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Degger [83]
3 years ago
7

What is enlarging scale?​

Engineering
1 answer:
8090 [49]3 years ago
8 0

Answer:

Explanation:

Enlargement, sometimes called scaling or dilation, is a kind of transformation that changes the size of an object. The image created is similar to the object.

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How is the difference between science and engineering Best stated?
stiv31 [10]

Answer:Science is the body of knowledge that explores the physical and natural world. Engineering is the application of knowledge in order to design, build and maintain a product or a process

Explanation:

8 0
2 years ago
What is the term RF exiciter?
Sati [7]
The exciter provides fully coherent receiver local oscillator signals at radar frequency band as well as requisite, auxiliary high frequency clock signals. The exciter function is divided into an internal frequency synthesizer and an upconverter.

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7 0
3 years ago
Electric heater wires are installed in a solid wall having a thickness of 8 cm and k=2.5 W/m.°C. The right face is exposed to an
Svet_ta [14]

Answer:

2.46 * 10⁵ W/m³

Explanation:

See attached pictures for detailed explanation.

6 0
4 years ago
Read 2 more answers
The ingredient weights for making 1 yd (cyd) of concrete by assuming aggregates in SSD state are given below. The volume of air
Pachacha [2.7K]

Answer:

Explanation:

Ans) Given batch weight of each component :

Cement = 700 lb

Water = 315 lb

Coarse aggregate = 1575 lb

Fine aggregate = 1100 lb

Part 1) Amount of water = 328.5 lb

Amount of water is needed to be increased if the aggregates has absorption capacity, To maintain constant water cement ratio, the mixing water is increased because some of the water is absorbed by aggregates.

Amount of water absorbed = 328.5 lb - 315 lb = 13.5 lb

Total amount of aggregates = 1575 + 1100 = 2675 lb

=> % Absorption capacity = 13.5 x 100 / 2675 = 0.5 %

Hence, new amount of Coarse aggregate = (1 - 0.005) x 1575 lb = 1567.125 lb

New amount of fine aggregate = (1 - 0.005) x 1100 = 1094.5 lb

Since, water cement ratio is maintained constant , amount of cement remains unchanged

=> Volume of water = 328.5 / 62.4 = 5.26 ft3

=> Volume of cement = 700 / (3.15 x 62.4) = 3.56 ft3

=> Volume of coarse aggregate = 1567.125 / (2.4 x 62.4) = 10.46 ft3

=> Volume of fine aggregate = 1100 / (2.4 x 62.4) = 7.34 ft3

Volume of air = 2% = 0.02 x 27 = 0.54 ft3

Total concrete volume = 5.26 + 3.56 + 10.46 + 7.34 + 0.54 \approx 27 ft3 = 1 yd3

Hence, calculated amount of each component is correct

Part 2) We know, minus sign indicated that the aggregate will absorb some moisture from concrete, hence mixing water amount needed to be corrected .

=> Amount of water absorbed by coarse aggregate = 0.01 x 1567.125 lb = 15.67 lb

=> Amount of water absorbed by fine aggregate = 0.02 x 1094.50 lb = 21.89 lb

Total amount of water absorbed = 15.67 + 21.89 = 37.56 lb

To maintain same water cement ratio, amount of mixing water is needed to be increased

=> Corrected amount of mixing water = 328.5 lb + 37.56 lb = 366 lb

=> Corrected amount of coarse aggregate = (1 - 0.01) x 1567.125 = 1551.45 lb

=> Corrected amount of fine aggregate = (1 - 0.02) x 1094.5 = 1072.6 lb

Part 3) We know,

Unit weight = Sum of weight of each material / Total volume

=> Sum of weight = 366 + 700 + 1551.45 + 1072.6 = 3690.05 lb

Total volume = 1 yd3 or 27 ft3

=> Expected Unit Weight = 3690.05 lb / 27 ft3 = 136.67 lb/ft3

Also, Concrete Yield = Weight of all components / Unit weight of concrete

=> Yield = 3690.05 / 136.67 = 27 ft3 or 1 yd3

4 0
4 years ago
When the 2.8-kg bob is given a horizontal speed of 1.5 m/s, it begins to rotate around the horizontal circular path A. The force
Rama09 [41]

Answer:

The speed is the same at 1.5 m/s while

The work done by the force F is 0.4335 J

Explanation:

Here we have angular acceleration α = v²/r

Force = ma = 2.8 × 1.5²/r₁

and ω₁ = v₁/r₁ = ω₂ = v₁/r₂

The distance moved by the force = 600 - 300 = 300 mm = 0.3 m

If the velocity is constant

The speed is 1.5 m/s while the work done is

2.8 × 1.5²1/(effective radius) ×0.3

r₁ = effective radius

2.8*9.81 = 2.8 × 1.5²/r₁

r₁ = 0.229

The work done by the force = 2.8 × 1.5²*1/r₁ *0.3 = 0.4335 J

4 0
3 years ago
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