Two pistons are in an enclosed volume with fluid in between them, such that movement in one piston affects the location of the other piston, as shown in
the image below. If an explosion occurs in the piston on the left such that it
moves 0.2 m, what is the distance that the piston on the right will move
(assuming that friction can be neglected)? (Recall that work on an object is
equal to the force on the object times the distance it is moved, and that work
is conserved.)
Cross-sectional area = 0.25 m
Cross-sectional area=0.5 m?
A. 0.2 m
B. 0.8 m
C. 0.4 m
O D. 0.1 m

the following is queen elizabeth’s purposes:

- to encourage the parliament that the technique of picking successors to the seat based merely on birthright needs to be reviewed.

- (not) to convince the parliament that her kid should not be measured as a likely choice for heir to the english power.

- to persuade the parliament that merely since she has a child does not assure that the kid will be a capable leader.

- to sway the parliament how shattering it could be if they proceeded discourteously concerning her or her child.

    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