Ascientist at a space research center is doing research on near-earth objects and determining the kinetic energy of a theoretical object. a meteorite with a mass of 12 grams falls to earth. if the meteorite’s velocity is 35 kilometers/second, what is its kinetic energy?

its kinetic energy is 7350kJ 


Kinetic energy is given as =

Now, m = 12 gms = 0.012 kg
 And, velocity = 35 kilometers/second = 35000 m/sec

Kinetic energy is given as =
                        = 6 ×1225 × m/ 
                        = 7350 kJ

Ireally just need to get finished with precal! i’m really sorry i have to do this! hope this !

    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