To identify situations when position, velocity, and /or acceleration change, realizing that change can be in direction or magnitude. If an object's position is described by a function of time, r⃗ (t) (measured from a nonaccelerating reference frame), then the object's velocity is described by the time derivative of the position, v⃗ (t)=dr⃗ (t)dt, and the object's acceleration is described by the time derivative of the velocity, a⃗ (t)=dv⃗ (t)dt=d2r⃗ (t)dt2. It is often convenient to discuss the average of the latter two quantities between times t1 and t2: v⃗ avg(t1,t2)=r⃗ (t2)−r⃗ (t1)t2−t1 and a⃗ avg(t1,t2)=v⃗ (t2)−v⃗ (t1)t2−t1.

Clike bone scan for find heart

    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