What unit of measurment would be used to measure current?

julsineya [31]

I attached a photo that explains and gives the answer to your questions. Had to add a border because the whole picture didn’t fit.

6 0

3 years ago

It has a piece of 1045 steel with the following dimensions, length of 80 cm, width of 30 cm, and a height of 15 cm. In this piec

Serggg [28]

Answer:

material remove in 3 min is 16790.4 mm³/s

Explanation:

given data

length L = 80 cm = 800 mm

width W = 30 cm

height H = 15 cm

make grove length = 80 cm

width = 8 cm

depth = 10 cm

mill toll diameter = 4 mm

axial cutting depth = 20 mm

to find out

How much material removed in 3 minutes

solution

first we find time taken for length of advance that is

time = $\frac{length}{advance}$

here advance is given as 0.001166 mts / sec

so time = $\frac{800}}{0.001166*1000}$

time = 686.106 seconds

now we find material remove rate that is

remove rate = mill toll rate × axial cutting depth × advance

remove rate = 4 × 20×0.001166 ×1000

remove rate = 93.28 mm³/s

so

material remove in 3 minute = 3 × 60 = 180 sec

so material remove in 3 min = 180 × 93.28

material remove in 3 min is 16790.4 mm³/s

7 0

3 years ago

Conclude from the scenario below which type of documentation Holly should use, and explain why this would be the best choice. Ho

NARA [144]

Answer:

Okay

Explanation:

7 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

what would the current through the Load curve look like when the alternating current source S has the following shape Draw the s

IgorLugansk [536]

Answer:

Following are the response to the given question:

Explanation:

In the given question, A curve one would be to "reject" them. Within this way, people are directly non-confrontationally off from their amorous interests or advances. The curve looks much like a current through the unidirectional load, and it has only a good maximum value. It seems like the total rating of the complete wave rectifier that is illustrated below, the present vas temporal curve please find it.

6 0

3 years ago