Units, Dimensions and Vectors

4

Step 1: Use the formula A·D = Ax·Dx + Ay·Dy. Step 2: Ax = 4, Ay = 5; Dx = 6, Dy = −4. Step 3: A·D = (4)(6) + (5)(−4) = 24 + (−20). Step 4: A·D = 24 − 20 = 4. Why others are wrong: 44 results from adding instead of algebraically combining (24 + 20 = 44 ignoring the negative sign); 24 considers only the î components; −20 considers only the ĵ components.

46 units in the −z direction (−46k̂)

Step 1: Expand C × D = (4î + 5ĵ) × (6î − 4ĵ). Step 2: = 4(6)(î×î) + 4(−4)(î×ĵ) + 5(6)(ĵ×î) + 5(−4)(ĵ×ĵ). Step 3: Use: î×î = 0, î×ĵ = k̂, ĵ×î = −k̂, ĵ×ĵ = 0. Result = 0 + (−16)k̂ + (30)(−k̂) + 0 = −16k̂ − 30k̂ = −46k̂. Step 4: Magnitude = 46 units, direction = −z direction. Why others are wrong: +46k̂ reverses the sign (confusing ĵ×î with î×ĵ); −4k̂ is the dot product result mistakenly used; −56k̂ results from arithmetic errors.

Both terms have dimensions ML²T⁻², so the equation is dimensionally consistent

Step 1: Find dimensions of (1/2)mv²: [m] = M, [v²] = (LT⁻¹)² = L²T⁻². So [mv²] = ML²T⁻². Step 2: Find dimensions of mgh: [m] = M, [g] = LT⁻², [h] = L. So [mgh] = M·LT⁻²·L = ML²T⁻². Step 3: Both terms have the same dimensions ML²T⁻² — this is also the dimension of energy (Work = Force × distance = MLT⁻² × L = ML²T⁻²). Step 4: The equation is dimensionally consistent. Why others are wrong: Options B and C contain arithmetic errors in working out dimensions of mgh; Option D incorrectly states dimensions as MLT⁻² (which is force, not energy).

25 units and 43.3 units

Step 1: For a vector of magnitude A making angle θ with x-axis: Ax = A·cosθ, Ay = A·sinθ. Step 2: Ax = 50 × cos60° = 50 × 0.5 = 25 units. Step 3: Ay = 50 × sin60° = 50 × (√3/2) = 50 × 0.866 = 43.3 units. Step 4: x-component = 25 units, y-component = 43.3 units. Why others are wrong: 43.3 and 25 reverses the roles of sin and cos (angle measured from x-axis uses cos for x); 35.4 and 35.4 corresponds to 45° not 60°; 50 and 50 ignores trigonometry entirely.