Explanation :
There are two types of collision i.e. elastic and elastic collision.
- Elastic collision : In this type of collision, the total momentum and the kinetic energy of the particles remains constant.
- Inelastic collision : In this type of collision, only the momentum remains constant while there is some loss of kinetic energy occurs.
From Newton's second law,
F = m a
a is the rate of change of velocity.
There is a inverse relation between the force and the time of collision.
The change in <em><u>momentum</u></em> will remain the same during a collision, the force needed to bring an object to a stop can be <em><u>increased</u></em> if the time of the collision is <u><em>decreased</em></u>.
kinetic energy is given as
KE = (0.5) m v²
given that : v = speed of the bottle in each case = 4 m/s
when m = 0.125 kg
KE = (0.5) m v² = (0.5) (0.125) (4)² = 1 J
when m = 0.250 kg
KE = (0.5) m v² = (0.5) (0.250) (4)² = 2 J
when m = 0.375 kg
KE = (0.5) m v² = (0.5) (0.375) (4)² = 3 J
when m = 0.0.500 kg
KE = (0.5) m v² = (0.5) (0.500) (4)² = 4 J
Answer:
5 atoms of Boron
Explanation:
Boron makes up approximately 15.944% of the mass and the rest of the 84.056% is Fluorine. There is 5 Atoms because Boron atomic mass is 10.811 in 1 molecule of BF3 and you wanted 5 Molecules.
Inertia is the tendency of an object to resist a change in its motion.
To develop this problem it is necessary to apply the concepts related to Work and energy conservation.
By definition we know that the work done by a particle is subject to the force and distance traveled. That is to say,
Where,
F= Force
d = Distance
On the other hand we know that the potential energy of a body is given based on height and weight, that is
The total work done would be given by the conservation and sum of these energies, that is to say
PART A) Applying the work formula,
PART B) Applying the height equation and considering that there is an angle in the distance of 25 degrees and the component we are interested in is the vertical, then
The net work would then be given by
Therefore the net work done by these 2 forces on the cannonball while it is in the cannon barrel is 3355J
