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kati45 [8]
2 years ago
15

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

Engineering
1 answer:
Aneli [31]2 years 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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An industrial load with an operating voltage of 480/0° V is connected to the power system. The load absorbs 120 kW with a laggin
Leni [432]

Answer:

Q=41.33 KVAR\ \\at\\\ 480 Vrms

Explanation:

From the question we are told that:

Voltage V=480/0 \textdegree V

Power P=120kW

Initial Power factor p.f_1=0.77 lagging

Final Power factor p.f_2=0.9 lagging

Generally the equation for Reactive Power is mathematically given by

Q=P(tan \theta_2-tan \theta_1)

Since

p.f_1=0.77

cos \theta_1 =0.77

\theta_1=cos^{-1}0.77

\theta_1=39.65 \textdegree

And

p.f_2=0.9

cos \theta_2 =0.9

\theta_2=cos^{-1}0.9

\theta_2=25.84 \textdegree

Therefore

Q=P(tan 25.84 \textdegree-tan 39.65 \textdegree)

Q=120*10^3(tan 25.84 \textdegree-tan 39.65 \textdegree)

Q=-41.33VAR

Therefore

The size of the capacitor in vars that is necessary to raise the power factor to 0.9 lagging is

Q=41.33 KVAR\ \\at\\\ 480 Vrms

6 0
3 years ago
What. is the capital city of panjab​
kotegsom [21]

Answer:

Chandigarh

Explanation:

Chandigarh city is the capital of the territory and of the states of Haryana and Punjab. Chandigarh's name, meaning “stronghold of the goddess Chandi,” is derived from the Chandi Mandir, a temple dedicated to the goddess that is located near the town of Mani Majra. Area union territory, 44 square miles (114 square km).

8 0
3 years ago
The velocity components u and v in a two-dimensional flow field are given by: u = 4yt ft./s, v = 4xt ft./s, where t is time. Wha
zmey [24]

Answer:

vec(a) = 16 i + 16 j

mag(a) = 22.63 ft/s^2

Explanation:

Given,

- The two components of velocity are given for fluid flow:

                                       u = 4*y ft/s

                                       v = 4*x ft/s

Find:

What is the time rate of change of the velocity vector V (i.e., the acceleration vector) for a fluid particle at x = 1 ft. and y = 1 ft. at time t = 1 second?

Solution:

- The rate of change of velocity is given to be acceleration. We will take derivative of each components of velocity with respect to time t:

                                      a_x = du / dt

                                      a_x = 4*dy/dt

                                      a_y = dv/dt

                                      a_y = 4*dx/dt

- The expressions dx/dt is the velocity component u and dy/dt is the velocity component v:

                                     a_x = 4*(4*y) = 16y

                                     a_y = 4*(4*x) = 16x

- The acceleration vector can be expressed by:

                                     vec(a) = 16y i + 16x j

- Evaluate vector (a) at x = 1 and y = 1:

                                     vec(a) = 16*1 i + 16*1 j = 16 i + 16 j

- The magnitude of acceleration is given by:

                                     mag(a) = sqrt ( a^2_x + a^2_y )

                                     mag(a) = sqrt ( 16^2 + 16^2 )

                                     mag(a) = 22.63 ft/s^2

7 0
3 years ago
Rearrange the formula to make “u” the subject. <br><br> v - u<br> ——— = t<br> a
solong [7]

Answer:

u = v - a * t

Explanation:

   v - u

t = ------

       a

v  - u  =  a * t

v - a * t = u

therefore,  u = v - a * t

5 0
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Why become an Android programmer?
Tpy6a [65]
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