Acar’s engine stalls going up a hill and slows down as a result. which statement describes the energies of the car?
a) the kinetic energy increases and the gravitational potential energy remains the same. c) the kinetic energy decreases and the gravitational potential energy remains the same. b) the kinetic energy increases and the gravitational potential energy decreases. d) the kinetic energy decreases and the gravitational potential energy increases.

D) The kinetic energy decreases and the gravitational potential energy increases.

Explanation:

Kinetic energy is the energy possessed by a body or an object while in motion. This energy is given by 1/2mv², where m is the mass of the object and v is the velocity. Kinetic energy will decrease as a car is going up hill as the speed will be decreasing.Potential gravitational energy is the energy possessed by a body while at rest and is given by mgh, where m is the mass, g is the gravitational acceleration and h is the height. In this case, the potential energy will be increasing as the car is going up hill as the height will be increasing.

    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