It is possible to generate a policy in which common points such as those mentioned above are agreed in order to hire or fire employees in their function of their psychological personality, that is, the character of knowledge and skills. Depending on the company, Test could be created in order to evaluate the psychological skills of the employees, as well as Test to periodically determine how their employees are kept up to date with regard to knowledge. The cumulative filter could be done every semester, for which each employee must exceed a minimum margin of score on these tests, otherwise his position could be at risk.
At the same time, incentives can be generated for the best scores that are rewarded not only with monetary values but also with rest days, coupons in restaurants or sports, which would cause the worker to strive to be constantly learning.
This policy agreement is outside the vision and mission of the company, and whose information must be given to the worker once he begins his work activities.
Answer:
7.7 kN
Explanation:
The capacity of a material having a crack to withstand fracture is referred to as fracture toughness.
It can be expressed by using the formula:
where;
fracture toughness K = 137 MPa
geometry factor Y = 1
applied stress 
= ???
crack length a = 2mm = 0.002
∴
Now, the tensile impact obtained is:
P = A × σ
P = 1728.289 × 4.5
P = 7777.30 N
P = 7.7 kN
Answer:
Relative density = 0.7 or 70%
Explanation:
The following information was provided by this question
Pd = 1.72mg/mg³
Pd max = 1.81 mg/mg³
Pd min = 1.54 mg/mg³
We substitute into the formula. This formula is contained in the attachment.
[(1/1.54)-(1/1.72)]/[1/1.54 - 1/1.81]
= 0.649350 - 0.581395 / 0.649350 - 0.552486
= 0.067955/0.096864
= 0.7015
= 0.7
The relative density is Therefore 0.7 or 70% when converted to percentage
Explanation:
150 divide by 150 and that how you do the is you what to divide together 15/ 150 you welcome have a good day is you need something else
Answer:
lead dioxide,sulfate and lead acid
