Answer:
Q=0.000604 m³/s
Explanation:
Given that
d₁=5 cm
d₂=1 cm
P= 30 KPa
Density of water ,ρ=1000 kg/m³
As we know that volume flow rate Q given as
A₁=0.0019 m²
A₂=0.000078 m²
Q=0.000604 m³/s
Answer:
sulfur dioxide
Explanation:
The scrubber is an apparatus installed in a coal-fired power plant to clean the passing gas through the smokestack. Due to the norm enacted through the clean air Act, almost all the scrubber used in the U.S is used to remove sulfur concentration from coal. it can remove approx 90-95% SO_2 from the smokestack.
Answer: I have answered the questions in rephrased sentences as below;
When implementing a safety and health program, management leadership need employee participation. Effective management of worker safety and health programs has improved employee productivity and morale in the workplace.
Nearly a third of all serious occupational injuries and illnesses stem from overexertion of repetitive motion.
Training is a way for employers to provide tools to enable employees to protect themselves and others from injuries.
Under OSHA, employees are protected from discrimination when reporting a work-related injury, illness, or fatality.
Explanation: All personnel including management & employees must be directly involved when workplace HSE policies are being made & reviewed. This is because everyone in the work environment is impacted one way or the other when incidents occur.
Training & Reporting are key responsibilities of managers, employers & supervisors, so it is mandatory to be done without discrimination so as to foster employees happiness which ultimately lead to zero incidents & increased productivity & profit.
Answer:
Computation of the load is not possible because E(test) >E(yield)
Explanation:
We are asked to ascertain whether or not it is possible to compute, for brass, the magnitude of the load necessary to produce an elongation of 7.0 mm (0.28 in.). It is first necessary/ important to compute the strain at yielding from the yield strength and the elastic modulus, and then the strain experienced by the test specimen. Then, if
E(test) is less than E(yield), deformation is elastic and the load may be computed. However is E(test) is greater than E(yield) computation/determination of the load is not possible even though defamation is plastic and we have neither a stress-strain plot or a mathematical relating plastic stress and strain. Therefore, we can compute these two values as:
Calculation of E(test is as follows)
E(test) = change in l/lo= Elongation produced/stressed tension= 7.0mm/267mm
=0.0262
Computation of E(yield) is given below:
E(yield) = σy/E=275Mpa/103 ×10^6Mpa= 0.0027
Therefore, we won't be able to compute the load because for computation to take place, E(test) <E(yield). In this case, E(test) is greater than E(yield).
