A perfectly elastic<span> collision is defined as one in which there is no loss of </span>kinetic energy<span> in the collision. Therefore, we just add the kinetic energies of each system. We calculate as follows:
KE = 0.5(</span>1.0 × 10^3)(12.5 )^2 + 0.5(1.0 × 10^3)(12.5 )^2
KE = 156250 J = 1.6 x 10^5 J -------> OPTION A
Answer:
Part of the question is missing but here is the equation for the function;
Consider the equation v = (1/3)zxt2. The dimensions of the variables v, x, and t are [L/T], [L], and [T] respectively.
Answer = The dimension for z = 1/T3 i.e 1/ T - raised to power 3
Explanation:
What is applied is the principle of dimensional homogenuity
From the equation V = (1/3)zxt2.
- V has a dimension of [L/T]
- t has a dimension of [T]
- from the equation, make z the subject of the relation
- z = v/xt2 where 1/3 is treated as a constant
- Substituting into the equation for z
- z = L/T / L x T2
- the dimension for z = 1/T3 i.e 1/ T - raised to power 3
Answer:
C. Have no change in electrical charge
Explanation:
If a element gains neutron it become an Isotope. The electrical charge do not change for this, only the atomic mass changes when an element gains neutrons.
The electrical charge is affected when there is a variation in the number of electrons or protons in the element.
