Centripetal acceleration, property of the motion of a body traversing a circular path. The acceleration is directed radially toward the centre of the circle and has a magnitude equal to the square of the body's speed along the curve divided by the distance from the centre of the circle to the moving body.
To solve the problem, we can use the Einstein's famous equivalence between energy and mass:
where
E is the energy
m is the mass
c is the speed of light
In this problem, the mass of the substance is
, so if all this mass would be converted in energy, we would have an energy of
You first subtract the speed at which the man is moving (11 m/s) from the rate the boat is moving (12.4 m/s). Which equals 1.4, then divide it by 6 meters, as the man is moving relative to the boat.
It therefore equals 4.29 s
Answer:
Following are the option are the correct option to the given question
- Plane landing on an aircraft carrier.
-
Rain sticking to a window.
-
Two train cars coupling together.
Explanation:
In the inelastic collisions is the kinetic energy is not preserved because of the some internal friction.In the elastic collision there is shortage of the kinetic capacity.
- When the airplane is landing it has always been in contact with the surface to relieve the inelastic collision condition.
- When then rain falls the clinging to the door once it intersect with as well so it is satisfied the condition of inelastic collision.
- When the two train or cars collide there is loss of kinetic energy so it is satisfied the condition of inelastic collision.
- All the other option is not the example of inelastic collision that's why they are incorrect option.
Distance fallen = 1/2 ( V initial + V final ) *t
We know
a = -9.8 m/s2
t=120s
To find distance fallen, we need to find V final
Use the equation
V final = V initial + a*t
Substitute known values
V final = 0 + (-9.8)(120)
V final = -1176 m/s
Then plug known values to distance fallen equation
Distance fallen = 1/2 ( 0 + 1176 )(120)
= 1/2(1776)(120)
=106,560 m
This way plugging into distance equation is actually the long way. A faster way is to plug the values into
Distance fallen = V initial * t + 1/2(a*t)
We won't need to find V final using another equation.
But anyways, good luck!