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viktelen [127]
3 years ago
10

What is the most common injury caused by working with machines unsafely?

Engineering
2 answers:
VladimirAG [237]3 years ago
8 0

Answer:

heyooo!!!!!!

Burns and scarring.

hope this helps!!!!

Explanation:

Lena [83]3 years ago
6 0

Answer:

<u>The most common injury caused by working with machinery are cuts and burns.</u>

Explanation:

Many times when working with machinery there are certain risks of damage to the personnel who handle them, so it is correct to request protective equipment for workers.

For example, if the most common injuries are <u>cuts and burns</u>, workers are asked to wear gloves or any protective equipment that covers them to avoid these damages when handling materials.

Safety shoes, glasses, gloves, helmet, mask and hearing protectors are the most recommended for any work with machinery.

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A cylinder with a 6.0 in. diameter and 12.0 in. length is put under a compres-sive load of 150 kips. The modulus of elasticity f
jeka94

Answer:

Final Length = 11.992 in

Final Diameter = 6.001 in

Explanation:

First we calculate the cross-sectional area:

Area = A = πr² = π(3 in)² = 28.3 in²

Now, we calculate the stress:

Stress = Compressive Load/Area

Stress = - 150 kips/28.3 in²

Stress = -5.3 ksi

Now,

Modulus of Elasticity = Stress/Longitudinal Strain

8000 ksi = -5.3 ksi/Longitudinal Strain

Longitudinal Strain = -6.63 x 10⁻⁴

but,

Longitudinal Strain = (Final Length - Initial Length)/Initial Length

-6.63 x 10⁻⁴ = (Final Length - 12 in)/12 in

Final Length = (-6.63 x 10⁻⁴)(12 in) + 12 in

<u>Final Length = 11.992 in</u>

we know that:

Poisson's Ratio = - Lateral Strain/Longitudinal Strain

0.35 = - Lateral Strain/(- 6.63 x 10⁻⁴)

Lateral Strain = (0.35)(6.63 x 10⁻⁴)

Lateral Strain = 2.32 x 10⁻⁴

but,

Lateral Strain = (Final Diameter - Initial Diameter)/Initial Diameter

2.32 x 10⁻⁴ = (Final Diameter - 6 in)/6 in

Final Diameter = (2.32 x 10⁻⁴)(6 in) + 6 in

<u>Final Diameter = 6.001 in</u>

8 0
3 years ago
Dean is buying a home for $170,000. The mortgage company he decided to use to finance the home requires a 10% down payment. What
ANTONII [103]
They answer is 3. $17,00
8 0
3 years ago
Read 2 more answers
As part of the overall systems engineering process, there are a variety of software development methods, but the three most comm
saw5 [17]

The three most common software development methods are the Waterfall Approach, the Incremental Approach, and the SPIRAL Approach. These methods depend on the team size and specific goals.

Software development is the sequential procedure that involves the division of the work into smaller and parallel stages in order to improve software design and product management.

The software development methods depend on both the team size and specific objectives.

The most common methodologies for software development include:

  • Waterfall
  • Spiral
  • Incremental
  • Agile
  • Continuous integration

Learn more about software development here:

brainly.com/question/14275830

8 0
2 years ago
When in regular operation, ash and other debris should be removed _______ from the combustion chamber of a wood-burning heater.
Klio2033 [76]

Answer:

    Daily from the combustion chamber of a wood-burning heater.

<h3>Explanation:</h3>
  • As a wood stove heats up, it radiates heat through the walls and top of the stove
  • This radiant heat warms the immediate area and can be carried into other parts of the home via the home's natural airflow.
  • Electric or convection-powered fans can help circulate this heat to warm a larger area.

To learn more about it, refer

to brainly.com/question/23275071

#SPJ4

8 0
1 year ago
Air is contained inside a vertical piston-cylinder assembly by a piston of mass 105 kg and face area of 0.05 m2 . The mass of ai
alexgriva [62]

Answer:

ΔQ = 1.06 KJ

Explanation:

The amount of heat transfer between the piston-cylinder system and the surrounding can easily be found by using the First Law of Thermodynamics. The first law of thermodynamics can be written as follows:

ΔQ = ΔU + W

ΔQ = mΔu + PΔV

where,

ΔQ = Heat transfer between system and surrounding = ?

Δu = specific change in internal energy of the system = - 175 KJ/kg

m = mass of air = 20 g = 0.02 kg

P = Constant Pressure = 101.3 KPa

ΔV = Change in Volume = 0.05 m³ - 0.005 m³ = 0.045 m³

Therefore,

ΔQ = (0.02 kg)(-175 KJ/kg) + (101.3 KPa)(0.045 m³)

ΔQ = -3.5 KJ + 4.56 KJ

<u>ΔQ = 1.06 KJ</u>

4 0
3 years ago
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