Ask Question
17 January, 16:19

How much sooner does the box reach the bottom of the incline than the disk? express your answer in terms of some or all of the variables m, h, θ, and r, as well as the acceleration due to gravity g. view available hint (s) ?

+2
Answers (1)
  1. 17 January, 17:25
    0
    The expression for time difference is

    Δt = td - tb

    Substitute 1/sinФ√3h/g for td and 1/sinФ√2h/g for tb

    Δt = 1/sinФ√3h/g - 1/sinФ√2h/g

    =1/sinФ√h/g (√3 - √2)

    = 0.32 [1/sinФ √h/g]

    The time where by the lock comes down to the bottom of inclined before the disk is 0.32 [ 1 / sinФ √h/g]
Know the Answer?
Not Sure About the Answer?
Find an answer to your question ✅ “How much sooner does the box reach the bottom of the incline than the disk? express your answer in terms of some or all of the variables m, ...” in 📘 Physics if you're in doubt about the correctness of the answers or there's no answer, then try to use the smart search and find answers to the similar questions.
Search for Other Answers