An object that is not accelerating or decelerating has zero net force acting on it. *

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

UNO [17]

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.

5 0

3 years ago

A train starts from rest at station A and accelerates at 0.6 m/s^2 for 60 s. Afterwards it travels with a constant velocity for

Aleks [24]

Answer:

The distance between the station A and B will be:

$x_{A-B}=55.620\: km$

Explanation:

Let's find the distance that the train traveled during 60 seconds.

$x_{1}=x_{0}+v_{0}t+0.5at^{2}$

We know that starts from rest (v(0)=0) and the acceleration is 0.6 m/s², so the distance will be:

$x_{1}=\frac{1}{2}(0.6)(60)^{2}$

$x_{1}=1080\: m$

Now, we need to find the distance after 25 min at a constant speed. To get it, we need to find the speed at the end of the first distance.

$v_{1}=v_{0}+at$

$v_{1}=(0.6)(60)=36\: m/s$

Then the second distance will be:

$x_{2}=v_{1}*1500$

$x_{2}=(36)(1500)=54000\: m$

The final distance is calculated whit the decelerate value:

$v_{f}^{2}=v_{1}^{2}-2ax_{3}$

The final velocity is zero because it rests at station B. The initial velocity will be v(1).

$0=36^{2}-2(1.2)x_{3}$

$x_{3}=\frac{36^{2}}{2(1.2)}$

$x_{3}=540\: m$

Therefore, the distance between the station A and B will be:

$x_{A-B}=x_{1}+x_{2}+x_{3}$

$x_{A-B}=1080+54000+540=55.620\: km$

I hope it helps you!

7 0

3 years ago

Three communications channels in parallel have independent failure modes of 0.1 failure per hour. These components must share a

ozzi

Answer:

the MTTF of the transceiver is 50.17

Explanation:

Given the data in the question;

failure modes = 0.1 failure per hour

system reliability = 0.85

mission time = 5 hours

Now, we know that the reliability equation for this situation is;

R(t) = [ 1 - ( 1 - $e^{-0.1t$ )³] $e^{-t/MTTF$

so we substitute

R(5) = [ 1 - ( 1 - $e^{-0.1(5)$ )³] $e^{-5/MTTF$ = 0.85

[ 1 - ( 1 - $e^{-0.5$ )³] $e^{-5/MTTF$ = 0.85

[ 1 - ( 0.393469 )³] $e^{-5/MTTF$ = 0.85

[ 1 - 0.06091 ] $e^{-5/MTTF$ = 0.85

0.9391 $e^{-5/MTTF$ = 0.85

$e^{-5/MTTF$ = 0.85 / 0.9391

$e^{-5/MTTF$ = 0.90512

MTTF = 5 / -ln( 0.90512 )

MTTF = 50.17

Therefore, the MTTF of the transceiver is 50.17

7 0

3 years ago

I took my dog for a walk, but he wants to go again.

Talja [164]

He probably just like the cool weather or being outside lol

7 0

3 years ago

Consider the same piping system. this time, the same pipe is buried underground. assuming that there is a constant heat flux of

charle [14.2K]

Complete Question

Complete Question is attached below

Answer:

Option A

Explanation:

From the question we are told that:

inner Diameter of pipe $d_i=100^c$

Thickness $t=50mm$

Outer diameter of pipe $d_o=1.1m$

Length $l=5m$

Temperature $T_i=130^oC$

Generally the equation for Heat Balance is mathematically given by

$q*\pi d_oL=mC_p(T_i-T_o)$

Therefore

$T_o=T_i+\frac{q*\pi d_oL}{mC_p}$

$T_o=130+\frac{100*3.142 *1.1*5}{0.5*4000}$

$T_o=129.136^oC$

Therefore the exit temperature of the water.is $T_o=129.136^oC$

Option A

7 0

3 years ago

An object that is not accelerating or decelerating has zero net force acting on it. *​

An object that is not accelerating or decelerating has zero net force acting on it. *