The acceleration (a) of a particle moving with uniform speed (v) in a circle of radius (r) is found to be propotional to r^n and v^m, where

Question

The acceleration (a) of a particle moving with uniform speed (v) in a circle of radius (r) is found to be propotional to r^n and v^m, where n and m are constants. Determine the values of n and m and write the simplest form of an equation of acceleration

in progress 0
Kim Chi 4 years 2021-08-07T19:07:37+00:00 2 Answers 20 views 0

Answers ( )

  1. Gerda
    0
    2021-08-07T19:09:03+00:00
    Reply

    Answer:

    n=-1 and m=2

    Explanation:

    A particle is moving uniformly with acceleration “a”

    Uniform speed is v

    And radius of circle is r

    The acceleration is proportional to

    r^n and v^m

    i.e

    a∝ rⁿ

    a ∝v^m

    Combing the two

    Then, a∝rⁿv^m

    Let k be constant of proportionality

    Then, a=krⁿv^m. Equation 1

    So, we know that the centripetal acceleration keeping an object in circular path is given as

    a=v²/r

    Rearranging

    a=v²r^-1. Equation 2

    So comparing this to the proportional

    Equating equation 1 and 2

    krⁿv^m = v²r^-1

    This shows that,

    k=1

    rⁿ = r^-1

    Then, n =-1

    Also, v^m =v²

    Then, m=2

    Therefore,

    n=-1 and m=2

  2. ngocdiep
    0
    2021-08-07T19:09:36+00:00
    Reply

    Answer:

    m=2

    n= -1

    a=Kv²/r

    Explanation:

    using dimensional analysis

    V = meter/second = m/s

    r = meter = m

    a = meter/second²=m/s²

    If a is proportional to r^n v^m

    we have a=r^n v^m =(m)^n (m/s)^m = m/s²

    from law of indices;

    m^n+m/s^m =m/s²

     using system of equations

    n+m=1

    m=2

    so n= -1

    then a=kr^n v^m

    a=Kv²/r

Leave an answer

Browse

Giải phương trình 1 ẩn: x + 2 - 2(x + 1) = -x . Hỏi x = ? ( )