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3 years ago

Which of the following is an example of seeking accreditation?

Engineering

1 answer:

USPshnik [31]3 years ago

3 0

Answer:

withdraw

blow

view

reviede

draw

throw

show

interview

allow

grow

know

borrow

narrow

renew

flow

bow

swallow

slow

Explanation:

withdraw

blow

view

review

draw

throw

show

interview

allow

grow

know

borrow

narrow

renew

flow

bow

swallow

slow

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According to OSHA standards, the air in the building that John works in is unsafe. The type of regulation that OSHA engages in i

ioda

Answer:

social regulation.

Explanation:

Social regulation are rules set aside to protect the environment or restrain activities that poses threat to public health and safety, examples includes environment pollution which includes lands, air, water etc, unhealthy work environment, etc. This rules identify activities that are allowed or under sanction for individuals, firms and government, breaking this rules most times comes with heavy fines or sanctions.

Social regulation help to see to the safety and well being of our environment, it serves as a guide for human activities.

7 0

3 years ago

Read 2 more answers

thier only motto and goal is to work for society and not make any profits A.small business entreprenuership B.scalable start up

sweet-ann [11.9K]

Answer:

social entrepreneurship

8 0

3 years ago

Read 2 more answers

Select the right answer<br>

Kruka [31]

Answer:

for 1st question the answer is 5th option.

for 2nd question the answer is 2nd option

hope it helps you mate

please mark me as brainliast

5 0

3 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

Calculate the viscosity(dynamic) and kinematic viscosity of airwhen

nikitadnepr [17]

Answer:

(a) dynamic viscosity = $1.812\times 10^{-5}Pa-sec$

(b) kinematic viscosity = $1.4732\times 10^{-5}m^2/sec$

Explanation:

We have given temperature T = 288.15 K

Density $d=1.23kg/m^3$

According to Sutherland's Formula dynamic viscosity is given by

${\mu} = {\mu}_0 \frac {T_0+C} {T + C} \left (\frac {T} {T_0} \right )^{3/2}$ , here

μ = dynamic viscosity in (Pa·s) at input temperature T,

$\mu _0$ = reference viscosity in(Pa·s) at reference temperature T0,

T = input temperature in kelvin,

$T_0$ = reference temperature in kelvin,

C = Sutherland's constant for the gaseous material in question here C =120

$\mu _0=4\pi \times 10^{-7}$

$T_0$ = 291.15

$\mu =4\pi \times 10^{-7}\times \frac{291.15+120}{285.15+120}\times \left ( \frac{288.15}{291.15} \right )^{\frac{3}{2}}=1.812\times 10^{-5}Pa-s$ when T = 288.15 K

For kinematic viscosity :

$\nu = \frac {\mu} {\rho}$

$kinemic\ viscosity=\frac{1.812\times 10^{-5}}{1.23}=1.4732\times 10^{-5}m^2/sec$

3 0

3 years ago