Explanation:
Given that,
Work done to stretch the spring, W = 130 J
Distance, x = 0.1 m
(a) We know that work done in stretching the spring is as follows :
(b) If additional distance is 0.1 m i.e. x = 0.1 + 0.1 = 0.2 m
So,
So, the new work is more than 130 J.
Answer: 0.067 s
Explanation:s = Ut + 1/2at^2
0.6 = 9t + 0.5 *10 *t^2
Where a = g =10m/s/s
Solving the quadratic equation
5t^2 + 9t - 0.6=0,
t= 0.067 s and - 1.7 s
Of which 0.067 s is a valid time
Answer:
(A) 60 J
Explanation:
At state 1
KE₁=100 J
At state 2
KE₂ = 0
U₂=80 J
Given that surface is rough so friction force will act in opposite to the direction of motion
Lets take work done by friction = Wfr
From work power energy
Work done by all forces = Change in kinetic energy
Wfr + U₂=ΔKE
Wfr+80 = 100
Wfr= 20 J
Now when book slides from top position then
Wfr+ U = KEf - KEi
-20 + 80 = KEf-0
KEf= 60 J
(A) 60 J
A. because everything is balanced.
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.
