C.) Laser. the light from the laser reflects off the shiny surface as the CD rotates
Answer: C. the rod gains mass and the fur loses mass.
Explanation:Atomic particles have mass. The electron has a mass that is approximately 1/1836 that of the proton and with exchange exchange of charge this is also factored in. The movement of effect described above is known as the triboelectic charging process—charging by friction—which results in a transfer of electrons between the two objects when they are rubbed together. Plastic having a much greater affinity for electrons than animal fur pulls electrons from the atoms of fur, leaving both objects with an imbalance of charge. The plastic rod would have an excess of electrons and the fur has a shortage of electrons. Having an excess of electrons, the plastic is charged negatively and has more mass. In the same vein, the shortage of electrons on the fur leaves it with a positive charge and consequently with lesser mass.
Answer:
C. crust, mantle, core
Explanation:
density increases as you travel from the crust to the inner core
the crust is on top
next is the mantle
and then the core
Answer:
a)
, b)
, c) 
Explanation:
a) The change in the gravitational potential energy of the marble-Earth system is:


b) The change in the elastic potential energy of the spring is equal to the change in the gravitational potential energy, then:

c) The spring constant of the gun is:




Look up pulleys problem through Khan academy and a video should pop up with a problem similar and you should be able to walk through it .