If your mass is 140kg , then your mass is <em>140kg</em>.
It doesn't make a bit of difference what time it is, whether you're happy or sad, sleeping or lifting weights or running, whether it's raining or shining, hot or cold, climbing a mountain or falling out of an airplane, on the surface of a planet, asteroid, comet or star, or floating or falling through empty lonely outer space. Your mass is your mass. The only way it can change is if YOU make changes in yourSELF, like eating a big steak, sweating through a long tough workout, or skipping dinner, or growing to maturity.
(I guess you're already pretty mature. If your mass is 140kg, then you weigh about 308 pounds when you're on Earth.)
2N- if it is sliding at a constant velocity, the net force (total force)=0 so the applied force and friction force are equal to make 0
Answer:
(a) 
(b) 
Explanation:
Given:
*p = charge on proton = 
*e = magnitude of charge on an electron =
*r = distance between the proton and the electron in the Rutherford's atom = 
Part (a):
Since two unlike charges attract each other.
According to Coulomb's law:
Hence, the attractive electrostatic force of attraction acting between an electron and a proton of Rutherford's atom is 
.
Part (b):
Potential energy between two charges separated by a distance r is given by:
So, the potential energy between the electron and the proton of the Rutherford's atom is given by:
Hence, the electrostatic potential energy of the atom is .
Answer:
Yes, because it would roll down slower
Explanation:
friction
To use standard enthalpies of formation to calculate the standard change in enthalpy for the melting of ice. kJ/mol
∆Hf° = standard heat of formation can be use to claculate the std enthalpy change as follows;
∆Hf°reaction = ∆Hf° p - ∆Hf° r
if the reaction is H₂O (s) ---- H₂O(l)
But
∆Hf°reaction = ∆Hf° H₂O(l) - ∆Hf° H₂O (s), which were -285kJ/mol and -291.8kJ/mol (obtained from the back of chemical process analysis text)
∆Hf°reaction = -285.8 kJ/mole - (-291.8 kJ/mole) = 6.0 kJ/mole = 6000 J/mole
heat gained by ice melting = heat lost by cooling water
(mass x ∆Hf°) ice = (mass x Cp x dT) liquid water
mass ice = (mass x Cp x dT) liquid water / (∆Hf°) ice
mass ice = 480 mL x 1.00 g/mL x 4.18 J/gC x (25-0 C) / 6000 J/mole
= 5.66 mole ice = 5.66 mole x 18.0 g /mole = 150 g of ice required to cool the 480mL of water.
