The photometer is a device that converts light to voltage which is read out by the digital multimeter (dmm). this is due to the photoelectric effect. (einstein won his nobel prize for his explanation of the photoelectric effect.) in the photoelectric effect, light hits a piece of metal. quantum mechanics tells us that light can be thought of as tiny packets of energy called photons. the photons are absorbed by electrons and increase the energy of the electrons. some of the electrons become so energetic that they escape from the metal. in other words electrons are emitted from the metal, producing a current that is amplified by the photometer. the current goes through a resistor in the photometer. ohm's law (v=ir) tells us that current i going through a resistor r is associated with a voltage drop v. the dmm measures the voltage drop across the resistor. saturation occurs when the metal emits the maximum flux of electrons that it can produce. (or it may be the maximum amplification of the photometer.)chrome-extension: ///newtab. html

Ibelieve your answer is b.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

W= fd
where f is an applied force and d is the distance over which it is applied. when a parachuting skydiver is at constant velocity, he is experiencing "terminal velocity", the point at which his acceleration down is canceled out by air resistance. that means that the resisting force of the air is equal to his weight, but in the opposite direction. with this, we can solve for the work done by the force.
w = fd
w = (92.0)(9.81 m/s^2) (325)
w= 2.93 x 10^5 j
the value for work in this case is positive because it is pushing up on the skydiver, who has a downward force of gravity.
therefore the answer is 2.93 x 10^5 j