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Sedaia [141]
3 years ago
7

Consider the equation . The dimensions of the variables v, x, and t are , , and , respectively. The numerical factor 3 is dimens

ionless. What must be the dimensions of the variable z, such that both sides of the equation have the same dimensions?
Physics
1 answer:
Vinvika [58]3 years ago
5 0

Answer:

Part of the question is missing but here is the equation for the function;

Consider the equation v = (1/3)zxt2. The dimensions of the variables v, x, and t are [L/T], [L], and [T] respectively.

Answer = The dimension for z = 1/T3 i.e 1/ T - raised to power 3

Explanation:

What is applied is the principle of dimensional homogenuity

From the equation V = (1/3)zxt2.

  • V has  a dimension of [L/T]
  • x has a dimension of [L]
  • t has a dimension of [T]
  • from the equation, make z the subject of the relation
  • z = v/xt2 where 1/3 is treated as a constant
  • Substituting into the equation for z
  • z = L/T / L x T2
  • the dimension for z = 1/T3 i.e 1/ T - raised to power 3
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