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Sauron [17]
3 years ago
14

Suppose A = B"Cm, where A has the dimensions LT, B has dimensions L²T-1, and C has dimensions LT2. Then the exponents n and have

the values
Physics
1 answer:
tresset_1 [31]3 years ago
8 0

Complete Question:

Suppose A = B^n C^m, where A has the dimensions LT, B has dimensions L²T⁻¹, and C has dimensions LT². Then the exponents n and m have the values

Answer:

The value of n = ¹/₅

The value of m = ³/₅

Explanation:

Given dimensions;

A = LT

B = L²T⁻¹

C = LT²

The values of n and m are calculated as follows;

LT = [L^2T^{-1}]^n[LT^2]^m\\\\L^1T^1 = [L^{2n}T^{-n}]\times [L^mT^{2m}]\\\\L^1 \times T^1 = [L^{(2n+m)}] \times [T^{(-n +2m)}]\\\\1 = 2n + m -----(1)\\\\1 = -n + 2m ----(2)\\\\from  \ (1); \ m = 1-2n, \ \ substitute \ the \ value \ of \ m \ in\  (2)\\\\1= -n +2(1-2n)\\\\1 = -n + 2-4n\\\\1-2 = -5n\\\\-1 = -5n\\\\1= 5n\\\\n = \frac{1}{5} \\\\m = 1 - 2n\\\\m = 1 - 2(\frac{1}{5} )\\\\m = 1- \frac{2}{5} \\\\m = \frac{3}{5}

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This is done by first multiplying the midpoints by the corresponding frequencies and then dividing the result by the total number of samples (n).

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