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Mekhanik [1.2K]
2 years ago
9

Pam Jones worked for 41 years at the same company and had positive performance ratings and personnel records. She needed a calcu

lator for work which she purchased with her own money but was not reimbursed because she lost the receipt. Later, a security guard stopped her as she was leaving work and discovered the calculator in her belongings. After a brief internal investigation, she was fired and it was announced through internal notices that she had committed a theft. The employee sued for libel, saying the company used her as an example to prevent other thefts.a) What are the key issues in this case?b) As an HR Director, how would you have handled this case?
Engineering
1 answer:
UNO [17]2 years ago
5 0

Answer:

a) The key issues are the sue for libel and the evidence.

b) I would make a deal with her and implement a security program in the company.

Explanation:

The main issue in this case is that Pam Jones sued the company for libel, and the company remains in a position in which it has to prove that the internal investigation followed the right steps and indeed, the proves reflected that she was guilty and the fact that she got fired was correct.

The important here is exactly that the theft can be proved.

As an HR Director, I would give the correct proves in order to win the case, and if that is impossible, because of the tools and evidence, I would make a deal with her where both parts can be adequate to the problem.

She can´t be working again in the company but she can get financed according to her working years; also I would use this case as a growing opportunity by implementing new security methods that give more confidence between the company and its employees.

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Discuss the nature of materials causing turbidity in
Anestetic [448]

Answer:

a

Explanation:

5 0
2 years ago
Read 2 more answers
Poles are values of Laplace transform variable, s, that make denominator of transfer function zero. Zeros are values of Laplace
Ostrovityanka [42]

Answer:

Zero 1 = -1

Zero 2 = -3

Pole 1 = 0

Pole 2 = -2

Pole 3 = -4

Pole 4 = -6

Gain = 4

Explanation:

For any given transfer function, the general form is given as

T.F = k [N(s)] ÷ [D(s)]

where k = gain of the transfer function

N(s) is the numerator polynomial of the transfer function whose roots are the zeros of the transfer function.

D(s) is the denominator polynomial of the transfer function whose roots are the poles of the transfer function.

k [N(s)] = 4s² + 16s + 12 = 4[s² + 4s + 3]

it is evident that

Gain = k = 4

N(s) = (s² + 4s + 3) = (s² + s + 3s + 3)

= s(s + 1) + 3 (s + 1) = (s + 1)(s + 3)

The zeros are -1 and -3

D(s) = s⁴ + 12s³ + 44s² + 48s

= s(s³ + 12s² + 44s + 48)

= s(s + 2)(s + 4)(s + 6)

The roots are then, 0, -2, -4 and -6.

Hope this Helps!!!

3 0
2 years ago
The legend that Benjamin Franklin flew a kite as a storm approached is only a legend—he was neither stupid nor suicidal. Suppose
Delicious77 [7]

Answer: 0.93 mA

Explanation:

In order to calculate the current passing through the water layer, as we have the potential difference between the ends of the string as a given, assuming that we can apply Ohm’s law, we need to calculate the resistance of the water layer.

We can express the resistance as follows:

R = ρ.L/A

In order to calculate the area A, we can assume that the string is a cylinder with a circular cross-section, so the Area of the water layer can be written as follows:

A= π(r22 – r12) = π( (0.0025)2-(0.002)2 ) m2 = 7.07 . 10-6 m2

Replacing by the values, we get R as follows:

R = 1.4 1010 Ω

Applying Ohm’s Law, and solving for the current I:

I = V/R = 130 106 V / 1.4 1010 Ω = 0.93 mA

7 0
2 years ago
An 80-L vessel contains 4 kg of refrigerant-134a at a pressure of 160kPa. Determine (a) the temperature, (b) the quality, (c) th
makvit [3.9K]

Answer:

temperature -15.6 C, quality x=0.646, enthalpy h=667.20 KJ, volume of vapor phase Vg= 79.8 L

Explanation:

property table for R-134a

https://www.ohio.edu/mechanical/thermo/property_tables/R134a/R134a_PresSat.html

at 160 KPa , temperature = -15.66 C

quality x=mass of vapour/ total mass of liq-vap mixture

alternaternately: x=(v-vf)/(vg-vf)    

v=total volume i.e. volume of container"80L"   80L=0.08 cubic meter

vf=vol of liquid phase  vg=vol of vapor phase vf, vg values at 160Kpa

x=(0.08-0.0007437)/(0.1235-0.0007437)=0.646

enthalpy

h=hf+xhfg          hf, hfg values at 160Kpa

h=hf+xhfg=31.2+0.646(209.9)=166.80 KJ/Kg

for 4Kg R-134a h=m(166.80 KJ/Kg )=667.20 KJ

volume of vapor phase

vg at 160Kpa=0.1235 cubic meter for quality=1.

in this case quality=0.646 , so it will occupy 64.6% space of the vapor phase at quality=1.

vol. of vapor phase=0.646*0.1235=0.0798 cubic meter=79.8 L

7 0
2 years ago
A single crystal of a metal that has the FCC crystal structure is oriented such that a tensile stress is applied parallel to the
bagirrra123 [75]

The magnitude of applied stress in the direction of 101 is 12.25 MPA and in the direction of 011, it is not defined.                                          

Explanation:        

Given:

tensile stress is applied parallel to the [100] direction

Shear stress is 0.5 MPA.

To calculate:

The magnitude of applied stress in the direction of [101] and [011].

Formula:

zcr=σ cosФ cosλ

Solution:

For in the direction of 101

cosλ = (1)(1)+(0)(0)+(0)(1)/√(1)(2)

cos λ = 1/√2

The magnitude of stress in the direction of 101 is 12.25 MPA

In the direction of 011

We have an angle between 100 and 011

cosλ = (1)(0)+(0)(-1)+(0)(1)/√(1)(2)

cosλ  = 0

Therefore the magnitude of stress to cause a slip in the direction of 011 is not defined.                                                                                                                                      

6 0
2 years ago
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