**Question:**in computing momentum, if the mass times the speed is 75 kilograms times 1 meter/second, how would you finish computing that?

and is the energy in a stack of papers that someone is holding equal to the energy of the stack of papers dropping to the floor? why or why not? PLEASE HELP!

**Answers:**

m = 75kg , v = 1m/s

so

momentum = p = mv = 75 x 1 = 75 kgm/s

Yes, conservation of energy.

A stack of papers held in the air has 'gravitational potential energy' (ie PE = mgh) initially.

When you drop the stack, it's decrease in potential energy converts into an increase in kinetic energy (ie KE = ½mv^2)

Energy is never lost, it just changes from one form to another

mathematically

initial energy = final energy

in this case

PE = KE

mgh = ½mv^2

.

for momentum p(momentum)=m(mass)*v(velocity...

so 75*1 = 75

momentum= 75 kg*m/s

the energy in a stack of papers while its held and its energy while its dropping is the same.

E(energy)= an objects kinetic energy + its gravitational potential energy

equation w/ symbols:( E=KE+MGH)

when you're holding the paper all the energy is potential. when its dropped the amount of energy doesn't change, only the form of energy. some of the energy just transferred to kinetic, but the total E is still the same.

hope this helps:)

So you have mass at 75kilo per second, so the energy in a stack of papers that weigh 75 kilo will drop onto the floor 1 meter per second.

That means it takes one second for a stack of papers weighing 75 kilograms to fall one meter.

same amount of energy just different kinds. like the papers that are being held have potential energy and i forgot what the other is called..thinking something with a k not sure.

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