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11 months ago

A machine has an efficiency of 15%. If the energy input is 300 joules, how much useful energy is generated?(1 point).

Engineering

1 answer:

Crazy boy [7]11 months ago

3 0

With an energy input of 300 J and an efficiency of 15%, the machine produces 45 J of useable energy.

<h3>What is energy?</h3>

Energy is characterized as a quantitative characteristic that is transmitted to a person or a physical system and can be observed in the manner in which labor is accomplished as well as in the generation of heat and light. The sun is one of the most significant energy sources.

The machine's useful energy output, or work output, can be calculated as follows:

Efficiency = work out / work in x 100

0.15 = work out / 300

Multiply by cross

Production = 0.15 x 300

Production: 45 J

Thus, with an energy input of 300 J and an efficiency of 15%, the machine produces 45 J of useable energy.

To learn more about energy, refer to the link below:

brainly.com/question/1932868

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The smallest crystal lattice defects is a) cracks b) point defects c) planar defects d) dislocations.

ss7ja [257]

Answer:b) Point defects

Explanation: The point defect is the tiny defect that occurs in the lattice. It usually occurs when there is the atoms or ions missing in the lattice structure that creates a irregularity in the structure.The name point defect itself describes that the occurring defect is having a size of point thus is the smallest defect. Therefore option(b) is the correct option.

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3 years ago

1. A cylindrical casting is 0.3 m in diameter and 0.5 m in length. Another casting has the same metal is rectangular in cross-se

Lorico [155]

Based on the Chvorinov's rule, the diference in the <em>solidification</em> times of the two castings is 14.092 times the <em>solidification</em> time of the prism casting.

<h3>How to apply the Chvorinov's rule for casting processes</h3>

The Chvorinov's rule is an empirical method to estimate the cooling time of a casting in terms of a <em>reference</em> time. This rule states that cooling time (<em>t</em>) is directly proportional to the square of the volume (<em>V</em>), in cubic meters, divided to the surface area (<em>A</em>), in square meters. Now we proceed to model each casting:

<h3>Cylindrical casting</h3>

t = C · [0.25π · D² · L/(0.5π · D² + π · D · L)]²

t = C · [0.25 · D · L/(0.5 · D + L)]² (1)

<h3>Prism casting</h3>

t' = C · [3 · T² · L/(6 · T · L + 2 · T · L + 6 · T²)]²

t' = C · [3 · T · L/(8 · L + 6 · T)]² (2)

<h3>Relationship between the cross sections of both castings</h3>

3 · T² = 0.25π · D² (3)

Where:

<em>t</em> - Cooling time of the cylindrical casting, in time unit.
<em>t'</em> - Cooling time of the prism casting, in time unit.
<em>C</em> - Cooling factor, in time unit per square meter.
<em>D</em> - Diameter of the cylinder, in meters.
<em>L</em> - Length of the casting, in meters.
<em>T</em> - Width of the cross section of the prism casting, in meters.

If we know that <em>D =</em> <em>0.3 m</em>, then the thickness of the prism casting is:

$T = \sqrt{\frac{\pi}{12} }\cdot D$

<em />

And (1) and (2) simplified into these forms:

<h3>Cylindrical casting</h3>

t = C · {0.25π · (0.3 m) · (0.5 m)/[0.5 · (0.3 m) + 0.5 m]}²

t = 0.0329 · C (1b)

<h3>Prism casting</h3>

t' = C · {3 · (0.153 m) · (0.5 m)/[8 · (0.5 m) + 6 · (0.153 m)]}²

t' = 0.00218 · C (2b)

Lastly we find the <em>percentual</em> difference in the solidification times of the two castings by using the following expression:

<em>r = (</em>1 <em>- t'/t) ×</em> 100 %

<em>r = (</em>1 <em>-</em> 0.00218<em>/</em>0.0329<em>) ×</em> 100 %

<em>r =</em> 93.374 %

The <em>cooling</em> time of the <em>prism</em> casting is 6.626 % of the <em>solidification</em> time of the <em>cylindrical</em> casting. The diference in the <em>solidification</em> times of the two castings is 14.092 times the <em>solidification</em> time of the <em>prism</em> casting. $\blacksquare$

To learn more on solidification times, we kindly invite to check this verified question: brainly.com/question/13536247

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2 years ago

Air is compressed by a 30-kW compressor from P1 to P2. The air temperature is maintained constant at 25°C during this process as

adell [148]

Answer:

-0.1006Kw/K

Explanation:

The rate of entropy change in the air can be reduced from the heat transfer and the air temperature. Hence,

ΔS = Q/T

Where T is the constant absolute temperature of the system and Q is the heat transfer for the internally reversible process.

S(air) = - Q/T(air) .......1

Where S.air =

Q = 30-kW

T.air = 298k

Substitute the values into equation 1

S(air) = - 30/298

= -0.1006Kw/K

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2 years ago

What is the maximum number of 12-2 with ground nonmetallic-sheathed cables permitted in an 18-cubic-inch device box if two singl

Novosadov [1.4K]

Answer:i think it is 35

Explanation:

i just guessed sorry im only in 5th grade

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3 years ago

Read 2 more answers

Technician A says that proper footwear may include both leather and steel-toed shoes. Technician B says that leather-soled shoes

Amanda [17]

Given:

We have given two statements.

Statement 1: Proper footwear may include both leather and steel-toed shoes.

Statement 2: Leather-soled shoes provide slip resistance.

Find:

Which statement is true.

Solution:

A slip-resistant outsole is smoother and more slip-resistant than other outsole formulations when exposed to water and oil. A smoother outsole in rubber ensures a slip-resistant shoe can handle a slippery floor more effectively.

Slip resistant shoes have an interlocked tread pattern that does not close the water in, enabling the slip resistant sole to touch the floor to provide better slip resistance.

Leather-soled shoes don't provide slop resistance.

Therefore, both the Technicians are wrong.

5 0

3 years ago