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Elis [28]
3 years ago
11

What relationship is there between the hydraulic system and power, force, energy, resources, and types of energy?

Engineering
2 answers:
VARVARA [1.3K]3 years ago
8 0

Answer:

Hydropower is fueled by water, so it's a clean fuel source, meaning it won't pollute the air like power plants that burn fossil fuels, such as coal or natural gas. ... In addition to a sustainable fuel source, hydropower efforts produce a number of benefits, such as flood control, irrigation, and water supply.

Explanation:

pshichka [43]3 years ago
7 0

Answer:

Well to me there all energy resources.

Explanation:

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Ymorist [56]

Answer:

what code for

Explanation:

8 0
4 years ago
g a heat engine is located between thermal reservoirs at 400k and 1600k. the heat engine operates with an efficiency that is 70%
Bond [772]

Answer:

<em>Heat rejected to cold body = 3.81 kJ</em>

Explanation:

Temperature of hot thermal reservoir Th = 1600 K

Temperature of cold thermal reservoir Tc = 400 K

<em>efficiency of the Carnot's engine = 1 - </em>\frac{Tc}{Th}<em> </em>

eff. of the Carnot's engine = 1 - \frac{400}{600}

eff = 1 - 0.25 = 0.75

<em>efficiency of the heat engine = 70% of 0.75 = 0.525</em>

work done by heat engine = 2 kJ

<em>eff. of heat engine is gotten as = W/Q</em>

where W = work done by heat engine

Q = heat rejected by heat engine to lower temperature reservoir

from the equation, we can derive that

heat rejected Q = W/eff = 2/0.525 = <em>3.81 kJ</em>

6 0
3 years ago
High strength steels are being used to reduce weight on cars. Explain why using a high strength steel would allow you to reduce
gulaghasi [49]

Engaging the frequently tough requirements of vehicle safety, weight reduction for combustible economy, and manufacturability has influenced the steel industry to create a unique variety of 'super steels' for the automobiles of the future.

<h3><u>Explanation</u>:</h3>

• That steel, though, is far more advanced than the materials of just a few years ago.

• At the forefront of these is Advanced High Strength Steel, AHSS, developed by World Auto Steel’s member companies, which is demonstrating to be something of a vision in automobile production.  

• The standard engineering trade-off in steel preference involves considering the need for ultimate strength against flexibility and work-ability – stronger steels tend to be stiffer and less ductile, making them more difficult to develop into cars and more laborious to weld.

• AHSS can retain greatest of the ductility and work-ability of lower grades of steel, while offering much greater strength.

• Where a typical mild steel might have a tensile strength of 300MPa, AHSS can exceed 1500MPa while retaining a highest elongation of 25%, compared to about 40% for mild steel. The intrigue is in the micro-structure, containing a martensite, bainite, austenite phase rather than ferrite, pearlite, or cementite.

4 0
3 years ago
g Consider a thin opaque, horizontal plate with an electrical heater on its backside. The front end is exposed to ambient air th
xxTIMURxx [149]

Answer:

The electrical power is 96.5 W/m^2

Explanation:

The energy balance is:

Ein-Eout=0

qe+\alpha sGs+\alpha skyGsky-EEb(Ts)-qc=0

if:

Gsky=oTsky^4

Eb=oTs^4

qc=h(Ts-Tα)

\alpha s=\frac{\int\limits^\alpha _0 {\alpha l Gl} \, dl }{\int\limits^\alpha _0 {Gl} \, dl }

\alpha s=\frac{\int\limits^\alpha _0 {\alpha lEl(l,5800 } \, dl }{\int\limits^\alpha _0 {El(l,5800)} \, dl }

if Gl≈El(l,5800)

\alpha s=(1-0.2)F(0-2)+(1-0.7)(1-F(0-2))

lt= 2*5800=11600 um-K, at this value, F=0.941

\alpha s=(0.8*0.941)+0.3(1-0.941)=0.77

The hemispherical emissivity is equal to:

E=(1-0.2)F(0-2)+(1-0.7)(1-F(0-2))

lt=2*333=666 K, at this value, F=0

E=0+(1-0.7)(1)=0.3

The hemispherical absorptivity is equal to:

qe=EoTs^{4}+h(Ts-T\alpha  )-\alpha sGs-\alpha oTsky^{4}=(0.3*5.67x10^{-8}*333^{4})+10(60-20)-(0.77-600)-(0.3*5.67x10^{-8}*233^{4})=96.5 W/m^{2}

3 0
3 years ago
The equilibrium fraction of lattice sites that are vacant in silver (Ag) at 700°C is 2 × 10-6. Calculate the number of vacancies
True [87]

Answer:

1.16*10^{23} vacancies/m^3

Explanation:

Data given

temperature =700c

Density=10.35g/cm^3

but

\frac{N_v}{N}=2*10^{-6}\\N_v=2*10^{-6}N\\

Nv is the number of vacant site and N is the number of lattice site.

Since the number of lattice site can also b computed as

N=\frac{p_{Ag} * N_A}{A_{AG}} \\N=\frac{6.022*10^{23} * 10.35*10^6g/m}{107.87g/mol} \\N=5.78*10^{28}atom/m^3

if we substitute the value of the number of lattice into the first equation, we arrive at

N_v=2*10^{-6} *5.78*10^28\\N_v=1.16*10^{23} vacancies/m^3

4 0
4 years ago
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