An elastic collision is one in which the system does not experience a net loss of kinetic energy as a result of the collision. In elastic collisions, momentum and kinetic energy are both conserved.
<h3>Explain about the Elastic Collision?</h3>
A collision between two bodies in physics is referred to as an elastic collision if their combined kinetic energy stays constant. There is no net conversion of kinetic energy into other forms, such as heat, noise, or potential energy, in an ideal, fully elastic collision
An example of an elastic collision is when two balls collide at a pool table. It is an elastic collision when you throw a ball on the ground and it bounces back into your hand because there is no net change in the kinetic energy.
If there is no kinetic energy lost in the impact, the collision is said to be perfectly elastic. A collision is considered to be inelastic if any of the kinetic energy is converted to another kind of energy during the collision.
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Answer: Attach it to clothing, personal flotation device or the person's person.
Explanation:
This particular Question or problem has to do with the rules and regulations or laws governing the use of Personal Watercraft(PWC) in the state of Florida in the United States of America. Hence, one of the rules that is applicable to the use of boats and waterways or the use of personal watercraft in Florida is that in florida, if ones pwc is equipped with an engine cut-off lanyard the person operating it must ATTACH IT TO THE CLOTHING, PERSONAL FLOATATION DEVICE IR THE OPERATORS' PERSON.
Another rule band the use of flammable personal flotation device with Personal Watercraft(PWC) in the state of Florida. All this rules and guildlines are made to limit or minimize hazards.
Answer:
The momentum of the photon is 1.707 x 10⁻²² kg.m/s
Explanation:
Given;
kinetic of electron, K.E = 100 keV = 100,000 eV = 100,000 x 1.6 x 10⁻¹⁹ J = 1.6 x 10⁻¹⁴ J
Kinetic energy is given as;
K.E = ¹/₂mv²
where;
v is speed of the electron

Therefore, the momentum of the photon is 1.707 x 10⁻²² kg.m/s
Answer:
W = -0.480 J
Explanation:
given,
q₁ = 4 μC
q₂ = -4.10 μC


b = 0.381
k = 8.99 × 10⁹ Nm²/C²

![W = [-147.436\times (5.88-2.62)\times 10^{-3}]J](https://tex.z-dn.net/?f=W%20%3D%20%5B-147.436%5Ctimes%20%285.88-2.62%29%5Ctimes%2010%5E%7B-3%7D%5DJ)
W = -0.480 J
Work done by the electric force W = -0.480 J