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:
6.75 seconds
Explanation:
t = Time taken
u = Initial velocity
v = Final velocity
s = Displacement
a = Acceleration = 16 m/s²
g = Acceleration due to gravity = 9.81 m/s²
Let y be the distance the rocket is accelerating
960-y is the distance traveled in free fall

In free fall

The distance the rocket will keep accelerating is 364.881828749 m
After which it will travel 960-364.881828749 = 595.118171251 m in free fall

The time the rocket is accelerating is 6.75 seconds
The force required to slow the truck was -5020 N
Explanation:
First of all, we find the acceleration of the truck, which is given by

where
v is the final velocity
u is the initial velocity
t is the time
For the truck in this problem,
v = 11.5 m/s
u = 21.9 m/s
t = 2.88 s
So the acceleration is

where the negative sign means that this is a deceleration.
Now we can find the force exerted on the truck, which is given by Newton's second law:

where
m = 1390 kg is the mass of the truck
is the acceleration
And substituting,

So the closest answer among the option is -5020 N.
Learn more about acceleration and forces:
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Here we will say that there is no external torque on the system so we will have

here we know that

where we know that

Also we know that

initial angular speed will be

now from above equation



now we have

so final speed will be 2.41 rad/s