Ask question

Ask question

All categories

2 years ago

5

What components are included in a basic engine block?

1 answer:

atroni [7]2 years ago

8 0

Houses the CYLINDERS, Water Jacket & Crankcase

You might be interested in

The greater the force applied to an object, the _____ the change in speed or direction of the object.

storchak [24]

Answer:

b

Explanation:

8 0

2 years ago

He is going ___ in the hot air ballon

Vladimir [108]

no artical shoul be used here

5 0

2 years ago

If coolant is mixed 50-50 with water, what is the freezing point?

sashaice [31]

Answer:

-35 degrees F

When mixed in equal parts with water (50/50), antifreeze lowers the freezing point to -35 degrees F and raises the boiling temperature to 223 degrees F. Antifreeze also includes corrosion inhibitors to protect the engine and cooling system against rust and corrosion.

8 0

2 years ago

A particle moves along a straight line with a velocity V=(200s) mm/s, where s is in millimeters. Determine acceleration of the p

iragen [17]

Answer:

200 mm/s²

Explanation:

See it in the pic

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