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Identify the dimensions of each variable.

Computer Science

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Answer

LHS=RHS\text{LHS} = \text{RHS}

Step-by-step solution

4 steps

  1. Identify the dimensions of each variable.

    • Velocity (vv): The dimension of velocity is length per unit time. [v]=[L][T]−1[v] = [L][T]^{-1}
    • Tension (TT): Tension is a force. Force is mass times acceleration. Acceleration is length per unit time squared. [T]=[M][L][T]−2[T] = [M][L][T]^{-2}
    • Length (LL): The dimension of length is length. [L]=[L][L] = [L]
    • Mass (MM): The dimension of mass is mass. [M]=[M][M] = [M]

  2. Substitute the dimensions into the right-hand side (RHS) of the given formula v=TLMv = \sqrt{\frac{TL}{M}}.

    RHS=[T][L][M]\text{RHS} = \sqrt{\frac{[T][L]}{[M]}} RHS=([M][L][T]−2)([L])[M]\text{RHS} = \sqrt{\frac{([M][L][T]^{-2})([L])}{[M]}}

  3. Simplify the dimensions on the RHS.

    RHS=[M][L]2[T]−2[M]\text{RHS} = \sqrt{\frac{[M][L]^2[T]^{-2}}{[M]}}

    Cancel out [M][M] from the numerator and denominator:

    RHS=[L]2[T]−2\text{RHS} = \sqrt{[L]^2[T]^{-2}}

    Take the square root:

    RHS=([L]2[T]−2)12\text{RHS} = ([L]^2[T]^{-2})^{\frac{1}{2}} RHS=[L]2×12[T]−2×12\text{RHS} = [L]^{2 \times \frac{1}{2}}[T]^{-2 \times \frac{1}{2}} RHS=[L][T]−1\text{RHS} = [L][T]^{-1}

  4. Compare the dimensions of the RHS with the dimensions of the LHS (vv).

    The dimension of the LHS is [v]=[L][T]−1[v] = [L][T]^{-1}.
    The dimension of the RHS is [L][T]−1[L][T]^{-1}.
    Since the dimensions on both sides of the equation are the same, the formula is dimensionally valid.

    The formula is dimensionally valid because:

    LHS=[L][T]−1\text{LHS} = [L][T]^{-1} RHS=[L][T]−1\text{RHS} = [L][T]^{-1} LHS=RHS\text{LHS} = \text{RHS}
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