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Olenka [21]
2 years ago
11

Explain three examples of workshop

Engineering
2 answers:
stealth61 [152]2 years ago
7 0

Answer:

Invitational Workshop

An invitational workshop is what many of us know. It’s what Lucy Calkins has made famous through the Reading and Writing Workshop. In the invitational workshop, the instructor usually hosts a minilesson. This minilesson is intended to meet the needs of the majority of children in the classroom. Afterward, the children are “invited” to employ the skills or strategy for the minilesson during workshop time, where students work independently or in small groups

Explanation:

Wewaii [24]2 years ago
6 0

Answer:

Here are 5 examples:

  1. Invitational Workshop
  2. Constructivist Workshop
  3. Reflection Workshop
  4. Conferencing Workshop
  5. Choice Workshop

Explanation:

I hope this helps!

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If the open circuit voltage of a circuit containing ideal sources and resistors is measured at 8 , while the current through the
solmaris [256]

Answer:

The power absorbed by the 60 ohm resistor is 1.064 W

Explanation:

When there's no load connected to the source (open circuit) its voltage output is ideal, since there won't be any voltage drop across it's internal resistance. When there's a shor circuit, the only load is the internal resistance of the source, so we can use Ohm's law to compute the internal resistance, as shown bellow:

Rinternal = Vopenload/Ishortcircuit

Rinternal = 8/200 = 0.04 Ohm

When we connect a load to this source, the total load we'll be the external resistance plus the internal resistance. We can now compute the curent flow when there's a 60 Ohm resistance connected to the terminals of the source by using Ohm's law again:

I = Vsource/(Rexternal + Rinternal)

I = 8/(60 + 0.04) = 8/(60.04) = 0.1332 A

The absorbed power is the product of the voltage across the terminals of the resistor and the current that goes through it. The current is the one we calculated above and the voltage across it's terminals is given by  I*R, so the power output is:

P = I*Vresistor

P = I*(I*R)

P = R*I^2 = 60*(0.1332)^2  = 1.064 W

8 0
3 years ago
Read 2 more answers
The water depths upstream and downstream of a hydraulic jump is 0.3 m and 1.2 m. Determine flow rate
Sergio [31]

This question is incomplete, the complete question is;

the water depths upstream and downstream of a hydraulic jump is 0.3 m and 1.2 m. Determine flow rate  ( in m³ ) if the rectangular channel is 20 m wide.

Answer:

the flow rate is 32.549 m³/sec

Explanation:

Given that

y₁ = 0.3 m

y₂ = 1.2 m

β = 20 m

Now for Rectangular Channel, we know that;

2q²/g = y₁y₂( y₁ + y₂)

where g = 9.81 m/s²

and q = Q/β

so

2(Q/β)²/g = y₁y₂( y₁ + y₂)

we substitute our given values

2(Q/20)²/9.81 = 0.3 × 1.2( 0.3 + 1.2)

2(Q²/400)/9.81 = 0.36(1.5)

2(Q²/400) = 0.54 × 9.81

Q²/400 = 5.2974 / 2

Q²/400 = 2.6587

Q² = 1059.48

Q = √1059.48

Q = 32.549 m³/sec

Therefore the flow rate is 32.549 m³/sec

3 0
3 years ago
Compare and contrast the three USDA quality grades available to consumers and describe how the grades affect labeling.
uysha [10]

Answer:

USDA has developed strict measures of quality for products such as grains, rice, corn, and beans. Quality grading is based on the standards developed for each product. Quality grades provide a common language among buyers and sellers, which in turn assures consistent quality for consumers.

Explanation:

7 0
3 years ago
A. 50
Licemer1 [7]

Answer:

love

Explanation: you

5 0
3 years ago
The velocity distribution for laminar flow between parallel plates is given by u umax = 1 − ( 2y h ) 2 Where h is the distance s
Lynna [10]

Answer:

Explanation:

For we to calculate the shear stress on the upper plate and give its direction. Sketch the variation of shear stress across the channel, I used hand in solving it, check attached file below

5 0
4 years ago
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