Available for ceiling and wall installation, can be covered with a finish covering material, plastered, or mounted directly to the ceiling or wall. TERE SESELT BP
A. hydronic heating systems
B. evaporative systems
C. electric radiant heating
D. panels unit cooling systems

Answers

Answer 1

Answer:

Hydronic heating system

Explanation:

From the available options, Hydronic heating systems can be covered with a finishing covering material, plastered or can be mounted directly to the celing or wall.

This system heats water and then moves it through pipes that are sealed pipes to radiators all around a house.

This sealed system is useful for heating towel rails, floor slabs, and also swimming pools, whenever it is useful

Hydronic Heating water is heated through the use of super energy


Related Questions

What test should be performed on abrasive wheels

Answers

Answer:

before wheel is put on it should be looked at for damage and a sound or ring test should be done to check for cracks, to test the wheel it should be tapped with a non metallic instrument (I looked it up)

The  test that should be performed on abrasive wheels is the ring test.

What is the purpose of the ring test on the  abrasive wheels?

The ring test can be regarded as one of the mechanical test that is used to know whether the wheel is cracked or damaged.

To carry out this test , the wheel will be arranged to be in the 45 degrees each side and it is then aligned to be at a specific diameter, this can be done by the expert in this field to know the state of that wheel.

Learn more about  ring test  on:

https://brainly.com/question/4621112

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A spherical Gaussian surface of radius R is situated in space along with both conducting and insulating charged objects. The net electric flux through the Gaussian surface is:______

Answers

Answer:

Ф = [tex]\frac{Q}{e_{0} } [ \frac{\frac{4\pi }{3 }(R)^3 }{\frac{4}{3}\pi (R)^3 } ][/tex]

Explanation:

Radius of Gaussian surface = R

Charge in the Sphere ( Gaussian surface ) = Q

lets take the radius of the sphere to be equal to radius of the Gaussian surface i.e. R

To determine the net electric flux through the Gaussian surface

we have to apply Gauci law

Ф = 4[tex]\pi r^2 E[/tex]

Ф = [tex]\frac{Q_{enc} |}{e_{0} }[/tex]

    = [tex]\frac{Q}{e_{0} } [ \frac{\frac{4\pi }{3 }(R)^3 }{\frac{4}{3}\pi (R)^3 } ][/tex]

You are playing guitar on a stool that is 22" tall. How tall is the stool if it is expressed as a combination of feet and inches?

Answers

Answer:

1 foot 10 inches

Explanation:

1 foot = 12 inches + 10 inches = 22 inches

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To provide some perspective on the dimensions of atomic defects, consider a metal specimen that has a dislocation density of 105 mm^-2 . Suppose that all the dislocations in 1000 mm^3 (1 cm^3) were somehow removed and linked end to end.

Required:
a. How far (in miles) would this chain extend?
b. Now suppose that the density is increased to 1010 mm^-2 by cold working. What would be the chain length of dislocations in 1000 mm^3 of material?

Answers

Answer:

[tex]62.14\ \text{miles}[/tex]

[tex]6213727.37\ \text{miles}[/tex]

Explanation:

The distance of the chain would be the product of the dislocation density and the volume of the metal.

Dislocation density = [tex]10^5\ \text{mm}^{-2}[/tex]

Volume of the metal = [tex]1000\ \text{mm}^3[/tex]

[tex]10^5\times 1000=10^8\ \text{mm}\\ =10^5\ \text{m}[/tex]

[tex]1\ \text{mile}=1609.34\ \text{m}[/tex]

[tex]\dfrac{10^5}{1609.34}=62.14\ \text{miles}[/tex]

The chain would extend [tex]62.14\ \text{miles}[/tex]

Dislocation density = [tex]10^{10}\ \text{mm}^{-2}[/tex]

Volume of the metal = [tex]1000\ \text{mm}^3[/tex]

[tex]10^{10}\times 1000=10^{13}\ \text{mm}\\ =10^{10}\ \text{m}[/tex]

[tex]\dfrac{10^{10}}{1609.34}=6213727.37\ \text{miles}[/tex]

The chain would extend [tex]6213727.37\ \text{miles}[/tex]

the pressure rise, across a pump can be expressed as where D is the impeller diameter, p, is the fluid density, w is the rotational speed, adn q is the flowrate. determine a suitable set of dimensionless parameters

Answers

Answer:

hello your question is incomplete below is the complete question

The pressure rise Δp across a pump can be expressed as Δp = f(D, p, w, Q) where D is the impeller diameter, p is the fluid density, w is the rotational speed, and Q is the flowrate. determine a suitable set of dimensionless parameters

answer : Δp / D^2pw^2 = Ф (Q / D^3w )

Explanation:

k ( number of variables ) = 5

r ( number of reference dimensions ) = 3

applying the pi theorem

hence the number of pi terms = k - r = 5 - 3 = 2

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