Answer:
If you see in the image above, there is an unbalance force applied while playing tug of war. Since it is 1 vs 2, there is a greater net force in the right side then the left side. If it was 2 vs 2 or 1 vs 1, then they are appling balance force. You can also see in the picture that the arrows are pointing outwards (--->) rather then inwards (<---) because you are pulling the rope not pushing the rope. If you add one person on the left side, then the newtons which is 20N will become to 35N and will be balanced, but since there in only 1 person, there is less force on the left side, the newtons gets subtracted having only 20N. Since you are pulling the rope, the friction is opposite (<---). Since you are pulling the rope, you are using Kinetic force and the rope stays in potential force since it stays constant.
Hope this helps, thank you :) and I am not sure about magnitude I think you can that since there is greater force on the right side, there is more magnitude there.
Answer : The enthalpy of combustion per mole of
is -2815.8 kJ/mol
Explanation :
Enthalpy change : It is defined as the difference in enthalpies of all the product and the reactants each multiplied with their respective number of moles. It is represented as 
The equation used to calculate enthalpy change is of a reaction is:
![\Delta H^o_{rxn}=\sum [n\times \Delta H^o_f(product)]-\sum [n\times \Delta H^o_f(reactant)]](https://tex.z-dn.net/?f=%5CDelta%20H%5Eo_%7Brxn%7D%3D%5Csum%20%5Bn%5Ctimes%20%5CDelta%20H%5Eo_f%28product%29%5D-%5Csum%20%5Bn%5Ctimes%20%5CDelta%20H%5Eo_f%28reactant%29%5D)
The equilibrium reaction follows:

The equation for the enthalpy change of the above reaction is:
![\Delta H^o_{rxn}=[(n_{(CO_2)}\times \Delta H^o_f_{(CO_2)})+(n_{(H_2O)}\times \Delta H^o_f_{(H_2O)})]-[(n_{(C_6H_{12}O_6)}\times \Delta H^o_f_{(C_6H_{12}O_6)})+(n_{(O_2)}\times \Delta H^o_f_{(O_2)})]](https://tex.z-dn.net/?f=%5CDelta%20H%5Eo_%7Brxn%7D%3D%5B%28n_%7B%28CO_2%29%7D%5Ctimes%20%5CDelta%20H%5Eo_f_%7B%28CO_2%29%7D%29%2B%28n_%7B%28H_2O%29%7D%5Ctimes%20%5CDelta%20H%5Eo_f_%7B%28H_2O%29%7D%29%5D-%5B%28n_%7B%28C_6H_%7B12%7DO_6%29%7D%5Ctimes%20%5CDelta%20H%5Eo_f_%7B%28C_6H_%7B12%7DO_6%29%7D%29%2B%28n_%7B%28O_2%29%7D%5Ctimes%20%5CDelta%20H%5Eo_f_%7B%28O_2%29%7D%29%5D)
We are given:

Putting values in above equation, we get:
![\Delta H^o_{rxn}=[(6\times -393.5)+(6\times -285.8)]-[(1\times -1260)+(6\times 0)]=-2815.8kJ/mol](https://tex.z-dn.net/?f=%5CDelta%20H%5Eo_%7Brxn%7D%3D%5B%286%5Ctimes%20-393.5%29%2B%286%5Ctimes%20-285.8%29%5D-%5B%281%5Ctimes%20-1260%29%2B%286%5Ctimes%200%29%5D%3D-2815.8kJ%2Fmol)
Therefore, the enthalpy of combustion per mole of
is -2815.8 kJ/mol
It’s a liquid cause liquid move past one another
The average speed of the horse for the entire motion is 66.67 m/s.
The given parameters;
- <em>distance traveled by the horse, d = 1 km = 1,000 m</em>
- <em>time of motion of the horse, t = 15 s</em>
The average speed of the horse is calculated by dividing the total distance traveled by the horse by the time of motion.
The average speed of the horse is calculated as follows;

Thus, the average speed of the horse for the entire motion is 66.67 m/s.
Learn more here: brainly.com/question/17289046