Force [kgms^-2] = mass [kg] x acceleration [ms^-2]
Work = force x distance
Work = [kgms^-2] x [m]
Work = [kgm^2s^-2]
Answer
given,
change in enthalpy = 51 kJ/mole
change in activation energy = 109 kJ/mole
when a reaction is catalysed change in enthalpy between the product and the reactant does not change it remain constant.
where as activation energy of the product and the reactant decreases.
example:
ΔH = 51 kJ/mole
E_a= 83 kJ/mole
here activation energy decrease whereas change in enthalpy remains same.
For the first part, we are looking for Vf when dy=11.0
Upward is positive, downward is negative.
So <span>Vf = square root [2(-9.8)(11.0) + (18.0)^2] </span>
<span>Vf = 10.4 m/s your answer is correct.
For the part b, t is equals to the time took to reach and dy is equals to 11.0
you did, </span>11= 18t m/s-(1/2) 9.8t^2 then <span>-11 + 18t- 9.8t^2. By quadratic formula, for the way down the answer is 2.9 s while on it's way up, the answer is 0.77 s</span><span>
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No. Motion is the thing that when you're moving, you're in it.
But it IS possible for one person to say you're moving and another person to say you're not moving, both at the same time, and both of them are correct !
In 1920, after returning from Army service, he produced a successful model and in 1923 turned it over to the Northeast Electric Company of Rochester for development.
