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d1i1m1o1n [39]
2 years ago
12

I need help with this question

Engineering
1 answer:
Roman55 [17]2 years ago
6 0

Answer:

last one 8;2per leg

Explanation:

sorry if I got it wrong

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Label and group products. One would think that a general cleanup would be the first step, but no, it's not. ...Clean up the area. ...Put up demarcation lines. ...Stack properly. ...Keep the aisles, paths and ramps clear. ...Have all the safety signs in place.

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At steady state, air at 200 kPa, 325 K, and mass flow rate
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Identify three operational controls and explain<br> how to use them?
ladessa [460]

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5.1 Personnel Security. ...

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8 0
3 years ago
An electric dipole is made of two charges of equal magnitudes and opposite signs. The positive charge, q=1.0 μC, is located at t
vfiekz [6]

Answer:

work done by electric field  is 0.06 J

Explanation:

Given data:

Two point charge is + 1\mu C  and -1 \mu C

0+1 charge positioned is (0 cm , 1 cm, 0.00 cm)

-1 charge positioned is (0 cm , -1 cm, 0.00 cm)

E = 3.0\times 10^6 N/C

From above information, the distance between  given two charges d = 2 cm

then d = 0.02m

 work needed is W = q E d

W = 1.0 \times 10^{-6} \times 3.0 \times 10^6 \times 0.02

W = 0.06 J  

Therefore work done by electric field  is 0.06 J

8 0
3 years ago
Rsidential Solar Solution:a. A type of photovoltaic solar panel manufactured in China receives a rating of 250W. The rating proc
aksik [14]

Answer:

a) \eta = 13.455\%, b) E_{day} = 812.716\,kJ, c) C_{month. total} = 19.505\, USD, d) t = 40.588\,years

Explanation:

a) The area of the solar panel is:

A = (20\,ft^{2})\cdot (\frac{0.3048\,m}{1\,ft} )^{2}

A = 1.858\,m^{2}

The energy potential is determined herein:

\dot E_{o} = (1000\,\frac{W}{m^{2}} )\cdot (1.858\,m^{2})

\dot E_{o} = 1858\,W

The efficiency of the solar panel is:

\eta = \frac{\dot E}{\dot E_{o}}\times 100\%

\eta = \frac{250\,W}{1858\,W}\times 100\%

\eta = 13.455\%

b) The energy generated by the solar panel is presented below:

E_{day} = (0.135)\cdot (150\,\frac{W}{m^{2}} )\cdot (20\,ft^{2})\cdot \left(\frac{0.3048\,m}{1\,ft} \right)^{2}\cdot (6\,h)\cdot (\frac{3600\,s}{1\,h} )\cdot (\frac{1\,kJ}{1000\,J} )

E_{day} = 812.716\,kJ

c) The energy generated per month and per panel is:

E_{month} = 30\cdot E_{day}

E_{month} = 30 \cdot (812.716\,kJ)\cdot \left(\frac{1\,kWh}{3600\,kJ}  \right)

E_{month} = 6.773\,kWh

Monthly energy savings due to the use of 18 panels are:

C_{month, total} = 18\cdot E_{month}\cdot c

C_{month, total} = 18\cdot (6.773\,kWh)\cdot (\frac{0.16\,USD}{1\,kWh} )

C_{month. total} = 19.505\, USD

d) The payback of the solar energy system is:

t = \frac{9500\,USD}{12\cdot (19.505\,USD)}

t = 40.588\,years

6 0
3 years ago
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