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KiRa [710]
3 years ago
11

What is a jack plane used for​

Engineering
1 answer:
kolezko [41]3 years ago
4 0

A jack plane is a general purpose woodworking bench plane, used for dressing timber down to the correct size in preparation for truing and/or edge jointing.

I hope this helps you :)

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4. Which type of duct undergoes more rigorous testing before it's labeled?
MakcuM [25]

Answer: Flexible duct because, They also developed more than 1,000 safety criteria, some of which are... labelled goods bearing a silver UL sticker on the duct that defines this listing. ... lie in the research carried out by UL and, as a result, the authorized use of the code. Air Ducts are required to perform 15 separate tests.

Source: https://dundasjafine.com/ul-classification/

3 0
3 years ago
What are the four categories of engineering materials used in manufacturing?
alexgriva [62]

Answer:

metals, composite, ceramics and polymers.

Explanation:

The four categories of engineering materials used in manufacturing are metals, composite, ceramics and polymers.

i) Metals: Metals are solids made up of atoms held by matrix of electrons. They are good conductors of heat and electricity, ductile and strong.

ii) Composite: This is a combination of two or more materials. They have high strength to weight ratio, stiff, low conductivity. E.g are wood, concrete.

iii) Ceramics: They are inorganic, non-metallic crystalline compounds with high hardness and strength as well as poor conductors of electricity and heat.

iv) Polymers: They  have low weight and are poor conductors of electricity and heat

8 0
3 years ago
:) 1. Outline two differences between a moving coil -coil and a moving-iron instrument.​
GalinKa [24]

Answer:

I do this for fun

Explanation:

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6 0
3 years ago
Hot air at an average temperature of 100 degC flows through a 3 m long tube with an inside diameter of 60 mm. The temperature of
Arte-miy333 [17]

Answer:

\dot Q = 912.017\,W

Explanation:

The heat transfer rate due to convection is calculated by this formula:

\dot Q = h_{air}\cdot A_{tube, in}\cdot (T_{air} - T_{tube})

The output is determined by replacing terms:

\dot Q = (20.1\,\frac{W}{m^{2}\cdot ^{\textdegree}C})\cdot (\pi )\cdot (0.06\,m)\cdot (3\,m)\cdot (100^{\textdegree}C-20^{\textdegree}C)

\dot Q = 912.017\,W

6 0
3 years ago
Read 2 more answers
A one-dimensional slab without heat generation has a thickness of 20 mm with surfaces main- tained at temperatures of 275 K and
vlada-n [284]

Answer:

a) 512.5 KW/m2

b) 40.75 KW/m2

c) 2 KW/m2

Explanation:

Given data;

T_2 = 325 K

T_1 = 275 K

dx = 0.20 mm

a) for aluminium   K = 205 W/m k

heat flux = k \frac{dt}{dx}

               = 205 \frac{325 - 275}{0.02}

               = 512.5 KW/m2

b) for AISI 316 stainless steel

k = 16.3 W/ m k

heat flux = k \frac{dt}{dx}

               = 16.3 \frac{325 - 275}{0.02}

               = 40.75 KW/m2

C) for Concrete

k = 0.8 W/ m k

heat flux = k \frac{dt}{dx}

               = 0.8 \times \frac{325 - 275}{0.02}

               = 2 KW/m2

6 0
3 years ago
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