not enough information is given to determine the velocity of the object at time to=0.00s
Using the given equation you get:
E = 1.99x10^-25 / 9.0x10^-6
Divide 1.99 by 9.0: 1.99/9.0 = 0.22
For the scientific notation, when dividing subtract the two exponents:
25 -6 = 19
So you now have 0.22 x 10^-19
Now you need to change the 0.22 to be in scientific notation form:
2.2 x 10^-20
The answer is B.
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°.