Arope with length l and mass density l(kg/m) is held in a pile, and you grab an end that sticks up a tiny bit. the pile is then released, and falls downward. as a function of time, what force must your hand apply to the top of the rope to keep it motionless? assume there is no friction of the rope with itself, so the pile is always in freefall.

F=  l  x ( L -  1/2 g t²) g

Explanation:

Given that

length =L m

mass density = l kg/m

Distance cover by time t

y = 1/2 g t²

The remaining length of the pile = L -  1/2 g t²

So the total weight of the remaining length of the pile ,m

m=  l  x ( L -  1/2 g t²)

The force apply to the top of the rope to keep it motionless is F

F= m g

F=  l  x ( L -  1/2 g t²) g

the amount of current in the coil

hello, i am papaguy.

i have found an answer to your question but use it as information.

information: because weak acids/bases have low concentration of hydrogen ions, the addition of water has a large impact on the ph. 2. a difference in 1 unit is a 10x difference in concentration for example; a liquid with ph of 3 is 10x more acidic than a liquid with a ph of 4. therefore, a liquid with a ph of 3 is stronger.

best of luck!