Measurement tools and vectors mini quiz

$[vt]=(L/T)T=L$ .

$[at]=(L/T^2)T=L/T$ .

They cannot be added as terms of a position because their dimensions differ.

72.0 $\mathrm{km/h}$ × (1000 m/1 km) × (1 h/3600 s).

20.0 $\mathrm{m/s}$ ; the units km and h cancel.

Vector equality requires the same magnitude and the same orientation.

Two arrows of equal length can point in different directions.

$\Delta x$ = Δr cos 32.0° and Δy = Δr sin 32.0°.

$\Delta x$ ≈ 6.36 m and Δy ≈ 3.97 m; both are positive because of the quadrant.

√(6.36²+3.97²) ≈ 7.50 m.

$\vec A\cdot\vec B=2(3)+\lambda(-2)+(-1)(4)=2-2\lambda$ .

$2-2\lambda=0$ , therefore λ = 1.

$|A+B|^2-|A-B|^2$ = 4 $\vec A\cdot\vec B$ , so $\vec A\cdot\vec B$ = 0.

$\vec A\cdot\vec B=3a-2=0$ .

a = $2/3$ .

The head is three units left and four units up: $A_{x}$ = −3 and $A_{y}$ = +4, in quadrant II.

The 3–4–5 triangle gives |A| = 5.

atan2(4,−3) gives θ ≈ 126.9° from $+x$ , equivalent to 53.1° above $-x$ .

The angle must lie between 90° and 180° because both components place the vector in quadrant II.

The first arrow represents A = (4,1), and the second, translated to its head, represents B = (−1,3).

R = (4−1, 1+3) = (3,4).

|R| = √(3²+4²) = 5.

The final head of the construction lies three units right and four units above the origin.

The digits written in a measured result communicate its precision.

4.50 contains three significant figures; the final zero is deliberate.

Because both vectors are nonzero, |A||B| ≠ 0.

Therefore cos θ = 0 and θ = 90°.

The head lies left of and above the origin.

Therefore $A_{x}$ < 0, $A_{y}$ > 0: quadrant II.

[T] = L^(p+q) T^(−2q).

−2q = 1 gives q = −1/2; p + q = 0 gives p = 1/2.

The dot product is linear in each argument.

A · (B + C) = A · B + A · C.