Aneutral kaon decays into two pions via the process k^0 → π + π − where m of k^0 =500mev/c2 and the pion masses are both 140 mev/c2. if the kaon is at rest, what are the energies (in mev), momenta (in mev/c), and velocities (in units of c) of the pions.

if the kaon moves in the lab with a kinetic energy of 600 mev, and the positive pion is emitted along the kaon’s direction of motion, find the kinetic energies (in mev) and momenta (in mev/c) of each pion.

the kaon has the same lab kinetic energy as in the previous problem, but the positive pion is emitted 45° relative to the kaon’s direction of motion (measured in the kaon’s rest frame). use conservation of energy and momentum to find the kinetic energies (in mev), momenta (in mev/c) and angles (in degrees) of each pion relative to the kaon initial direction.

Ithink it is c ,i hope i you

The modern-day generator works on the principle of electromagnetic induction discovered by michael faraday in 1831-32. faraday discovered that the above flow of electric charges could be induced by moving an electrical conductor, such as a wire that contains electric charges, in a magnetic field.

    m₁v₁ + m₂v₂ = (m₁ + m₂)vmomentum is conserved.the left-hand side of the equation represents the momentum in the system before the collision.the right-hand side represents the momentum in the system after the collisionm₁ and v₁ represents the mass and the velocity of the physics blockm₂ and v₂ represents the mass and velocity of the ball of puttyv represents the final velocity of the combined mass.the question states the two masses stick together. they combined mass has its own velocity with the body mass the sum of the two objects' massestheir combined momentum is (m₁ + m₂)vto find the magitude of (m₁ + m₂)v, we need to add the vectors on the left-hand side of the equation: m₁v₁ + m₂v₂.note the diagram. i have positioned the vectors in accordance with their directions.the magnitude of the resultant can be calculated using the pythagorean theorem since this is a right triangle situation (north and west are perpendicular)|(m₁ + m₂)v| =  √[ (m₁v₁)² + (m₂v₂)² ]|(m₁ + m₂)v| =  √[ (1.5kg·8.0m/s)² + (0.40kg·20m/s)² ]|(m₁ + m₂)v| =  14.4 kg m/sfor two significant figures (assuming that 20m/s is two), this rounds to 14 kg m/s

What the heck r u talkin bout? ?