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fiasKO [112]
2 years ago
5

In order to keep an automobile operating, it is necessary to keep adding fuel as it is used up. Explain why this doesn't contrad

ict the law of conservation of energy.
Engineering
1 answer:
Semmy [17]2 years ago
3 0

Answer:

There is conversion of energy from one form to another and this justifies the law of conservation of energy, which states that energy can neither be created nor destroyed but can be converted from one form to another.

Explanation:

The fuel added to automobile is a chemical energy, which provides thermal energy of the combustion engine of the automobile, which is then converted to mechanical energy of the moving parts of the automobile. Thus, there is conversion of energy from one form to another. This justifies the law of conservation of energy, which states that energy can neither be created nor destroyed but can be converted from one form to another.

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What is the key to being a good engineer?
wariber [46]

Answer:

Approaching a problem and finding a solution within given guidelines.

Explanation:

To be a good engineer, you simply need to know how to create a solution to a problem with a given set of restraints or guidelines.  

For instance, let's say your boss wants you to build a machine that can automate some trivial task so he can use the worker elsewhere.  He wants the machine to be low maintenance and under $40,000.

As an engineer, your first thought should be, let's see what this "trivial task" is, and then after your observation, you should begin to pull upon your experience and resources to build a solution that can be low maintenance and under $40,000.

Put simply, when you approach a problem, consider all aspects of the problem and then build a solution that satisfies all requirements.

Cheers.

4 0
2 years ago
Read 2 more answers
An aircraft component is fabricated from an aluminum alloy that has a plane strain fracture toughness of 30 . It has been determ
UNO [17]

Answer:

fracture will occur since ( 31.8 Mpa√m ) is greater than the K_{Ic of the material ( 30 Mpa√m )

Explanation:

Given the data in the question;

To determine whether the aircraft component will fracture, given a fracture toughness of 30 Mpa√m, stress level of 355 and maximum internal crack length of 1.39 mm.

On a similar component, it has been said that fracture results at a stress of 237 MPa when the maximum (or critical) internal crack length is 2.78 mm.

so we first of all solve for the parameter Y in the condition where fracture occurred.

K_{Ic = 30 Mpa√m

σ = 237 MPa

2α = 2.78 mm = 2.78 × 10⁻³ m  

so

Y = K_{Ic / σ√πα

we substitute

Y = (30 Mpa√m) / (237 MPa)√(π(2.78 × 10⁻³ m / 2 ) )

Y =  (30 Mpa) / (237)( 0.06608187 )

Y = 30 / 15.6614

Y = 1.9155

Next we solve for Yσ√πα for the second case;

σ = 355 Mpa, 2α = 1.39 mm = 1.39 × 10⁻³ m

so

Yσ√πα = 1.9155 × 355 Mpa × √( π × (1.39 × 10⁻³ m / 2) )

= 1.9155 × 355 × 0.0467269

= 31.8 Mpa√m

so

( 31.8 Mpa√m ) > K_{Ic ( 30 Mpa√m )

Therefore, fracture will occur since ( 31.8 Mpa√m ) is greater than the K_{Ic of the material ( 30 Mpa√m )

4 0
2 years ago
In an experiment, the local heat transfer over a flat plate were correlated in the form of local Nusselt number as expressed by
Stells [14]

Answer:

\dfrac{\bar{h}}{h}=\dfrac{5}{4}

Explanation:

Given that

Nu_x=0.035Re_x^{0.8} Pr^{1/3}

We know that

Rex=ρvx/μ

So

Nu_x=0.035Re_x^{0.8} Pr^{1/3}

Nu_x=0.035\times\left(\dfrac{\rho vx}{\mu}\right)^{0.8}Pr^{1/3}

All other quantities are constant only x is a variable in the above equation .so lets take all other quantities as a constant C

Nu_x=C.x^{0.8}=C.x^{4/5}

We also know that

Nux=hx/K

C.x^{4/5}=\dfrac{hx}{k}

m is the constant

h=mx^{-1/5}

This is local heat transfer coefficient

The average value of h given as

\bar{h}=\dfrac{\int_{0}^{L}hdx}{L}

\bar{h}=\dfrac{5m}{4}\times\dfrac{L^{4/5}}{L}

\bar{h}=\dfrac{5m}{4}L^{-1/5}             ---------1

The value of local heat transfer coefficient at x=L

h=mx^{-1/5}

h=mL^{-1/5}            -----------2

From 1 and 2 we can say that

\dfrac{\bar{h}}{h}=\dfrac{5}{4}

8 0
3 years ago
Mike is involved in developing the model building codes that various states and local authorities in the United States adopt. He
Lyrx [107]
<h3>Answer:</h3>

Mike is involved in developing the model building codes that various states and local authorities in the United States adopt. He works with the <u>Workers</u> , which consists of members who are building code officials and building safety professionals.

8 0
3 years ago
A school is playing $0.XY per kWh for electric power. To reduce its power bill, the school installs a wind turbine with a rated
Semmy [17]

Answer: Your question has some missing figures so kindly plug in the values into the solution provided to get the exact amount of money saved

answer : Electric power generated = 216 * 10^6 kJ

             money saved = $0.XY * 60000 kwh

Explanation:

<u>Calculating  the amount of electric power generated by wind turbine</u>

power generated = ( 30 * 2000 ) kWh  = 60000 kWh

Electric energy generated = 60000 kWh * 3600 kJ = 216 * 10^6 kJ

<u>Calculate money saved by school per year </u>

$0.XY * 60000 kwh

5 0
2 years ago
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