Answer:
Damping ratio 
Explanation:
Given that
m=4.2 kg,K=85.9 N/m,C=1.3 N.s/m
We need to find damping ratio
We know that critical damping co-efficient
N.s/m
Damping ratio(
) is the ratio of damping co-efficient to the critical damping co-efficient
So 
So damping ratio
Answer:
The correct answer is letter "C": Both.
Explanation:
Industrial seals are used at interfaces between components to prevent leakage, to maintain heat, and to avoid contamination. The design, construction, and materials they use vary depending on industrial use but the most common are Polytetrafluoroethylene (PTFE), Nitrile Buna Rubber (NBR), and fluorocarbon.
Thus, using a sharp chisel could pry a seal out of a hole and a regular socket can often be used to force smaller metal-backed seals into place. Thus, technicians "A" and "B" are correct.
A: Inexhaustible resources are limited
is true
Option C (making lessons learned a regular part of meetings) is the correct approach.
- As nothing more than a general rule, typically construction companies only plan lessons that have been learned exercises or initiatives towards the end of a particular endeavor or segment.
- As almost a result of the team knowing, valuable lessons are intended to increase the comprehensive implementation of quality management practices as well as deadlines.
Aside from this, none of the choices are viable methods to learning lessons or gaining knowledge from the past. As a result, the methodology outlined above is the appropriate one.
Learn more about project teamwork here:
brainly.com/question/14279121
Answer:
See step by step explanations for answer.
Explanation:
600 megawatts =
568 690.272 btu / second
thermal eficiency=work done/Heat supllied
0.38=568690.272/Heat supplied
Heat supplied=1496553.35btu /s
heat emmitted to the atmosphere=heat supplied -work done=(1496553.35-568690.272)=927863.1 btu/s
feed rate=(1496553.35)/12000=124.71 lb/s =10775184.1056 lb/day=5 387.472 ton / day
sulphur content released=(0.03*124.71)/(1.496553)=2.5 lb SO2/million Btu of heat input
so
the degree (%) of sulfur dioxide control needed to meet an emission standard=(2.5/0.15)*100=1666.67 %
the CO2 emission rate=220*(1.496553) =329.241 lb/s =12 903.0802 metric ton / day
