Answer:
H = 3.9 m
Explanation:
mass (m) = 48 kg
initial velocity (initial speed) (U) = 8.9 m/s
final velocity (V) = 1.6 m/s
acceleration due to gravity (g) = 9.8 m/s^{2}
find the height she raised her self to as she crosses the bar (H)
from energy conservation, the change in kinetic energy = change in potential energy
0.5m(V^{2} - [test]U^{2}[/tex]) = mg(H-h)
where h = initial height = 0 since she was on the ground
the equation becomes
0.5m(V^{2} - [test]U^{2}[/tex]) = mgH
0.5 x 48 x (1.6^{2} - [test]8.9^{2}[/tex]) = 48 x 9.8 x H
-1839.6 = 470.4 H (the negative sign indicates a decrease in kinetic energy so we would not be making use of it further)
H = 3.9 m
First, balance the reaction:
_ KClO₃ ==> _ KCl + _ O₂
As is, there are 3 O's on the left and 2 O's on the right, so there needs to be a 2:3 ratio of KClO₃ to O₂. Then there are 2 K's and 2 Cl's among the reactants, so we have a 1:1 ratio of KClO₃ to KCl :
2 KClO₃ ==> 2 KCl + 3 O₂
Since we start with a known quantity of O₂, let's divide each coefficient by 3.
2/3 KClO₃ ==> 2/3 KCl + O₂
Next, look up the molar masses of each element involved:
• K: 39.0983 g/mol
• Cl: 35.453 g/mol
• O: 15.999 g/mol
Convert 10 g of O₂ to moles:
(10 g) / (31.998 g/mol) ≈ 0.31252 mol
The balanced reaction shows that we need 2/3 mol KClO₃ for every mole of O₂. So to produce 10 g of O₂, we need
(2/3 (mol KClO₃)/(mol O₂)) × (0.31252 mol O₂) ≈ 0.20835 mol KClO₃
KClO₃ has a total molar mass of about 122.549 g/mol. Then the reaction requires a mass of
(0.20835 mol) × (122.549 g/mol) ≈ 25.532 g
of KClO₃.
Answer: For ideal machine efficiency = 1. Hence M.A = V. R. The V. R of an ideal machine and the practical machine is a constant or is the same for both
