Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

boomerang285

  • 3 years ago

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.

  • This Question is Closed
  1. boomerang285
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    I know for B i just plug in the numbers...

  2. shivaniits
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    you know \[v ^{2}-u ^{2}=2as\] and here s=x and u=0 since it started from rest so we have:- \[v ^{2}-0=2ax\] \[a=v ^{2}/2x\]

  3. shivaniits
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    now for second part just put the values for the equation.. hope this helps..!!

  4. boomerang285
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    so the equation for part A is ...

  5. boomerang285
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    v^2-0=2ax?

  6. shivaniits
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    yup that'w what you get after putting in the values x=s and u=0 in more general eqn \[v ^{2}-u ^{2}=2as\]

  7. boomerang285
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    ok, then for B it comes out like ...

  8. boomerang285
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    [1.8 x 10^7]^2/ {2*0.1) ?

  9. shivaniits
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    yes..

  10. shivaniits
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    hope this sorted you out..!!

  11. boomerang285
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    so it equals ...

  12. boomerang285
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    1.62e + 15 m/s^2 ?

  13. boomerang285
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    thats what comes up on a calculator.....

  14. shivaniits
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    oh..the answer is \[16.2 \times 10^{14}\]

  15. boomerang285
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    ok.......

  16. boomerang285
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    hmm wonder why mine is so off...

  17. shivaniits
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    everything alright..!!

  18. Not the answer you are looking for?
    Search for more explanations.

    • Attachments:

Ask your own question

Sign Up
Find more explanations on OpenStudy
Privacy Policy