Recent content by HareJare

  1. H

    Engineering Highest velocity of a girl running a 60m dash

    Okay thank you for your help but i don't understand how to do it this way.
  2. H

    Engineering Highest velocity of a girl running a 60m dash

    Okay i did that and now calculate the area?
  3. H

    Engineering Highest velocity of a girl running a 60m dash

    8.6 = sqrt(30/a) + ? i don't know how to do the second term
  4. H

    Engineering Highest velocity of a girl running a 60m dash

    So i know the total time is 8.6 s t1 + t2 = 8.6s And from the (v^2 - v0^2 = 2as) i will get that after 15m the velocity which is also the highest velocity v^2 = 30a And then after that the velocity will be constant (s = v*t) 45/t2 Now i am stuck here because i have 3 equations but 4...
  5. H

    Engineering Tension force in two strings supporting a sign

    Yes okay i will think about this next time! Again thank you for your help
  6. H

    Engineering Tension force in two strings supporting a sign

    Thank your for your reply but i don't really understand yet Are you saying that when the force F1x is increased the wire will eventually break? And F1 will always stay the same right? its just the F1y and F1x that will change?
  7. H

    Engineering Tension force in two strings supporting a sign

    First i calculated the sign Force which was 4*9.81 = 39.24N which meant that the force already exceeds the traction force. What i don t understand in the question is how the distance h will avoid this?
  8. H

    Engineering Tips for Starting a Small Business

    OMG i can not believe i had trouble with this :H i solved it now thank you!
  9. H

    Engineering Tips for Starting a Small Business

    It had to be equal to 120 or am i wrong here?
  10. H

    Engineering Tips for Starting a Small Business

    Well i learned about torque and to achieve equilibrium it has to be the same on the left of the torque point and to the right. My problem with this question is i don't know how to deal with I know what to do on the right. Torque RIGHT = 500 * F but the rest i am confused about
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