**Question 30**

(a)

(b)

(c)

BC = 8 sin 18° = 2.5 ms

^{-1}BD = 5.0 ms

^{-1}in direction W18° S

** Question 31**

(a) F

_{H}= 100 sin 60° = 87 N(b) F

_{V}= 100 cos 60° = 50N

** Question 32**

v

_{PG}= 250/0.5 = 500 km^{-1}Ev

_{AG}= 50 kmh^{-1}N (not S); hence v_{GA}= 50 kmh^{-1}Sv

_{PA}= v_{PG}+ v_{GA}= 500 kmh^{-1}E + 50 kmh^{-1}S

v

_{PA }= 502.5 kmh^{-1}; angle q = tan^{-1}50/500 = 5.7° ; hence direction is E 5.7° S

** Question 33**

d

_{P}= 4 d_{Q }and r_{P}= 4 r_{Q}(a) A

_{P}= 4p r_{P}^{2}A

_{Q}= 4p (r_{P}/4)^{2}= 4p r_{P}^{2}/16 = A_{Q}/16Hence A

_{P}= A_{Q}/16(b) V

_{P}= 4/3 p r_{P}^{3}V

_{Q}= 4/3 p r_{Q}(1/4)^{3}Hence V

_{P}= V_{Q}/64(c) m

_{P}= r V_{P}m

_{Q}= r V_{Q}m

_{P }= m_{Q}/64(d) r

_{P}= r_{Q}

**Question 34**

(a) V

_{1}= 35 V_{2}4/3 p r

(b)_{1}^{3}= 35 x 4/3 p r_{2}^{2}

**Question 35**

F

_{P}= F_{W }sinQ = 1000 x 3/5 = 600N

F_{N}= F_{W }cosQ = 1000 x 4/5 = 800N

**Question 36**

PART 1: Pointing directly across

Time to cross: t = s/v = 1/6 = 0.167 hr

v_{BW}= 1.6km/0.167hr = 9.6 km/h

Distance travelled by boat relative to ground (using Pythagoras) = √(1^{2}+ 1.6^{2}) = 1.887 km

Velocity of boat relative to ground v_{BG}= 1.887km/0.167hr = 11.3 km/hr_{ }v_{BG}= v_{BW}+ v_{WG }11.3 = 9.6 + v_{WG }v_{WG}= 6 km/h (remember, this is a vector sum)

PART 2: Pointing upstream

v_{BG}= v_{BW}+ v_{WG }v_{BG}= 9.6 + 6 (vectorially)

v_{BG}= 7.5 km/h (from vector sum)

time = s/v = 1.6km/7.5km/h = 0.213 hr

sinq = 6/9.6

q = 38° upstream

Square both sides of the equation to give: Z^{2}= R^{2}+ 4p^{2}L^{2}f^{2 }

Rearrange the equation to: Z^{2}= 4p^{2}L^{2}f^{2}+ R^{2 }

This is now in the form y = mx + c

where the slope (m) = 4p^{2}L^{2}and the intercept on the y-axis (c) = R^{2}.

Plot the graph:

Slope = (44100 - 10000)/90000 = 0.379

4p^{2}L^{2}= 0.379

L = 0.098 metres

Intercept = 10000 = R^{2 }

Therefore R = 100 ohm

**Question 38**

* Return to Physics Text Home
Page*