Explanation:
It is given that,
Mass of the tackler, m₁ = 120 kg
Velocity of tackler, u₁ = 3 m/s
Mass, m₂ = 91 kg
Velocity, u₂ = -7.5 m/s
We need to find the mutual velocity immediately the collision. It is the case of inelastic collision such that,
v = -1.5 m/s
Hence, their mutual velocity after the collision is 1.5 m/s and it is moving in the same direction as the halfback was moving initially. Hence, this is the required solution.
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.
That is because there are other forces like the friction forces that apply differently on both of them. The frictional forces applied to the sled are smaller than they are on the father, for example, so it's possible for him to pull it.
Change in pressure : 772-753 = 19 mm Hg
Since 1 mm Hg = 133 Pa Hg
Then, 19 mm Hg = 2527 Pa Hg
Area of rectangular pool = L* B = 11 x 21 = 231 m^2
To find the force, we multiply the area by the change in pressure: 2527 (231) = 583737N
