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Svetlanka [38]
3 years ago
9

What should you do before you start welding?

Engineering
1 answer:
Deffense [45]3 years ago
8 0

Explanation:

1. Weld only in authorized areas. Make sure the area is dry, chemical free, and well ventilated.

2. Inspect the equipment before starting to use it.

3. Keep other people away, unless they are authorized to be there and are wearing the appropriate personal protective equipment.

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The Role of Fuel Cells in Renewable Energy Solutions
gogolik [260]

Answer:

chemical energy directly

4 0
2 years ago
A gas stream flowing at 1000 cfm with a particulate loading of 400 gr/ft3 discharges from a certain industrial plant through an
Makovka662 [10]

<u>Solution and Explanation:</u>

Volume of gas stream = 1000 cfm (Cubic Feet per Minute)

Particulate loading = 400 gr/ft3 (Grain/cubic feet)

1 gr/ft3 = 0.00220462 lb/ft3

Total weight of particulate matter = 1000 \mathrm{cfm} \times 400 \mathrm{gr} / \mathrm{tt} 3 \times .000142857 \mathrm{lb} / \mathrm{ft} 3 \times 60=3428.568 \mathrm{lb} / \mathrm{hr}

Cyclone is to 80 % efficient

So particulate remaining = 0.20 \times 3428.568 \mathrm{lb} / \mathrm{hr}=685.7136

emissions from this stack be limited to = 10.0 lb/hr

Particles to be remaining after wet scrubber = 10.0 lb/hr

So particles to be removed = 685.7136- 10 = 675.7136

Efficiency = output multiply with 100/input = 98.542 %

4 0
2 years ago
If a worksite includes more than one set of management and workers, who should have access to the information, training, and con
Elden [556K]

Answer:

In a work site with more than one set of management and workers the Health and  safety officers in each set  should have access to the information, training and controls needed to avoid workplace accidents

Explanation:

The primary aim of a health and  safety officer in a workplace is to prevent accidents,injuries and work-related sickness from occurring in the work site by creating and implementing health and safety policies according to international standards and also ensure that these policies are implemented  by the sets of management and workers/staffs of the work site.  to achieve these they therefore should have access to the information,training and controls needed to avoid workplace accidents

4 0
3 years ago
Read 2 more answers
Suppose that you can throw a projectile at a large enough v0 so that it can hit a target a distance R downrange. Given that you
viktelen [127]

Answer:

\theta_1=15^o\\\theta_2=75^o

Explanation:

<u>Projectile Motion</u>

In projectile motion, there are two separate components of the acceleration, velocity and displacement. The horizontal component has zero acceleration (assuming no friction), and the acceleration in the vertical direction is always the acceleration of gravity. The basic formulas are shown below:

V_x=V_{ox}=V_ocos\theta

Where \theta is the angle of launch respect to the positive horizontal direction and Vo is the initial speed.

V_y=V_{oy}-gt=V_osin\theta-gt

The  horizontal and vertical distances are, respectively:

x=V_{o}cos\theta t

\displaystyle y=y_o+V_{o}sin\theta t-\frac{gt^2}{2}

The total flight time can be found as that when y = 0, i.e. when the object comes back to ground (or launch) level. From the above equation we find

\displaystyle t_f=\frac{2V_osin\theta}{g}

Using this time in the horizontal distance, we find the Range or maximum horizontal distance:

\displaystyle R=\frac{V_o^2sin2\theta}{g}

Let's solve for \theta

\displaystyle sin2\theta=\frac{R.g}{V_o^2}

This is the general expression to determine the angles at which the projectile can be launched to hit the target. Recall the angle can have to values for fixed positive values of its sine:

\displaystyle \theta_1=\frac{asin\left(\frac{R.g}{V_o^2}\right)}{2}

\displaystyle \theta_2=\frac{180^o-asin\left(\frac{R.g}{V_o^2}\right)}{2}

Or equivalently:

\theta_2=90^o-\theta_1

Given Vo=37 m/s and R=70 m

\displaystyle \theta_1=\frac{asin\left(\frac{70\times 9.8}{37^2}\right)}{2}

\theta_1=15^o

And

\theta_2=90^o-15^o=75^o

5 0
3 years ago
The mass fractions of total ferrite and total cementite in an iron–carbon alloy are 0.91 and 0.09, respectively. Is this a hypoe
WINSTONCH [101]

Answer:

hypoeutectoid

Explanation:

ferrite:  pure form of iron

cementite: It is iron carbide with 93.3% iron and 6.67% carbon

hypoeutectoid: Eutectoid steel with carbon fraction less than 0.8%

hypereutectoid: Eutectoid Steel with carbon content more than 0.8%

For the mentioned iron-carbide alloy,

% of carbon in iron-carbide alloy= percentage of cementite × percentage of carbon in cementite

% of carbon in iron-carbide alloy= 0.09× 0.0667

                                                     = 0.6%

so the alloy is hypoeutectoid

4 0
3 years ago
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