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3 years ago

13

Joan needs to eliminate some employees for a short while. She and her managers identify those employees who are not meeting perf

ormance expectations and explain that this termination is temporary but they are encouraged to seek other positions elsewhere. How is Jane trying to reduce the size of the workforce here?

1 answer:

tigry1 [53]3 years ago

4 0

Answer: layoff

Explanation:

From the information in the question, we can see that Jane is trying to reduce the size of the workforce here through layoff.

Since Joan explains that the termination is temporary, then it's a layoff. If it were to be firing, the termination won't be temporary but permanent as they can't be recalled by the company. But since the employees are discharged temporarily, it's a layoff.

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A car is travelling to the right with a speed of 40 m/s when the driver slams on the brakes. The car skids for 4 s with constant

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aliya0001 [1]

Answer:

Lower than it should be.

Explanation:

Hello.

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If some water is splashed out of the cylinder, the volume of water will be lower than originally measured, it means that the volume of the solid will be higher than real. In such a way, since the density is in an inversely proportional relationship with volume, as the volume of the solid is wrongly increased, therefore the density of the solid will be lower than in should be.

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An astronaut finds herself in a predicament in which she has become untethered from her shuttle. She figures that she could get

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where,

m=Total mass

$m_O =$ Mass of Object

$V_i =$ Velocity before throwing

$V_f =$ Final Velocity

$V_O =$ Velocity of Object

Our values are:

$m_1=5.3kgm_2=7.9kg\\m_3=10.5kg\\m_A=75kg\\m_{Total}=m=98.7Kg$

Solving to find the final speed, after throwing the object we have

$V_f=\frac{mV_0+m_TV_O}{m-m_O}$

We have three objects. For each object a launch is made so the final mass (denominator) will begin to be subtracted successively. In addition, during each new launch the initial speed will be given for each object thrown again.

That way during each section the equations should be modified depending on the previous one, let's start:

A) $5.3Kg\rightarrow 15m/s$

$V_{f1}=\frac{mV_0+m_TV_O}{m-m_O}$

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$V_{f2}=\frac{(98.7)(0.8511)+(7.9)(11.2)}{98.7-5.3-7.9}$

$V_{f2} = 2.0173m/s$

C) $10.5Kg\rightarrow 7m/s$

$V_{f3}=\frac{mV_{f2}+m_TV_O}{m-m_O}$

$V_{f3}=\frac{(98.7)(2.0173)+(10.5)(7)}{98.7-5.3-7.9-10.5}$

$V_{f3} = 3.63478m/s$

Therefore the final velocity of astronaut is 3.63m/s

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