and u=0 in more general eqn \[v ^{2}-u ^{2}=2as\]
anonymous
  • anonymous
ok, then for B it comes out like ...
anonymous
  • anonymous
[1.8 x 10^7]^2/ {2*0.1) ?
shivaniits
  • shivaniits
yes..
shivaniits
  • shivaniits
hope this sorted you out..!!
anonymous
  • anonymous
so it equals ...
anonymous
  • anonymous
1.62e + 15 m/s^2 ?
anonymous
  • anonymous
thats what comes up on a calculator.....
shivaniits
  • shivaniits
oh..the answer is \[16.2 \times 10^{14}\]
anonymous
  • anonymous
ok.......
anonymous
  • anonymous
hmm wonder why mine is so off...
shivaniits
  • shivaniits
everything alright..!!

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