Answer: ![56.87m/s^{2}](https://tex.z-dn.net/?f=56.87m%2Fs%5E%7B2%7D)
Explanation:
If we make an analysis of the net force
of the rock that was thrown upwards, we will have the following:
(1)
Where:
is the force with which the rock was thrown
is the weight of the rock
Being the weight the relation between the mass
of the rock and the acceleration due gravity
:
(2)
(3)
Substituting (3) in (1):
(4)
(5) This is the net Force on the rock
On the other hand, we know this force is equal to the multiplication of the mass with the acceleration, according to Newton's 2nd Law:
(6)
Finding the acceleration
:
(7)
(8)
Finally:
Acceleration due to gravity
I think it would be d because the spring or whatever was pushing until it reached the farthest it could then it would pull down but idrk
Answer:
The pressure is ![P= - 6.39*10^8Pa](https://tex.z-dn.net/?f=P%3D%20-%20%206.39%2A10%5E8Pa)
The temperature is ![T =1218.63 K](https://tex.z-dn.net/?f=T%20%3D1218.63%20K)
Explanation:
Generally Gibbs free energy is mathematically represented as
![G = E + PV -TS](https://tex.z-dn.net/?f=G%20%3D%20E%20%2B%20PV%20-TS)
Where E is the enthalpy
PV is the pressure volume energy (i.e PV energy)
S is the entropy
T is the temperature
For stability to occur the Gibbs free energy must be equal to zero
Considering Diamond
So at temperature of T = 300 K
![E + PV - TS = 0](https://tex.z-dn.net/?f=E%20%2B%20PV%20-%20TS%20%3D%200)
making P the subject
![P = \frac{TS-E}{V}](https://tex.z-dn.net/?f=P%20%3D%20%5Cfrac%7BTS-E%7D%7BV%7D)
Now substituting 300 K for T , 2900 J for E ,
for V and
for S
![P = \frac{(300 * 2.38)- 2900}{3.42*10^{-6}}](https://tex.z-dn.net/?f=P%20%3D%20%5Cfrac%7B%28300%20%2A%202.38%29-%202900%7D%7B3.42%2A10%5E%7B-6%7D%7D)
![P= - 6.39*10^8Pa](https://tex.z-dn.net/?f=P%3D%20-%20%206.39%2A10%5E8Pa)
The negative sign signifies the direction of the pressure
Given that ![P = 1*0^5Pa](https://tex.z-dn.net/?f=P%20%3D%201%2A0%5E5Pa)
making T the subject
![T = \frac{PV+E}{S}](https://tex.z-dn.net/?f=T%20%3D%20%5Cfrac%7BPV%2BE%7D%7BS%7D)
Substituting into the equation
![T = \frac{1*10^5 * 3.42 *10^{-6}+2900}{2.38}](https://tex.z-dn.net/?f=T%20%3D%20%5Cfrac%7B1%2A10%5E5%20%2A%203.42%20%2A10%5E%7B-6%7D%2B2900%7D%7B2.38%7D)
![T =1218.63 K](https://tex.z-dn.net/?f=T%20%3D1218.63%20K)