Structures and Forces - What is force and external forces. Question 1) Jacob and Ravleen are moving furniture. They need to move the filing cabinet over 6 feet and it is very heavy. Jacob wants to apply force and push it by putting his hands on point A because he thinks it would be easier to move it from the top as there is less weight on the top. Ravleen says it's better to push at point B because the center of gravity is lower. Which person is correct and EXPLAIN why and what may happen if it is pushed in the wrong spot.

The restoring force that prevents the furniture from tipping over when pushing the furniture towards the right, F= W × D

Where;

W = The weight of the furniture

D = The horizontal distance from the center of gravity of the furniture, CG and the to the lower right corner of the furniture

Given that A > D, for the furniture to tip over when pushing above the CG, we have;

F1 × A ≤ W × D

∴ F1 << W or for a force much smaller than W, the weight of the furniture applied above the center of gravity, the furniture will tip over

However when the force F2 is applied, where C < D, we have for no tipping over;

F2 × C ≤ W × D, therefore, given that C < D, when a force much higher than the weight, W, is applied at F2, the furniture will remain upright

Therefore, it is better push at a point lower than or equal to the center of gravity, whereby the point B is lower than the center of gravity, it is better to push at point B, Ravleen is correct and it is better to push at point B because the center of gravity is lower

If the furniture is pushed at the wrong spot it may tip over

Please see attached drawing

Explanation:

answer: the enthalpy of the overall chemical reaction is +87.88 kj or +88 kj.

according to hess's law, the overall enthalpy of a reaction is the difference between the sum of enthalpy values of all the products and the sum of enthalpy values of all the reactants.

δh(reaction) = ∑nδh(products) - ∑nδh(reactants)

δh(reaction) = [δh +δh] - [δh]

δh(reaction) = [1(-287.02) + 1(0) ] - [1(-374.9)] kj = +87.88 kj

here, δh (pcl3) = -287.02 kj, δh(cl2) = 0, δh(pcl5) = -374.9 kj

therefore, enthalpy of the overall reaction is +87.88 kj or +88 kj.

? you expect us to answer this? ? this is your schoolwork, i don't think we can with this, i think it would considering being