Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

dondon93

  • 3 years ago

How to prove that (2/5)(sqrt(H/2g)) is equal to (1/5)(sqrt(2H/g)) ?

  • This Question is Closed
  1. dondon93
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    \[prove that \frac{ 2 }{ 5 } \sqrt{\frac{ H }{ 2g }} is equal \to \frac{ 1 }{ 5 }\sqrt{\frac{ 2H }{ g }}\]

  2. RadEn
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    2/5 sqrt(H/2g) = 2/5 sqrt(H)/(sqrt(2) * sqrt(g)) multiply by sqrt(2)/sqrt(2) = 2/5 sqrt(H)/(sqrt(2) * sqrt(g)) * sqrt(2)/sqrt(2) = 2/5 sqrt(2H)/((2)*sqrt(g)) = 2/10 * sqrt(2H)/sqrt(g) = 1/5 * sqtr(2H/g)

  3. dondon93
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Thanks :) That was really helpful :)

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

    \[25\times \left( \frac{ 2 }{ 5 } \times \sqrt{\frac{ H }{ 2g }} \right) = 25\times \left( \frac{ 1 }{ 5 } \times \sqrt{\frac{ 2H }{ g }} \right)\] \[10\times \sqrt{\frac{ H }{ 2g }}=5\times \sqrt{\frac{ 2H }{ g }}\] \[\left( 10\times \sqrt{\frac{ H }{ 2g }} \right)^{2} = \left( 5\times \sqrt{\frac{ 2H }{ g }} \right)^{2}\] \[\frac{ 100H }{ 2g } = \frac{ 25\times2H}{ g }\] \[\frac{ 50H }{ g } = \frac{ 50H }{ g }\] \[\frac{ H }{ g } = \frac{ H }{ g }\]

  5. 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