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1 year ago

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

1 answer:

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.

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A spherical container made of steel has 20 ft outer diameter and wal thickness of 1/2 inch. Knowing the internal pressure is 50

anastassius [24]

Answer:

maximum normal stress = 5975 psi

maximum shear stress = 2987.50 psi

Explanation:

Given data

dia = 20 ft

wall thickness = 1/2 inch

internal pressure = 50 psi

To find out

the maximum normal stress and the maximum shearing stress

Solution

By the Mohr's circle we will find out shear stress

first we calculate inner radius

i.e. r = (diameter/2) - t

r = (20 × 12 in )/2 - ( 1/2 )

r = 120 - 0.5 = 119.5 inch

Now we find out maximum normal stress by given formula

normal stress = ( internal pressure× r ) / 2 t

normal stress = ( 50×119.5 ) / 2 × 0.5

maximum normal stress = 5975 psi

and minimum normal stress is 0, due to very small radius

and maximum shear stress will be

shear stress = ( maximum normal stress - minimum normal stress ) / 2

shear stress = ( 5975- 0 ) / 2

maximum shear stress = 2987.50 psi

5 0

3 years ago

When selecting the appropriate gear for intended direction of travel, automatic transmission equipped vehicles should be placed

True [87]

Answer:

Automatic transmissions should be in Drive and Manual transmissions should be in first gear.

6 0

3 years ago

A sedimentation basin in a water treatment plant has a length = 48 m, width = 12 m, and depth = 3 m. The flow rate = 4 m 3 /s; p

Slav-nsk [51]

Answer:

The minimum particle diameter that is removed at 85% is 1.474 * 10 ^⁻4 meters.

Solution

Given:

Length = 48 m

Width = 12 m

Depth = 3m

Flow rate = 4 m 3 /s

Water density = 10 3 kg/m 3

Dynamic viscosity = 1.30710 -3 N.sec/m

Now,

At the minimum particular diameter it is stated as follows:

The Reynolds number= 0.1

Thus,

0.1 =ρVTD/μ

VT = Dp² ( ρp- ρ) g/ 10μ²

Where

gn = The case/issue of sedimentation

VT = Terminal velocity

So,

0.1 = Dp³ ( ρp- ρ) g/ 10μ²

This becomes,

0.1 = 1000 * dp³ (1100-1000) g 0.1/ 10 *(1.307 * 10 ^⁻3)²

= 3.074 * 10 ^⁻6 = dp³ (.g01 * 10^6)

dp³=3.1343 * 10 ^⁻12

Dp minimum= 1.474 * 10 ^⁻4 meters.

8 0

3 years ago

A fluid with a relative density of 0.9 flows in a pipe which is 12 m long and lies at an angle of 60° to the horizontal At the t

Minchanka [31]

Answer:

$Q=7.3\times 10^{-3} m^3/s$

Explanation:

Given that

At top $d_2=30 mm,P_2=860 KPa ,P_1=1000 KPa,d_1=85 mm$

$\rho =900\dfrac{Kg}{m^3}$

We know that

$\dfrac{P_1}{\rho g}+\dfrac{V_1^2}{2g}+Z_1=\dfrac{P_2}{\rho g}+\dfrac{V_2^2}{2g}+Z_2$

$A_1V_1=A_2V_2$

$\frac{V_1}{V_2}=\left(\dfrac{d_2}{d_1}\right)^2$

$\frac{V_1}{V_2}=\left(\dfrac{30}{85}\right)^2$

$V_2=8.02V_1$

$Z_2=12 sin60^{\circ}$

$\dfrac{1000\times 1000}{900\times 9.81}+\dfrac{V_1^2}{2\times 9.81}+0=\dfrac{860\times 1000}{900\times 9.81 }+\dfrac{V_2^2}{2\times 9.81}+12 sin60^{\circ}$