A fully charged new battery will have a low conductance reading.

When a process is in a state of statistical control, all of the points on a control chart should fall within the control limits.

Mars2501 [29]

it is undesirable that all of the points should fall extremely near, or exactly on, the centerline of the control chart Because:

It can make it difficult to interpret patterns and draw conclusions in a run chart.
It can lead hinder the process, and make it unstable.

<h3>When a data point falls outside the control limits of a run chart ?</h3>

If a data point falls is said to outside the control limits, we can say that the process is said to be out of one's control and it is good that an investigation is done to ascertain and remove the cause or causes.

So, it is undesirable that all of the points should fall extremely near, or exactly on, the centerline of the control chart Because:

It can make it difficult to interpret patterns and draw conclusions in a run chart.
It can lead hinder the process, and make it unstable.

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4 0

2 years ago

An angle is observed repeatedly using the same equipment and procedures producing the data below:35 ∘ 40'00",35 ∘ 40'10",35 ∘ 40

Helen [10]

Answer: (a) +/- 7.5° (b) +/- 3.75°

Explanation:

See attachment

6 0

3 years ago

In a wind-turbine, the generator in the nacelle is rated at 690 V and 2.3 MW. It operates at a power factor of 0.85 (lagging) at

Juli2301 [7.4K]

To solve this problem we will apply the concepts related to real power in 3 phases, which is defined as the product between the phase voltage, the phase current and the power factor (Specifically given by the cosine of the phase angle). First we will find the phase voltage from the given voltage and proceed to find the current by clearing it from the previously mentioned formula. Our values are

$V = 690V$

$P_{real} = 2.3MW$

Real power in 3 phase

$P_{real} = 3V_{ph}I_{ph} Cos\theta$

Now the Phase Voltage is,

$V_{ph} = \frac{V}{\sqrt{3}}$

$V_{ph} = \frac{690}{\sqrt{3}}$

$V_{ph} = 398.37V$

The current phase would be,

$P_{real} = 3V_{ph}I_{ph} Cos\theta$

Rearranging,

$I_{ph}=\frac{P_{real}}{3V_{ph}Cos\theta}$

Replacing,

$I_{ph}=\frac{2.3MW}{3( 398.37V)(0.85)}$

$I_{ph}= 2.26kA/phase$

Therefore the current per phase is 2.26kA

6 0

3 years ago

You are preparing to work with Chemical A. You open the appropriate storage cabinet, and notice Chemical B, as well as Chemical

PtichkaEL [24]

The correct answer is; Stability and reactivity.

Further Explanation:

The stability and reactivity section of the SDS sheets is where to check for the possibility of hazardous reactions for the chemicals. This also lists the chemical stability of each chemical that people may be using. This can be found in section 10 of the OSHA Quick Card.

The SDS sheets has 16 sections for employees to use. Since 2015, the sections can be found in uniform format for easier and faster ways to find the section needed. The 16 sections for the SDS sheets are:

Identification
Hazard(s) identification
Composition/information on ingredients
First-aid measures
First-aid measures
First-aid measures
Handling and storage
Exposure controls/personal protection
Physical and chemical properties
Stability and reactivity
Toxicological information
Ecological information
Disposal considerations
Transport information
Regulatory information
Other information

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5 0

4 years ago

The worst time you have had with a mechanical issue

Paul [167]

Answer:

When I was taking my sat exam online and my phone battery died

8 0

3 years ago

Read 2 more answers