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3 years ago

8

Your new team is working hard, but they are all less experienced than you and don't complete their tasks as quickly.What would y

ou be most and least likely to do? (A). Reach out to your manager to discuss the situation.Try to identify the best way to support the new associates. (B).Remind your coworkers of performance expectations and that they need to be working harder to complete tasks. (C). Offer to help others complete their tasks; observe their work process and provide some tips that might help them. (D). Give your coworkers more time to figure out how to do the task; they likely just need more practice with what they've already learned.

1 answer:

irina1246 [14]3 years ago

8 0

I believe most likely: (C) least likely: (B)

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An AC generator supplies an rms voltage of 120 V at 50.0 Hz. It is connected in series with a 0.650 H inductor, a 4.80 μF capaci

Serggg [28]

Answer:

Explanation:

f = 50.0 Hz, L = 0.650 H, π = 3.14

C = 4.80 μF, R = 301 Ω resistor. V = 120volts

XL = wL = 2πfL

= 2×3.14×50* 0.650

= 204.1 Ohm

Xc= 1/wC

Xc = 1/2πfC

Xc = 1/2×3.14×50×4.80μF

= 1/0.0015072

= 663.48Ohms

1. Total impedance, Z = sqrt (R^2 + (Xc-XL)^2)= √ 301^2+ (663.48Ohms - 204.1 Ohm)^2

√ 90601 + (459.38)^2

√ 90601+211029.98

√ 301630.9844

= 549.209

Z = 549.21Ohms

2. I=V/Z = 120/ 549.21Ohms =0.218Ampere

3. P=V×I = 120* 0.218 = 26.16Watt

Note that

I rms = Vrms/Xc

= 120/663.48Ohms

= 0.18086A

4. I(max) = I(rms) × √2

= 0.18086A × 1.4142

= 0.2557

= 0.256A

5. V=I(max) * XL

= 0.256A ×204.1

=52.2496

= 52.250volts

6. V=I(max) × Xc

= 0.256A × 663.48Ohms

= 169.85volts

7. Xc=XL

1/2πfC = 2πfL

1/2πfC = 2πf× 0.650

1/2×3.14×f×4.80μF = 2×3.14×f×0.650

1/6.28×f×4.8×10^-6 = 4.082f

1/0.000030144× f = 4.082×f

1 = 0.000030144×f×4.082×f

1 = 0.000123f^2

f^2 = 1/0.000123048

f^2 = 8126.922

f =√8126.922

f = 90.14 Hz

8 0

3 years ago

Conduct online research and write a short report on the origin and evolution of the meter as a measurement standard. Discuss how

valina [46]

Answer:

People have come up with all sorts of inventive ways of measuring length. The most intuitive are right at our fingertips. That is, they are based upon the human body: the foot, the hand, the fingers or the length of an arm or a stride.

In ancient Mesopotamia and Egypt, one of the first standard measures of length used was the cubit. In Egypt, the royal cubit, which was used to build the most important structures, was based on the length of the pharaoh’s arm from elbow to the end of the middle finger plus the span of his hand. Because of its great importance, the royal cubit was standardized using rods made from granite. These granite cubits were further subdivided into shorter lengths reminiscent of centimeters and millimeters.

piece of black rock with white Egyptian markings

Fragment of a Cubit Measuring Rod

Credit: Gift of Dr. and Mrs. Thomas H. Foulds, 1925

Later length measurements used by the Romans (who had taken them from the Greeks, who had taken them from the Babylonians and Egyptians) and passed on into Europe generally were based on the length of the human foot or walking and multiples and subdivisions of that. For example, the pace—one left step plus one right step—is approximately a meter or yard. (On the other hand, the yard did not derive from a pace but from, among other things, the length of King Henry I of England’s outstretched arm.) Mille passus in Latin, or 1,000 paces, is where the English word “mile” comes from.

And thus, the meter has and likely will remain so elegantly defined in these terms for the foreseeable future.

Explanation:

is this short enough

5 0

2 years ago

One kilogram of water fills a 150-L rigid container at an initial pressure of 2 MPa. The container is then cooled to 40∘C. Deter

lukranit [14]

The pressure of water is 7.3851 kPa

<u>Explanation:</u>

Given data,

V = 150× $10^{-3}$ $m^{3}$

m = 1 Kg

$P_{1}$ = 2 MPa

$T_{2}$ = 40°C

The waters specific volume is calculated:

$v_{1}$ = V/m

Here, the waters specific volume at initial condition is $v_{1}$ , the containers volume is V, waters mass is m.

$v_{1}$ = 150× $10^{-3}$ $m^{3}$ /1

$v_{1}$ = 0.15 $m^{3}$ / Kg

The temperature from super heated water tables used in interpolation method between the lower and upper limit for the specific volume corresponds 0.15 $m^{3}$ / Kg and 0.13 $m^{3}$ / Kg.

$T_{1}$ = 350+(400-350) $\frac{0.15-0.13}{0.1522-0.1386}$

$T_{1}$ = 395.17°C

Hence, the initial temperature is 395.17°C.

The volume is constant in the rigid container.

$v_{2}$ = $v_{1}$ = 0.15 $m^{3}$ / Kg

In saturated water labels for $T_{2}$ = 40°C.

$v_{f}$ = 0.001008 $m^{3}$ / Kg

$v_{g}$ = 19.515 $m^{3}$ / Kg

The final state is two phase region $v_{f}$ < $v_{2}$ < $v_{g}$ .

In saturated water labels for $T_{2}$ = 40°C.

$P_{2}$ = $P_{Sat}$ = 7.3851 kPa

$P_{2}$ = 7.3851 kPa

7 0

3 years ago

A turbine operates at steady state, and experiences a heat loss. 1.1 kg/s of water flows through the system. The inlet is mainta

strojnjashka [21]

Answer:

$\dot W_{out} = 399.47\,kW$

Explanation:

The turbine is modelled after the First Law of Thermodynamics:

$-\dot Q_{out} -\dot W_{out} + \dot m\cdot (h_{in}-h_{out}) = 0$

The work done by the turbine is:

$\dot W_{out} = \dot m \cdot (h_{in}-h_{out})-\dot Q_{out}$

The properties of the water are obtained from property tables:

Inlet (Superheated Steam)

$P = 10\,MPa$

$T = 520\,^{\textdegree}C$

$h = 3425.9\,\frac{kJ}{kg}$

Outlet (Superheated Steam)

$P = 1\,MPa$

$T = 280\,^{\textdegree}C$

$h = 3008.2\,\frac{kJ}{kg}$

The work output is:

$\dot W_{out} = \left(1.1\,\frac{kg}{s}\right)\cdot \left(3425.9\,\frac{kJ}{kg} -3008.2\,\frac{kJ}{kg}\right) - 60\,kW$

$\dot W_{out} = 399.47\,kW$

5 0

3 years ago

Air entrainment is used in concrete to: __________.

Dahasolnce [82]

The answer is number 2) Increase the resistance of the concrete to freeze-thaw damage.

5 0

2 years ago