Weight equals mass*gravity
W = mg
Given m = 3.1 kg, g = 9.8 m/s^2
W = (3.1)(9.8)
W = 30.38
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°.
This gots to be the answer, average, seawater in the world's oceans has a salinity of approximately 3.5%, or 35 parts per thousand.
1st derivative gives velocity;
d r(t)/ dt = 2t i + 6 j + 4/t k
2nd derivative gives acceleration;
d^2 r(t)/ dt^2 = 2 i - 4/ t^2
Speed ;
Square root of (4 t^2 + 36 + 16/ t^2)
For a given time, like 2 seconds, t will be 2. And answer of speed will be scalar.
