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3 years ago

What are the three categories of tools?

Engineering

1 answer:

noname [10]3 years ago

6 0

Answer:

1. Low power hand tools/small

2. Light to medium industrial tools

3. Large industrial tools

There are definitely a lot more categories than three, but this is what I have.

You might be interested in

Which metal is used in planes.

wel

Answer:

<h2>Steel</h2>

Explanation:

Steel is the metal that using in planes.

Aluminum and titanium also used in this aircraft industry.

Aluminum is ideal for aircraft manufacture because it's lightweight and strong.

hope this helps!!

have a nice day!

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5 0

3 years ago

Read 2 more answers

5) Calculate the LMC wal thickness of a pipe and tubing with OD as 35 + .05 and ID as 25 + .05 A) 4.95 B) 5.05 C) 10 D) 15.025

padilas [110]

Answer:

B) 5.05

Explanation:

The wall thickness of a pipe is the difference between the diameter of outer wall and the diameter of inner wall divided by 2. It is given by:

Thickness of pipe = (Outer wall diameter - Inner wall diameter) / 2

Given that:

Inner diameter = ID = 25 ± 0.05, Outer diameter = OD = 35 ± 0.05

Maximum outer diameter = 35 + 0.05 = 35.05

Minimum inner diameter = 25 - 0.05 = 24.95

Thickness of pipe = (maximum outer wall diameter - minimum inner wall diameter) / 2 = (35.05 - 24.95) / 2 = 5.05

Thickness = (35 - 25) / 2 + 0.05 = 10/2 + 0.05 = 5 + 0.05 = 5.05

Therefore the LMC wall thickness is 5.05

6 0

3 years ago

Read 2 more answers

A container filled with a sample of an ideal gas at the pressure of 150 Kpa. The gas is compressed isothermally to one-third of

lyudmila [28]

Answer: c) 450 kPa

Explanation:

Boyle's Law: This law states that pressure is inversely proportional to the volume of the gas at constant temperature and number of moles.

$P\propto \frac{1}{V}$ (At constant temperature and number of moles)

$P_1V_1=P_2V_2$

where,

$P_1$ = initial pressure of gas = 150 kPa

$P_2$ = final pressure of gas = ?

$V_1$ = initial volume of gas = v L

$V_2$ = final volume of gas = $\frac{v}{3}L$

$150\times v=P_2\times \frac{v}{3}$

$P_2=450kPa$

Therefore, the new pressure of the gas will be 450 kPa.

7 0

3 years ago

A cone penetration test was carried out in normally consolidated sand, for which the results are summarized below: Depth (m) Con

Cerrena [4.2K]

Answer:

hello your question is incomplete attached below is the missing equation related to the question

answer : 40.389° , 38.987° , 38° , 39.869° , 40.265°

Explanation:

Determine the friction angle at each depth

attached below is the detailed solution

To calculate the vertical stress = depth * unit weight of sand

also inverse of Tan = Tan^-1

also qc is in Mpa while σ0 is in kPa

Friction angle at each depth

2 meters = 40.389°

3.5 meters = 38.987°

5 meters = 38.022°

6.5 meters = 39.869°

8 meters = 40.265°

6 0

3 years ago

All holly plants are dioecious—a male plant must be planted within 45 to 55 feet of the

mel-nik [20]

Answer:

The answer is 0.727

Explanation:

lemme know if that's right

8 0

3 years ago