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Alja [10]
4 years ago
14

The angle of twist can be computed using the material’s shear modulus if and only if: (a)- The shear stress is still in the elas

tic region (b)- The material is ductile (c)- The Young's modulus has also been determined
Engineering
1 answer:
ollegr [7]4 years ago
3 0

Answer:

The angle of twist can be computed using the material’s shear modulus if and only if the shear stress is still in the elastic region

Explanation:

The shear modulus (G) is the ratio of shear stress to shear strain. Like the modulus of elasticity, the shear modulus is governed by Hooke’s Law: the relationship between shear stress and shear strain is proportional up to the proportional limit of the material. The angle of twist can be computed using the material’s shear modulus if and only if the shear stress is still in the elastic region.

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During an expansion process, the pressure of a gas changes from 15 to 140 psia according to the relation P = aV + b, where a = 5
Dominik [7]

The work done during the process is 359 btu

<u></u>

<u>Explanation:</u>

Given-

P1 = 15psia

P2 = 140 psia

V1 = 7ft³

a = 5 psia/ft³

b = C

P = aV +b

Work done, W = ?

P1 = aV1 + b

15 = 5 (7) + b

b = -20 psia

P2 = aV2 + b

140 = 5 ( V2) - 20

V2 = 32 ft³

The work done by the process is the area under the curve which is trapezoidal.

Therefore,

Work done, W = area of trapezoid

= (P2 + P1 / 2) (V2 - V1)

= ( 140 + 15 / 2 ) ( 32 - 7)

= 1937.5 psia ft³

= 1937.5/ 5.4039 = 359 btu

Therefore, the work done during the process is 359 btu

5 0
4 years ago
In a black box experiment, when the amount of material exiting a closed system is less than the amount of material entering the
Oksi-84 [34.3K]

When the material that exits is lesser in amount than that of the entering material in a black box experiment, the parts of the system need to be changed.

<h3>What happens in a black box experiment?</h3>

In a black box experiment, the experimenters need to make assumptions regarding the drawing of conclusions. One such conclusion is the amount of material that exits.

If such amount is lesser than the one that enters the system, such experiment concludes that it is the time to change the parts of the system.

Hence, option D holds true regarding the black box experiment.

Learn more about black box experiment here:

brainly.com/question/13403296

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4 0
2 years ago
The switch has been in position a for a long time. At t = 0, the switch moves from position a to position b. The switch is a mak
SashulF [63]

Answer:

Hello there, please check explanations for step by step procedures to get answers.

Explanation:

Given that;

mkasblog

College Engineering 10+5 pts

The switch has been in position a for a long time. At t = 0, the switch moves from position a to position b. The switch is a make-before-break type so there is no interruption of the inductor current. a. Find the expression for i(t) for t ≥ 0. b. What is the initial voltage across the inductor after the switch has been moved to position b? c. Does this initial voltage make sense in terms of circuit behavior? d. How many milliseconds after the switch has been put in position b does the inductor voltage equal 24V? e. Plot both i(t) and v(t) versus t

See attachment for more clearity answer.

6 0
4 years ago
Why are you asked to draw a sketch of a Bridge-Design on a graph sheet/A4 sheet?
elena-s [515]

Engineers or architects can be asked to draw a sketch of a bridge-design on a graph sheet or A4 paper to communicate their ideas and scale their drawings.

Architectural sketches are about the communication of ideas. Engineers and architects use hand sketching to clarify elements of the design to the contractor or client.

It should be noted that drawings can also be drawn on a graph sheet or A4 paper to represent the detail before it's eventually constructed.

Scale drawings are used for the illustration of items that are not convenient to draw their actual size. In the construction industry, scales are used depending on the nature of the drawing. Drawings can be made on a graph or A4 paper due to the ease of printing, reproducing, and resizing the drawings.

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4 0
2 years ago
Read 2 more answers
2. A thin vertical panel L = 3 m high and w = 1.5 m wide is thermally insulated on one side and exposed to a solar radiation flu
n200080 [17]

Answer: 383.22K

Explanation:

L = 3m, w = 1.5m

Area A = 3 x 1.5 = 4.5m2

Q' = 750W/m2 (heat from sun) ,

& = 0.87

Q = &Q' = 0. 87x750 = 652.5W/m2

E = QA = 652.5 x 4.5 = 2936.25W

T(sur) = 300K, T(panel) = ?

Using E = §€A(T^4(panel) - T^4(sur))

§ = Stefan constant = 5.7x10^-8

€ = emmisivity = 0.85

2936.25 = 5.7x10^-8 x 0.85 x 4.5 x (T^4(panel) - 300^4)

T(panel) = 383.22K

See image for further details.

5 0
3 years ago
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