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3 years ago

Where Does a Solar Engineer Work? (2 sentences or more please)

2 answers:

5 0

Where Does a Solar Engineer Work? Most solar engineers work in offices, but may also travel frequently to different work sites, including overseas locations. They often must climb ladders onto rooftops to inspect installations, and may need to carry heavy loads for short distances.

Vsevolod [243]3 years ago

3 0

Answer:

in a service mask...they also work there to mount mask and take care of the sites

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Motion is defined as a change in an object's position when compared to other objects around it. Mary Ann was watching a slug cra

lara [203]

Answer:

Explanation:

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4 years ago

(30 pts) A simply supported beam with a span L=20 ft and cross sectional dimensions: b=14 in; h=20 in; d=17.5 in. is reinforced

Nat2105 [25]

Answer:

Zx = 176In³

Explanation:

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5 0

3 years ago

When the grounded conductor is to be spliced in a 12-inch by 12-inch junction box, there shall be at least ? of free conductor l

slamgirl [31]

There should be a 6" free conductor left for splicing

Why there should be 6" splicing?

You must leave the junction box with at least six inches of free conductor wiring when running electrical cables from the box to the box for connecting needs.

Actually, 6-8 inches can be left when running the electric cables, additionally even if the grounded conductor is to be spliced in a 12-inch by 12-inch.

Conductor splicing: A splice, which can be finished using either the crimping or soldering method, is the joining of two or more conductors in a way that produces a permanent electrical termination and mechanical link.

Hence we can conclude that 6 inches is fine for the conductor splicing

To learn more about splicing
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7 0

1 year ago

50 for brainliest HELP ASAP<br> absurd answers will be recorded

s2008m [1.1K]

Answer:

1) This is because too much fuel is needed to get a payload from the surface to orbital altitude an accelerated to orbital speed.

2) This is because space travel present extreme environment that affect machines operations and survival.

Explanation:

Hope it helps

3 0

3 years ago

Refrigerant R-12 is used in a Carnot refrigerator

Sphinxa [80]

Answer:

Heat transferred from the refrigerated space = 95.93 kJ/kg

Work required = 18.45 kJ/kg

Coefficient of performance = 3.61

Quality at the beginning of the heat addition cycle = 0.37

Explanation:

From figure

$Q_H$ is heat rejection process

$Q_L$ is heat transferred from the refrigerated space

$T_H$ is high temperature = 50 °C + 273 = 323 K

$T_L$ is low temperature = -20 °C + 273 = 253 K

$W_{net}$ is net work of the cycle (the difference between compressor's work and turbine's work)

Coefficient of performance of a Carnot refrigerator $(COP_{ref})$ is calculated as

$COP_{ref} = \frac{T_L}{T_H - T_L}$

$COP_{ref} = \frac{253 K}{323 K - 253 K}$

$COP_{ref} = 3.61$

From figure it can be seen that heat rejection is latent heat of vaporisation of R-12 at 50 °C. From table

$Q_H = 122.5 kJ/kg$

From coefficient of performance definition

$COP_{ref} = \frac{Q_L}{Q_H - Q_L}$

$Q_H \times COP_{ref} = (COP_{ref} + 1) \times Q_L$

$Q_L = \frac{Q_H \times COP_{ref}}{(COP_{ref} + 1)}$

$Q_L = \frac{122.5 kJ/kg \times 3.61}{(3.61 + 1)}$

$Q_L = 95.93 kJ/kg$

Energy balance gives

$W_{net} = Q_H - Q_L$

$W_{net} = 122.5 kJ/kg - 95.93 kJ/kg$

$W_{net} = 26.57 kJ/kg$

Vapor quality at the beginning of the heat addition cycle is calculated as (f and g refer to saturated liquid and saturated gas respectively)

$x = \frac{s_1 - s_f}{s_g - s_f}$

From figure

$s_1 = s_4 = 1.165 kJ/(K kg)$

Replacing with table values

$x = \frac{1.165 kJ/(K \, kg) - 0.9305 kJ/(K \, kg)}{1.571 kJ/(K \, kg) - 0.9305 kJ/(K \, kg)}$

$x = 0.37$

Quality can be computed by other properties, for example, specific enthalpy. Rearrenging quality equation we get

$h_1 = h_f + x \times (h_g - h_f)$

$h_1 = 181.6 kJ/kg + 0.37 \times 162.1 kJ/kg$

$h_1 = 241.58 kJ/kg$

By energy balance, $W_{t}$ turbine's work is

$W_{t} = |h_1 - h_4|$

$W_{t} = |241.58 kJ/kg - 249.7 kJ/kg|$

$W_{t} = 8.12 kJ/kg$

Finally, $W_{c}$ compressor's work is

$W_{c} = W_{net} + W_{t}$

$W_{c} = 26.57 kJ/kg + 8.12 kJ/kg$

$W_{c} = 34.69 kJ/kg$

6 0

3 years ago