Answer: At 34°c
Explanation:
Using The Arrhenius Equation:
k = Ae − Ea/RT
k represents rate constant
A represents frequency factor and is constant
R represents gas constant which is = 8.31J/K/mol
Ea represents the activation energy
T represents the absolute temperature.
By taking the natural log of both sides,
ln k = ln A- Ea/RT
Reactions at temperatures T1 and T2 can be written as;
ln k1= ln A− Ea/RT1
ln k2= ln A− Ea/RT2
Therefore,
ln(k1/k2) = −Ea/RT1 + Ea/RT2
Since k2=2k1 this becomes:
ln(1/2) = Ea/R*[1/T2 − 1/T1]
Theefore,
-0.693 = 37.2 x 10^3/8.31 * [ 1/T2 - 1/293]
1/T2 - 1/293 = -1.55 x 10^-4
1/T2 = -1.55 x 10^-4 + 34.13x 10^-4
1/T2 = 32.58 x 10^-4
Therefore T2 = 307K
T2 = 307 - 273 = 34 °c
What a delightful little problem !
Here's how I see it:
When 'C' is touched to 'A', charge flows to 'C' until the two of them are equally charged. So now, 'A' has half of its original charge, and 'C' has the other half.
Then, when 'C' is touched to 'B', charge flows to it until the two of <u>them</u> are equally charged. How much is that ? Well, just before they touch, 'C' has half of an original charge, and 'B' has a full one, so 1/4 of an original charge flows from 'B' to 'C', and then each of them has 3/4 of an original charge.
To review what we have now: 'A' has 1/2 of its original charge, and 'B' has 3/4 of it.
The force between any two charges is:
F = (a constant) x (one charge) x (the other one) / (the distance between them)².
For 'A' and 'B', the distance doesn't change, so we can leave that out of our formula.
The original force between them was 3 = (some constant) x (1 charge) x (1 charge).
The new force between them is F = (the same constant) x (1/2) x (3/4) .
Divide the first equation by the second one, and you have a proportion:
3 / F = 1 / ( 1/2 x 3/4 )
Cross-multiply this proportion:
3 (1/2 x 3/4) = F
F = 3/2 x 3/4 = 9/8 = <em>1.125 newton</em>.
That's my story, and I'm sticking to it.
We can solve the problem by using the law of conservation of energy.
Initially, the rock has only gravitational potential energy, which is given by
where mg is the weigth of the rock (2200 N), while h is the height at which the rock has been released (h=15 m). If we calculate it, we get
Just before hitting the ground, the rock height is zero, so its potential energy is now zero. So the total mechanical energy of the rock now is just kinetic energy:
however, the mechanical energy of the rock must be conserved, so
and so we have that the kinetic energy of the rock just before hitting the ground is equal to its initial potential energy:
The branch of science which deals with celestial objects, space, and the physical universe as a whole.
Online definition so if you’re using it for school work I recommend re-wording!