Qn2 Supposing that [period I] = [mx][iy][9z]. a) Find the value of: i) x, ii) y, iii) z. b) Find the relationship for the above expression on Q2.

Here's the solution to your question: a) To find the values of x, y, and z, we use dimensional analysis. The dimensions of the quantities are: • Period [T] = [T] • Mass [m] = [M] • Length [l] = [L] • Acceleration due to gravity [g] = [LT^-2] The given relationship is [period] = [m^x][l^y][g^z]. Substituting the dimensions: [T] = [M^x][L^y][(LT^-2)^z] [M^0 L^0 T^1] = [M^x L^y L^z T^-2z] [M^0 L^0 T^1] = [M^x L^y+z T^-2z] Now, we equate the exponents of M, L, and T on both sides: For M: x = 0 For L: y+z = 0 For T: 1 = -2z Step 1: Solve for x. From the exponent of M: x = 0 The value of x is 0. Step 2: Solve for z. From the exponent of T: 1 = -2z z = -(1)/(2) The value of z is -(1)/(2). Step 3: Solve for y. From the exponent of L: y+z = 0 Substitute the value of z: y + (-(1)/(2)) = 0 y = (1)/(2) The value of y is (1)/(2). b) The relationship for the above expression is found by substituting the values of x, y, and z back into the original equation. Period = C · m^x l^y g^z Period = C · m^0 l^1/2 g^-1/2 Period = C · 1 · sqrt(l) · (1)/(sqrt(g)) Period = C sqrt((l)/(g)) The relationship is Period = C sqrt((l)/(g)), where C is a dimensionless constant. Send me the next one 📸