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11111nata11111 [884]

3 years ago

7

If OSHA determines that an employer's response to a non-formal complaint is adequate, what options does the employee filing the

non-formal complaint have?

1 answer:

Anuta_ua [19.1K]3 years ago

4 0

Fill it out without telling ur employer as that may cause backlash and have an osha certified employee come check out ur work or job site

You might be interested in

A piston–cylinder device contains a mixture of 0.5 kg of H2 and 1.2 kg of N2 at 100 kPa and 300 K. Heat is now transferred to th

Taya2010 [7]

Answer:

(a) The heat transferred is 2552.64 kJ

(b) The entropy change of the mixture is 1066.0279 J/K

Explanation:

Here we have

Molar mass of H₂ = 2.01588 g/mol

Molar mass of N₂ = 28.0134 g/mol

Number of moles of H₂ = 500/2.01588 = 248 moles

Number of moles of N₂ = 1200/28.0134 = 42.8 moles

P·V = n·R·T

V₁ = n·R·T/P = 290.8×8.3145×300/100000 = 7.25 m³

Since the volume is doubled then

V₂ = 2 × 7.25 = 14.51 m³

At constant pressure, the temperature is doubled, therefore

T₂ = 600 K

If we assume constant specific heat at the average temperature, we have

Heat supplied = m₁×cp₁×dT₁ + m₂×cp₂×dT₂

cp₁ = Specific heat of hydrogen at constant pressure = 14.50 kJ/(kg K

cp₂ = Specific heat of nitrogen at constant pressure = 1.049 kJ/(kg K

Heat supplied = 0.5×14.50×300 K+ 1.2×1.049×300 = 2552.64 kJ

b) $\Delta S = - R(n_A \times lnx_A + n_B \times ln x_B)$

Where:

$x_A$ and $x_B$ are the mole fractions of Hydrogen and nitrogen respectively.

Therefore, $x_A$ = 248 /(248 + 42.8) = 0.83

$x_B$ = 42.8/(248 + 42.8) = 0.1472

∴ $\Delta S = - 8.3145(248 \times ln0.83 + 42.8 \times ln 0.1472)$ = 1066.0279 J/K

5 0

3 years ago

A smooth ceramic sphere (SG 5 2.6) is immersed in a fl ow of water at 208C and 25 cm/s. What is the sphere diameter if it is enc

Aleks [24]

Answer:

a. 4 $\mu m$

b. 1 m

Explanation:

According to the question, the data is as follows

The Density of water at 20 degrees celcius is 1000 kg/m^3

Viscosity is 0.001kg/m/.s

Velocity V = 25 cm/s

V = 0.25 m/s

Now

a. The creeping motion is

As we know that

Reynold Number = (Density of water × V × d) ÷ (Viscosity)

1 = (1,000 × 0.25 × d) ÷ 0.0001

d = (1 × 0.001) ÷ (1,000 × 0.25)

= 4E - 06^m

= 4 $\mu m$

b. Now the sphere diameter is

Reynold Number = (Density of water × V × d) ÷ (Viscosity)

250,000 = (1,000 × 0.25 × d) ÷ 0.0001

d = (250,000 × 0.001) ÷ (1,000 × 0.25)

= 1 m

6 0

3 years ago

Which metal is used in planes.

wel

Answer:

<h2>Steel</h2>

Explanation:

Steel is the metal that using in planes.

Aluminum and titanium also used in this aircraft industry.

Aluminum is ideal for aircraft manufacture because it's lightweight and strong.

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5 0

3 years ago

Read 2 more answers

A gasoline engine has a piston/cylinder with 0.1 kg air at 4 MPa, 1527◦C after combustion, and this is expanded in a polytropic

Roman55 [17]

Answer:

The expansion work is 71.24 kJ and heat transfer is -16.89 kJ

Explanation:

From ideal gas law,

Initial volume (V1) = nRT/P

n is the number of moles of air in the cylinder = mass/MW = 0.1/29 = 0.00345 kgmol

R is gas constant = 8314.34 J/kgmol.K

T is initial temperature = 1527 °C = 1527+273 = 1800 K

P is initial pressure = 4 MPa = 4×10^6 Pa

V1 = 0.00345×8314.34×1800/(4×10^6) = 0.013 m^3

V2 = 10×V1 = 10×0.013 = 0.13 m^3

The process is a polytropic expansion process

polytropic exponent (n) = 1.5

P2 = P1(V1/V2)^n = 4×10^6(0.013/0.13)^1.5 = 1.26×10^5 Pa

Expansion work = (P1V1 - P2V2) ÷ (n - 1) = (4×10^6 × 0.013 - 1.26×10^5 × 0.13) ÷ (1.5 - 1) = 35620 ÷ 0.5 = 71240 J = 71240/1000 = 71.24 kJ

Heat transfer = change in internal energy + expansion work

change in internal energy (∆U) = Cv(T2 - T1)

T2 = PV/nR = 1.26×10^5 × 0.13/0.00345×8314.34 = 571 K

Cv = 20.785 kJ/kgmol.K

∆U = 20.785(571 - 1800) = -25544.765 kJ/kgmol × 0.00345 kgmol = -88.13 kJ

Heat transfer = -88.13 + 71.24 = -16.89 kJ

5 0

3 years ago

A cylindrical specimen of steel has an original diameter of 12.8 mm. It is tested in tension its engineering fracture strength i

Mama L [17]

Answer:

a) The ductility = -30.12%

the negative sign means reduction

Therefore, there is 30.12% reduction

b) the true stress at fracture is 658.26 Mpa

Explanation:

Given that;

Original diameter $d_{o}$ = 12.8 mm

Final diameter $d_{f}$ = 10.7

Engineering stress $\alpha _{E}$ = 460 Mpa

a) determine The ductility in terms of percent reduction in area;

Ai = π/4( $d_{o}$ )² ; Ag = π/4( $d_{f}$ )²

% = π/4 [ ( ( $d_{f}$ )² - ( $d_{o}$ )²) / ( π/4 ( $d_{o}$ )²) ]

= ( ( $d_{f}$ )² - ( $d_{o}$ )²) / ( $d_{o}$ )² × 100

we substitute

= [( (10.7)² - (12.8)²) / (12.8)² ] × 100

= [(114.49 - 163.84) / 163.84 ] × 100

= - 0.3012 × 100

= -30.12%

the negative sign means reduction

Therefore, there is 30.12% reduction

b) The true stress at fracture;

True stress $\alpha _{T}$ = $\alpha _{E}$ ( 1 + $E_{E}$ )

$E_{E}$ is engineering strain

$E_{E}$ = dL / Lo

= (do² - df²) / df² = (12.8² - 10.7²) / 10.7² = (163.84 - 114.49) / 114.49

= 49.35 / 114.49

$E_{E}$ = 0.431

so we substitute the value of $E_{E}$ into our initial equation;

True stress $\alpha _{T}$ = 460 ( 1 + 0.431)

True stress $\alpha _{T}$ = 460 (1.431)

True stress $\alpha _{T}$ = 658.26 Mpa

Therefore, the true stress at fracture is 658.26 Mpa

6 0

2 years ago