Which tool is used for cutting bricks and other masonry materials with precision?

Answers

Answer 1

Answer:

A turbo blade has the best features from both other types of blade. The continuous, serrated edge makes for fast cutting while remaining smooth and clean. They are mostly used to cut a variety of materials, such as tile, stone, marble, granite, masonry, and many other building materials.

Explanation:

Answer 2

Answer:

Masonry Saw

Explanation:

Masonry Saws can be used to cut brick, ceramic, tile and or stone.


Related Questions

The quantity of bricks required increases with the surface area of the wall, but the thickness of a masonry wall does not affect the total quantity of bricks used in the wall

True or False

Answers

Answer:

false

Explanation:

False is the answer:)!

Forcing a solid piece of heated aluminum through a die forms:
A. a stamped part
B.a cast part
C.an extruded part.
D.a forged part

Answers

Answer:

B a cast part

Explanation:

Extrusion is defined as the process of shaping material, such as aluminum, by forcing it to flow through a shaped opening in a die. Extruded material emerges as an elongated piece with the same profile as the die opening.

Forcing a solid piece of heated aluminum through a die forms: a cast part. Hence, option B is correct.

What is cast part?

A liquid element is more often filled with concrete that has a hollow chamber in the correct form during the casting manufacturing process, and the item is then let to harden.

A casting, which is the term for the solidified component, is ejected or broken out of the mould to complete the procedure.

Thus, option B is correct.

For more details about cast part, click here:

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trevor moves a magnetic toy train away from a magnet that cannot move. what happens to the potential energy in the system of magnets during the movement?

Answers

Answer:a

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How do you describe sound? (SELECT ALL THAT APPLY.) PLEASE HELP AND SELECT ALL THAT APPLY PLEASE!! A. Sound waves have to have a medium to travel through. B. The volume of a sound is known as amplitude. C. Loud sounds have high amplitude and vibrate with more energy than soft sounds. D. Sound waves are compression waves that cause energy transfer in air molecules.

Answers

Answer:

Sound waves are compression waves that cause energy transfer in air molecules

Sound waves have to have a medium to travel through

Loud sounds have high amplitude and vibrate with more energy than soft sounds

Explanation:

Sound waves is a form of energy composed of compression and rare factions. Sound waves are compression waves that cause energy transfer in air molecules.

Sound is an example of a mechanical wave hence it requires a material medium for propagation.

The amplitude of a sound wave determines its loudness or volume. A larger amplitude implies that we will have a louder sound, and a smaller amplitude means that we will have a softer sound.

An open-circuit wind tunnel draws in air from the atmosphere through a well-contoured nozzle. In the test section, where the flow is straight and nearly uniform, a static pressure tap is drilled into the tunnel wall. A manometer connected to the tap shows that static pressure within the tunnel is 45 mm of water below atmospheric. Assume that the air is incompressible, and at 25 C, 100 kPa absolute. Calculate the air speed in the wind-tunnel test section

Answers

Answer:

Air speed in the wind-tunnel [tex]v_{2}[/tex] = 27.5 m/s

Explanation:

Given data :

Manometer reading ; p1 - p2 = 45 mm of water

Pressure at section ( I ) p1 = 100 kPa ( abs )

temperature ( T1 ) = 25°C

Pw ( density of water ) = 999 kg/m3

g = 9.81 m/s^2

next we apply Bernoulli equation at section 1 and section 2

p1 - p2 = [tex]\frac{PairV^{2} _{2} }{2}[/tex]     ----------  ( 1 )

considering  ideal gas equation

Pair ( density of air ) = [tex]\frac{P}{RT}[/tex] ------------------- ( 2 )

R ( constant ) = 287 NM/kg.k

T = 25 + 273.15 = 298.15 k

P1 = 100 kN/m^2 = 100 * 10^3 or N/m^2

substitute values into equation ( 2 )

= 100 * 10^3 / (287 * 298.15)

= 1.17 kg/m^3

Also note ; p1 - p2 = PwgΔh  ------- ( 3 )

finally calculate the Air speed in the wind-tunnel test section by equating equation ( 1 ) and ( 3 )

[tex]\frac{PairV^{2} _{2} }{2}[/tex]   =  PwgΔh  

[tex]V^{2} _{2}[/tex] = [tex]\frac{2*999* 9.81* 0.045}{1.17}[/tex]  =  753.86

[tex]v_{2}[/tex] = 27.5 m/s

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