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1 year ago

Which of the following situations best describes student engaged in active learning

1 answer:

ioda1 year ago

3 0

A situation which best describes a student that is engaged in active learning is: C. a student records questions to ask the teacher about a concept.

<h3>What is a question?</h3>

A question can be defined as a group of words or sentence that are developed, so as to elicit an information in the form of answer from an individual such as a student or teacher.

This ultimately implies that, questions are primarily formulated by professionals to test the understanding of an individual. Also, questions should be formulated by a good reader or student so as to enable him or her connect to a story and understand a concept better.

In this context, we can reasonably infer and logically deduce that a situation which best describes a student that is engaged in active learning is when a student records questions to ask the teacher about a concept.

Read more on questions here: brainly.com/question/9044572

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Complete Question:

Which of the following situations best describes a student engaged in active learning?

A. A student uses only one of the resources provided by the teacher.

B. A student skips parts of the lesson that she finds difficult.

C. A student records questions to ask the teacher about a concept.

D. A student does not discuss math outside of the lesson.

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What is the first test you should do when checking the charging system?

yan [13]

Answer:

Connect the test light in series with the negative post, and start pulling feed wires. The first to check is the heavy charging wire from the alternator. A bad or leaky diode in an alternator is a very common source of overnight battery drain. Connect wires one at a time to see what lead is drawing current.

8 0

2 years ago

What is a smooth flow of air over a surface is called?

pochemuha

Answer:

laminar flow

Explanation:

In fluid dynamics, laminar flow is characterized by fluid particles following smooth paths in layers, with each layer moving smoothly past the adjacent layers with little or no mixing.

4 0

3 years ago

What is the least count of screw gauge?<br> (a) 0.01 cm<br> (b) 0.001 cm<br> (c) 0.1 cm<br> (d) 1 mm

Nonamiya [84]

Its 0.001

0.01 x100 = 1mm
0.001x100=0.1mm
0.1=10mm
1m

3 0

3 years ago

A medium-sized jet has a 3.8-mm-diameter fuselage and a loaded mass of 85,000 kg. The drag on an airplane is primarily due to th

SCORPION-xisa [38]

Answer:

$F_{thrust}$ ≅ 111 KN

Explanation:

Given that;

A medium-sized jet has a 3.8-mm-diameter i.e diameter (d) = 3.8

mass = 85,000 kg

drag co-efficient (C) = 0.37

(velocity (v)= 230 m/s

density (ρ) = 1.0 kg/m³

To calculate the thrust; we need to determine the relation of the drag force; which is given as:

$F_{drag}$ = $\frac{1}{2}$ × CρAv²

where;

ρ = density of air wind.

C = drag co-efficient

A = Area of the jet

v = velocity of the jet

From the question, we can deduce that the jet is in motion with a constant speed; as such: the net force acting on the jet in the air = 0

SO, $F_{drag}-F_{thrust} = 0$

We can as well say:

$F_{drag}= F_{thrust}$

We can now replace $F_{thrust} with F_{drag}$ in the above equation.

Therefore, $F_{thrust}$ = $\frac{1}{2}$ × CρAv²

The A which stands as the area of the jet is given by the formula:

$A=\frac{\pi d^2}{4}$

We can now have a new equation after substituting our A into the previous equation as:

$F_{thrust}$ = $\frac{1}{2}$ × Cρ $(\frac{\pi d^2}{4})v^2$

Substituting our data from above; we have:

$F_{thrust}$ = $\frac{1}{2}$ × $(0.37)(1.0kg/m^3)(\frac{\pi(3.8m)^2 }{4})(230m/s)^2$

$F_{thrust}$ = $\frac{1}{8} (0.37)(1.0kg/m^3)({\pi(3.8m)^2 })(230m/s)^2$

$F_{thrust}$ = 110,990N

$F_{thrust}$ in N (newton) to KN (kilo-newton) will be:

$F_{thrust}$ = $(110,990N)*\frac{1KN}{1,000N}$

$F_{thrust}$ = 110.990 KN

$F_{thrust}$ ≅ 111 KN

In conclusion, the jet engine needed to provide 111 KN thrust in order to cruise at 230 m/s at an altitude where the air density is 1.0 kg/m³.

5 0

3 years ago

(3) Calculate the heat flux through a sheet of brass 7.5 mm (0.30 in.) thick if the temperatures at the two faces are 150°Cand 5

bezimeni [28]

Answer:

a.) 1.453MW/m2, b.) 2,477,933.33 BTU/hr c.) 22,733.33 BTU/hr d.) 1,238,966.67 BTU/hr

Explanation:

Heat flux is the rate at which thermal (heat) energy is transferred per unit surface area. It is measured in W/m2

Heat transfer(loss or gain) is unit of energy per unit time. It is measured in W or BTU/hr

1W = 3.41 BTU/hr

Given parameters:

thickness, t = 7.5mm = 7.5/1000 = 0.0075m

Temperatures 150 C = 150 + 273 = 423 K

50 C = 50 + 273 = 323 K

Temperature difference, T = 423 - 323 = 100 K

We are assuming steady heat flow;

a.) Heat flux, Q" = kT/t

K= thermal conductivity of the material

The thermal conductivity of brass, k = 109.0 W/m.K

Heat flux, $Q" = \frac{109 * 100}{0.0075} = 1,453,333.33 W/m^{2} \\ Heat flux, Q" = 1.453MW/m^{2} \\$

b.) Area of sheet, A = 0.5m2

Heat loss, Q = kAT/t

Heat loss, $Q = \frac{109*0.5*100}{0.0075} = 726,666.667W$

Heat loss, Q = 726,666.667 * 3.41 = 2,477,933.33 BTU/hr

c.) Material is now given as soda lime glass.

Thermal conductivity of soda lime glass, k is approximately 1W/m.K

Heat loss, $Q=\frac{1*0.5*100}{0.0075} = 6,666.67W$

Heat loss, Q = 6,666.67 * 3.41 = 22,733.33 BTU/hr

d.) Thickness, t is given as 15mm = 15/1000 = 0.015m

Heat loss, $Q=\frac{109*0.5*100}{0.015} =363,333.33W$

Heat loss, Q = 363,333.33 * 3.41 = 1,238,966.67 BTU/hr

5 0

3 years ago