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boomerang285
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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.
 one year ago
 one year ago
boomerang285 Group Title
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.
 one year ago
 one year ago

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boomerang285 Group TitleBest ResponseYou've already chosen the best response.0
I know for B i just plug in the numbers...
 one year ago

shivaniits Group TitleBest ResponseYou've already chosen the best response.0
you know \[v ^{2}u ^{2}=2as\] and here s=x and u=0 since it started from rest so we have: \[v ^{2}0=2ax\] \[a=v ^{2}/2x\]
 one year ago

shivaniits Group TitleBest ResponseYou've already chosen the best response.0
now for second part just put the values for the equation.. hope this helps..!!
 one year ago

boomerang285 Group TitleBest ResponseYou've already chosen the best response.0
so the equation for part A is ...
 one year ago

boomerang285 Group TitleBest ResponseYou've already chosen the best response.0
v^20=2ax?
 one year ago

shivaniits Group TitleBest ResponseYou've already chosen the best response.0
yup that'w what you get after putting in the values x=s and u=0 in more general eqn \[v ^{2}u ^{2}=2as\]
 one year ago

boomerang285 Group TitleBest ResponseYou've already chosen the best response.0
ok, then for B it comes out like ...
 one year ago

boomerang285 Group TitleBest ResponseYou've already chosen the best response.0
[1.8 x 10^7]^2/ {2*0.1) ?
 one year ago

shivaniits Group TitleBest ResponseYou've already chosen the best response.0
hope this sorted you out..!!
 one year ago

boomerang285 Group TitleBest ResponseYou've already chosen the best response.0
so it equals ...
 one year ago

boomerang285 Group TitleBest ResponseYou've already chosen the best response.0
1.62e + 15 m/s^2 ?
 one year ago

boomerang285 Group TitleBest ResponseYou've already chosen the best response.0
thats what comes up on a calculator.....
 one year ago

shivaniits Group TitleBest ResponseYou've already chosen the best response.0
oh..the answer is \[16.2 \times 10^{14}\]
 one year ago

boomerang285 Group TitleBest ResponseYou've already chosen the best response.0
ok.......
 one year ago

boomerang285 Group TitleBest ResponseYou've already chosen the best response.0
hmm wonder why mine is so off...
 one year ago

shivaniits Group TitleBest ResponseYou've already chosen the best response.0
everything alright..!!
 one year ago
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