<span>Traveled Downstream a distance of 33 Mi and then came right back. If the speed of the current was 12 mph and the total trip took 3 hours and 40 minutes.
Let S = boat speed in still water then (s + 12) = downstream speed (s -12) = upstream speed
Given Time = 3 hours 40 minutes = 220 minutes = (220/60) h = (11/3) h Time = Distance/Speed
33/(s +12) + 33/(s-12) = 11/3 3{33(s-12) + 33(s +12)} = 11(s+12) (s -12) 99(s -12 + s + 12) = 11(</span> s^{2} + 12 s -12 s -144) 99(2 s) = 11(s^{2} -144) 198 s/11 = (s^{2} -144) 18 s = (s^{2} -144) (s^{2} - 18 s - 144) = 0 s^{2} - 24 s + 6 s -144 =0 s(s- 24) + 6(s -24) =0 (s -24) (s + 6) = 0 s -24 = 0, s + 6 =0 s = 24, s = -6 Answer) s = 24 mph is the average speed of the boat relative to the water.
Answer:
53.08 ft
Step-by-step explanation:
The pressure at the bottom of the tank, P = ρgh where ρ = density of water = 1000 kg/m³, g = acceleration due to gravity = 9.8 m/s² and h = depth of water in tank.
Since the pressure at the bottom of the tank, P = ρgh, making h subject of the formula, we have
h = P/ρg
Since P = 23 psig = 23 × 1 psig = 23 × 6894.76 Pa = 158579.48 Pa
Substituting the values of the other variables into h, we have
h = P/ρg
h = 158579.48 Pa/(1000 kg/m³ × 9.8 m/s²)
h = 158579.48 Pa/(9800 kg/m²s²)
h = 16.18 m
h = 16.18 × 1 m
h = 16.18 × 3.28 ft
h = 53.08 ft
Answer:
d
Step-by-step explanation:
This is point-slope form, and these coordinates make both sides equal to 0.
