- #1
RedX
- 970
- 3
[tex]\delta(\frac{p_f^2}{2m}-E_i^o-\hbar\omega)=\frac{m}{p_f} \delta(p_f-[2m(E_i^o+\hbar\omega)]^{\frac{1}{2}})[/tex]
Shouldn't the right hand side be multiplued by 2?
Shouldn't the right hand side be multiplued by 2?
marlon said:i am sorry but according to me the fraction before the delta should disappear.
marlon
marlon said:thanks for the polite correction.
marlon
da_willem said:It's always nice to encounter the two of you in a post. All this harmony and warmth...
The energy equation is used to calculate the total energy of a system, taking into account both kinetic and potential energy.
Momentum is a component of kinetic energy in the energy equation. It represents the mass and velocity of an object and is used to calculate the kinetic energy of the system.
The formula for solving for momentum in the energy equation is p = mv, where p represents momentum, m represents mass, and v represents velocity.
In a closed system, momentum is always conserved, meaning the total momentum of the system before an event is equal to the total momentum after the event. This is reflected in the energy equation as the change in kinetic energy is equal to the change in potential energy.
The energy equation and solving for momentum are used in various fields such as physics, engineering, and mechanics. They are essential for understanding the motion and energy of objects in systems, such as in the design of roller coasters, cars, and rockets.