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95123

  • 2 years ago

physics 2 see the photo please and help Me :(

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  1. 95123
    • 2 years ago
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    this is the Q

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  2. hosein
    • 2 years ago
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    i think these formulas for that : F=qe &F=ma & this equation :\[y=(1/2)at ^{2}\] if simplify these we have: \[y=(1/2)(qE/m)t ^{2}\] solve this with t=1 E is electric field act on drop

  3. hosein
    • 2 years ago
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    sorry in first formula F=qE not f=qe

  4. 95123
    • 2 years ago
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    why t=1

  5. hosein
    • 2 years ago
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    first say me by the times means any 1 second?

  6. hosein
    • 2 years ago
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    if no put t=L/V in the y=(1/2)at^2

  7. hosein
    • 2 years ago
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    for m use this\[\rho=m/(4/3\Pi r ^{3})\]

  8. hosein
    • 2 years ago
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    did you get it?

  9. 95123
    • 2 years ago
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    wait a minite

  10. hosein
    • 2 years ago
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    sure

  11. 95123
    • 2 years ago
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    not correct

  12. hosein
    • 2 years ago
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    y?

  13. hosein
    • 2 years ago
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    can you explain

  14. 95123
    • 2 years ago
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    i used mastering physics and it tell me its wrong

  15. 95123
    • 2 years ago
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    i have another Q can u help me

  16. hosein
    • 2 years ago
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    i don't solve please wait what's your Q?

  17. 95123
    • 2 years ago
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    this it

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  18. hosein
    • 2 years ago
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    m=3.35*10^-10 right?

  19. 95123
    • 2 years ago
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    wtich one

  20. 95123
    • 2 years ago
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    please see the second Q

  21. hosein
    • 2 years ago
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    ohh this is a another ask am iright?

  22. 95123
    • 2 years ago
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    please help me

  23. hosein
    • 2 years ago
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    second attachment is another ask ?

  24. 95123
    • 2 years ago
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    yes it is another Q

  25. hosein
    • 2 years ago
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    i think x=vt

  26. hosein
    • 2 years ago
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    these asks solve with above formulas

  27. 95123
    • 2 years ago
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    wrong answer

  28. hosein
    • 2 years ago
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    asks end is magnitude of deflection or path that charge move ? if deflection is right use \[y=0.5(qE/m)t ^{2}\]

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