Complete Question:
Suppose 
, 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}](https://tex.z-dn.net/?f=LT%20%3D%20%5BL%5E2T%5E%7B-1%7D%5D%5En%5BLT%5E2%5D%5Em%5C%5C%5C%5CL%5E1T%5E1%20%3D%20%5BL%5E%7B2n%7DT%5E%7B-n%7D%5D%5Ctimes%20%5BL%5EmT%5E%7B2m%7D%5D%5C%5C%5C%5CL%5E1%20%5Ctimes%20T%5E1%20%3D%20%5BL%5E%7B%282n%2Bm%29%7D%5D%20%5Ctimes%20%5BT%5E%7B%28-n%20%2B2m%29%7D%5D%5C%5C%5C%5C1%20%3D%202n%20%2B%20m%20-----%281%29%5C%5C%5C%5C1%20%3D%20-n%20%2B%202m%20----%282%29%5C%5C%5C%5Cfrom%20%20%5C%20%281%29%3B%20%5C%20m%20%3D%201-2n%2C%20%5C%20%5C%20substitute%20%5C%20the%20%5C%20value%20%5C%20of%20%5C%20m%20%5C%20in%5C%20%20%282%29%5C%5C%5C%5C1%3D%20-n%20%2B2%281-2n%29%5C%5C%5C%5C1%20%3D%20-n%20%2B%202-4n%5C%5C%5C%5C1-2%20%3D%20-5n%5C%5C%5C%5C-1%20%3D%20-5n%5C%5C%5C%5C1%3D%205n%5C%5C%5C%5Cn%20%3D%20%5Cfrac%7B1%7D%7B5%7D%20%5C%5C%5C%5Cm%20%3D%201%20-%202n%5C%5C%5C%5Cm%20%3D%201%20-%202%28%5Cfrac%7B1%7D%7B5%7D%20%29%5C%5C%5C%5Cm%20%3D%201-%20%5Cfrac%7B2%7D%7B5%7D%20%5C%5C%5C%5Cm%20%3D%20%5Cfrac%7B3%7D%7B5%7D)
We know, U = mgh
Substitute the known values,
19.6 = m * 9.8 * 8
m = 19.6 / 78.4
m = 1/4
In short, Your Answer would be 1/4 Kg
Hope this helps!
field is force at a distnce = falling under grav ... contact is force at point of contact
