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LenaWriter [7]
2 years ago
11

Think about the science you have studied in the past or are currently studying. Give an example of something you have learned in

science that an engineer might use to solve a problem.
Engineering
1 answer:
Oksana_A [137]2 years ago
8 0

Answer:

A topic learned in science that an engineer might use to solve a problem is the nature of le=ight interaction with metals

Light energy made up of packets of photons when impinging on the surface of metal elevates the electrons to a higher energy level which later falls back to lower energy levels to release packets of photons, seen as light

Engineers can use the knowledge of the nature of reflection and refraction of light by metallic surface to develop a special class of unbreakable mirror that can be used as side mirrors in cars

Explanation:

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Write what you already know about college majors. What are they? Can you think of any examples? When do you have to pick one? Ca
Mandarinka [93]
College majors are specific fields of study that help people prepare for career paths and learn content related to that subject. Some college majors would criminal justices, forensic science, gender studies, engineering, chemistry, and more. You should generally begin to research in your junior year of high school and select one by the beginning or your senior year for college applications. Once you get to college you can change your major.
5 0
2 years ago
Read 2 more answers
Consider a thermal energy reservoir at 1500 K that can supply heat at a rate of 150,000 kJ/h. Determine the exergy of this suppl
ANEK [815]

Answer:

exergy = 33.39 kW

Explanation:

given data

thermal energy reservoir T2 = 1500 K

heat at a rate = 150,000 kJ/h = \frac{15000}{3600} kW =  41.67 kW

environment temperature T1 = 25°C = 298 K

solution

we get here maximum efficiency that is  reversible efficiency is express as

reversible efficiency = 1 - \frac{T1}{T2}    ...............1

reversible efficiency = 1 - \frac{298}{1500}  

reversible efficiency =  0.80133

and

the exergy of this supplied energy that is

exergy  = efficiency × hat supply   ................2

exergy = 0.80133 × 41.67 kW

exergy = 33.39 kW

5 0
3 years ago
Consider a circular grill whose diameter is 0.3 m. The bottom of the grill is covered with hot coal bricks at 961 K, while the w
soldi70 [24.7K]

Answer:

Step 1

Given

Diameter of circular grill,   D = 0.3m

Distance between the coal bricks and the steaks,  L = 0.2m

Temperatures of the hot coal bricks,  T₁ = 950k

Temperatures of the steaks, T₂ = 5°c

Explanation:

See attached images for steps 2, 3, 4 and 5

4 0
2 years ago
Define, in words the following i relative humidity ii) dew point temperature
defon

Explanation:

i) When we divide water vapor's partial pressure by the water's equilibrium vapor pressure at a given temperature we get relative humidity.

It is denoted by ∅

\phi ={\frac {p_ {H_{2}O} }{p_{H_{2}O} ^{*}}}

ii) When air is cooled so that it becomes saturated with water vapor the temperature where this occurs is called the dew point.

7 0
3 years ago
An AC generator supplies an rms voltage of 120 V at 50.0 Hz. It is connected in series with a 0.650 H inductor, a 4.80 μF capaci
Serggg [28]

Answer:

Explanation:

f = 50.0 Hz, L = 0.650 H, π = 3.14

C = 4.80 μF, R = 301 Ω resistor. V = 120volts

XL = wL = 2πfL

= 2×3.14×50* 0.650

= 204.1 Ohm

Xc= 1/wC

Xc = 1/2πfC

Xc = 1/2×3.14×50×4.80μF

= 1/0.0015072

= 663.48Ohms

1. Total impedance, Z = sqrt (R^2 + (Xc-XL)^2)= √ 301^2+ (663.48Ohms - 204.1 Ohm)^2

√ 90601 + (459.38)^2

√ 90601+211029.98

√ 301630.9844

= 549.209

Z = 549.21Ohms

2. I=V/Z = 120/ 549.21Ohms =0.218Ampere

3. P=V×I = 120* 0.218 = 26.16Watt

Note that

I rms = Vrms/Xc

= 120/663.48Ohms

= 0.18086A

4. I(max) = I(rms) × √2

= 0.18086A × 1.4142

= 0.2557

= 0.256A

5. V=I(max) * XL

= 0.256A ×204.1

=52.2496

= 52.250volts

6. V=I(max) × Xc

= 0.256A × 663.48Ohms

= 169.85volts

7. Xc=XL

1/2πfC = 2πfL

1/2πfC = 2πf× 0.650

1/2×3.14×f×4.80μF = 2×3.14×f×0.650

1/6.28×f×4.8×10^-6 = 4.082f

1/0.000030144× f = 4.082×f

1 = 0.000030144×f×4.082×f

1 = 0.000123f^2

f^2 = 1/0.000123048

f^2 = 8126.922

f =√8126.922

f = 90.14 Hz

8 0
3 years ago
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