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kati45 [8]
1 year ago
15

1. Discuss the benefits of observing good safety measures in relation to an increase in

Engineering
1 answer:
Aneli [31]1 year ago
8 0

The benefits of observing good safety measures in relation to an increase in productivity within a pharmaceutical laboratory is that the act of adhering to all these policies helps a lot of employees to hinder the  spills of chemicals and other kinds of  accidents, as well as reduce the damage to the environment that is outside of the lab.

<h3>What are the benefits of practicing safety in the laboratory?</h3>

A laboratory is known to be one that is known to have a lot of potential risks that is said to often arise due to  a person's exposure to chemicals that are corrosive and toxic, flammable solvents, high pressure gases and others.

Therefore,  A little care and working in line to all the prescribed safety guidelines will help a person to be able to avoid laboratory mishaps.

Therefore, The benefits of observing good safety measures in relation to an increase in productivity within a pharmaceutical laboratory is that the act of adhering to all these policies helps a lot of employees to hinder the  spills of chemicals and other kinds of  accidents, as well as reduce the damage to the environment that is outside of the lab.

Learn more about safety measures  from

brainly.com/question/26264740

#SPJ1

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Shielding gases are used to protect the molten metal from what?
Gre4nikov [31]

Answer:Oxygen,Carbon dioxide,Nitrogen

Explanation:

4 0
2 years ago
Air enters a compressor steadily at the ambient conditions of 100 kPa and 22°C and leaves at 800 kPa. Heat is lost from the comp
telo118 [61]

Answer:

a) 358.8K

b) 181.1 kJ/kg.K

c) 0.0068 kJ/kg.K

Explanation:

Given:

P1 = 100kPa

P2= 800kPa

T1 = 22°C = 22+273 = 295K

q_out = 120 kJ/kg

∆S_air = 0.40 kJ/kg.k

T2 =??

a) Using the formula for change in entropy of air, we have:

∆S_air = c_p In \frac{T_2}{T_1} - Rln \frac{P_2}{P_1}

Let's take gas constant, Cp= 1.005 kJ/kg.K and R = 0.287 kJ/kg.K

Solving, we have:

[/tex] -0.40= (1.005)ln\frac{T_2}{295} ln\frac{800}{100}[/tex]

-0.40= 1.005(ln T_2 - 5.68697)- 0.5968

Solving for T2 we have:

T_2 = 5.8828

Taking the exponential on the equation (both sides), we have:

[/tex] T_2 = e^5^.^8^8^2^8 = 358.8K[/tex]

b) Work input to compressor:

w_in = c_p(T_2 - T_1)+q_out

w_in = 1.005(358.8 - 295)+120

= 184.1 kJ/kg

c) Entropy genered during this process, we use the expression;

Egen = ∆Eair + ∆Es

Where; Egen = generated entropy

∆Eair = Entropy change of air in compressor

∆Es = Entropy change in surrounding.

We need to first find ∆Es, since it is unknown.

Therefore ∆Es = \frac{q_out}{T_1}

\frac{120kJ/kg.k}{295K}

∆Es = 0.4068kJ/kg.k

Hence, entropy generated, Egen will be calculated as:

= -0.40 kJ/kg.K + 0.40608kJ/kg.K

= 0.0068kJ/kg.k

3 0
3 years ago
Can a 1½ " conduit, with a total area of 2.04 square inches, be filled with wires that total 0.93 square inches if the maximum f
Papessa [141]

Answer:

it is not possible to place the wires in the condui

Explanation:

given data

total area = 2.04 square inches

wires total area = 0.93 square inches

maximum fill conduit =  40%

to find out

Can it is possible place wire in conduit conduit

solution

we know maximum fill is 40%

so here first we get total area of conduit that will be

total area of conduit = 40% × 2.04

total area of conduit = 0.816 square inches

but this area is less than required area of wire that is 0.93 square inches

so we can say it is not possible to place the wires in the conduit

4 0
3 years ago
HELP!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
trasher [3.6K]

Answer:

The publication of a parody for commercial gain does not fall within the protection afforded by Section 107, as it is used for commercial gain.

Explanation:

<h2><u><em>PLEASE MARK AS BRAINLIEST!!!!!</em></u></h2>
4 0
2 years ago
How much does 5 acres of land cost in the US?
Dvinal [7]

Answer:

Larson, an economist, who placed the total value of $23 trillion for the entire 1.9 billion acres of land in the United States. This means that the average cost for an acre of land is $12,000 or $60,000 for 5 acres of land. Almost half of the land in the US is used for agricultural purposes.

Explanation:

6 0
3 years ago
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