First, let's put 22 km/h in m/s:

Now the radial force required to keep an object of mass m, moving in circular motion around a radius R, is given by

The force of friction is given by the normal force (here, just the weight, mg) times the static coefficient of friction:

Notice we don't use the kinetic coefficient even though the bike is moving. This is because when the tires meet the road they are momentarily stationary with the road surface. Otherwise the bike is skidding.
Now set these equal, since friction is the only thing providing the ability to accelerate (turn) without skidding off the road in a line tangent to the curve:
Answer:
2.63 cm
Explanation:
Hooke's law gives that the force F is equal to cy where c is spring constant and x is extension
Making c the subject of the formula then

Since F is gm but taking the given mass to be F

By substitution now considering F to be 3.3 kg

Answer:
5235.84 kg
Explanation:
There is one theorem - whose proof I will never remember without having to drag calculus in there - that says that the variation of momentum is equal to the force applied times the time the application last.
As long as the engine isn't ejecting mass - at this point it's a whole new can of worm - we know the force, we know the variation in speed, time to find the mass. But first, let's convert the variation of speed in meters per second. The ship gains 250 kmh,
;

Answer:
The magnitude of gravitational force between two masses is
.
Explanation:
Given that,
Mass of first lead ball, 
Mass of the other lead ball, 
The center of a large ball is separated by 0.057 m from the center of a small ball, r = 0.057 m
We need to find the magnitude of the gravitational force between the masses. It is given by the formula of the gravitational force. It is given by :

So, the magnitude of gravitational force between two masses is
. Hence, this is the required solution.