The point in which it originates.
The maximum velocity in a banked road, ignoring friction, is given by;
v = Sqrt (Rg tan ∅), where R = Radius of the curved road = 2*1000/2 = 1000 m, g = gravitational acceleration = 9.81 m/s^2, ∅ = Angle of bank.
Substituting;
30 m/s = Sqrt (1000*9.81*tan∅)
30^2 = 1000*9.81*tan∅
tan ∅ = (30^2)/(1000*9.81) = 0.0917
∅ = tan^-1(0.0917) = 5.24°
Therefore, the road has been banked at 5.24°.
Answer:
x = 1.6 + 1.7 t^2 omitting signs
a) at t = 0 x = 1.6 m
b) V = d x / d t = 3.4 t
at t = 0 V = 0
c) A = d^2 x / d t^2 = 3.4 (at t = 0 A = 3.4 m/s^2)
d) x = 1.6 + 1.7 * (4.4)^2 = 34.5 (position at 4.4 sec = 34.5 m)
Answer:
1.24 m/s
Explanation:
Metric unit conversion:
9.25 mm = 0.00925 m
5 mm = 0.005 m
The volume rate that flow through the single pipe is

This volume rate should be constant and divided into the 4 narrower pipes, each of them would have a volume rate of

So the flow speed of each of the narrower pipe is:
