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mrs_skeptik [129]
3 years ago
13

Coving is a curved edge between a floor and a wall. O True O False

Engineering
2 answers:
wolverine [178]3 years ago
7 0

Answer:

True.

Explanation:

ser-zykov [4K]3 years ago
3 0
Uhmmmmmmm it’s True..
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A family in Florida heard about evaporative coolers as a way to cool their home without the expenses of a refrigerated air condi
I am Lyosha [343]

Their HVACR technician advises against it because of the residential energy costs from HVAC units.

hat is the cost energy?

Energy Cost is known to be a term that connote the cost of electricity and this is one that is often related to fuel oil, gasoline, heating oil, natural gas, as well as other kinds or source of energy linked to any operation.

Hence, Since family in Florida heard about evaporative coolers as a way to cool their home without the expenses of a refrigerated air conditioning system. Their HVACR technician advises against it because of the residential energy costs from HVAC units.

Learn more about energy costs from

brainly.com/question/9821162

#SPJ1

7 0
2 years ago
A sandy soil has a total unit weight of 120 pcf, a specific gravity of solids of 2.64, and a water content of 16 percent. Comput
olchik [2.2K]

Answer:

A). Dry unit weight = 1657.08Kg/m3

B). Porosity  = 0.37

C). Void ratio  = 0.593 

D). 0.712

Explanation:

Total unit weight, Y = 120pcf =1922.2 Kg/m3

Specific gravity of solids, Gs = 2.64

Water content, w = 16%

A). Dry unit weight

Yd = Y/(1+w)

= 1922.2/(1+0.16) = 1657.08Kg/m3

B). Porosity

However void ratio, e = Gs×Yw/Yd, where Yw = 1000Kg/m3

Void ratio = 2.64×1000/1657.08 = 0.593

 

And porosity = e/(1+e) =0.593/(1+0.593) = 0.37

C). void ratio, e = 0.593

D). Degree of saturation, S = m×Gs/e where m =water content

S = 0.16×2.64/0.593 = 0.712

5 0
4 years ago
_____ are used to control the flow of electricity in a circuit.
Travka [436]

Answer:

Switches control the flow of electricity in a circuit.

8 0
3 years ago
Read 2 more answers
An insulated piston-cylinder device contains 0.15 of saturated refrigerant-134a vapor at 0.8 MPa pressure. The refrigerant is no
Stells [14]

Answer:

Assumption:

1. The kinetic and potential energy changes are negligible

2. The cylinder is well insulated and thus heat transfer is negligible.

3. The thermal energy stored in the cylinder itself is negligible.

4. The process is stated to be reversible

Analysis:

a. This is reversible adiabatic(i.e isentropic) process and thus s_{1} =s_{2}

From the refrigerant table A11-A13

P_{1} =0.8MPa   \left \{ {{ {{v_{1}=v_{g}  @0.8MPa =0.025645 m^{3/}/kg } } \atop { {{u_{1}=u_{g}  @0.8MPa =246.82 kJ/kg } -   also  {{s_{1}=s_{g}  @0.8MPa =0.91853 kJ/kgK } } \right.

sat vapor

m=\frac{V}{v_{1} } =\frac{0.15}{0.025645} =5.8491 kg\\and \\\\P_{2} =0.2MPa  \left \{ {{x_{2} =\frac{s_{2} -s_{f} }{s_{fg }}=\frac{0.91853-0.15449}{0.78339}   = 0.9753 \atop {u_{2} =u_{f} +x_{2} }(u_{fg}) =  38.26+0.9753(186.25)= 38.26+181.65 =219.9kJ/kg \right. \\s_{1} = s_{2}

T_{2} =T_{sat @ 0.2MPa} = -10.09^{o}  C

b.) We take the content of the cylinder as the sysytem.

This is a closed system since no mass leaves or enters.

Hence, the energy balance for adiabatic closed system can be expressed as:

E_{in} - E_{out}  =ΔE

w_{b, out}  =ΔU

w_{b, out} =m([tex]u_{1} -u_{2)

w_{b, out}  = workdone during the isentropic process

=5.8491(246.82-219.9)

=5.8491(26.91)

=157.3993

=157.4kJ

4 0
3 years ago
Calculate the unit length of a roof using common rafters if the unit run is 7 inches. Round your answer to two decimal places.
elena-14-01-66 [18.8K]

Answer:

Explanation:

https://www.pole-barn.info/roof-rafter-calculations.html

4 0
3 years ago
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