F = (mass)(acceleration) = ma
m = 55 kg
Vi = 20 m/s
t = 0.5 s
Vf = 0 m/s (since she was put to rest)
a=(Vf-Vi)/t
a=(0-20)/5
a = 40 m/s^2 (decelerating)
F = ma = (55 kg)(40 m/s^2)
F = 2200 N
Sir! I would have to tell that this is a educational website/app so please don’t type random letters in the question bar if you don’t need help with nothing.
Answer: 0.0014 atm
Explanation:
Given that,
Original pressure of air (P1) = 1.08 atm
Original volume of air (T1) = 145mL
[Convert 145mL to liters
If 1000mL = 1l
145mL = 145/1000 = 0.145L]
New volume of air (V2) = 111L
New pressure of air (P2) = ?
Since pressure and volume are given while temperature is held constant, apply the formula for Boyle's law
P1V1 = P2V2
1.08 atm x 0.145L = P2 x 111L
0.1566 atm•L = 111L•P2
Divide both sides by 111L
0.1566 atm•L/111L = 111L•P2/111L
0.0014 atm = P2
Thus, the new pressure of air when the volume is decreased to 111 L is 0.0014 atm
simple molecules contain two atom bonded together
Answer:
Phosphorylation
Explanation:
Phosphorylation is the term used to describe the transfer of a phosphate group (PO34-) from a donor to a receptor molecule or compound. The process of phosphorylation is usually catalysed by a biological enzyme called KINASE.
Phosphorylation can either be addition of an inorganic phosphate to a molecule. For example, addition of phosphate group to ADP to form ATP, or the donation of a phosphate group to another molecule, e.g transfer of phosphate from ATP to glucose to form glucose-6-phosphate during glycolysis.
