Answer:
2ib
Explanation:
if you divide 10 divided by 2 it gives you 5 and then subtract it by 2.2 = 2.8
there goes your answer.
Answer:
Yes, it is possible to maintain a pressure of 10 kPa in a condenser that is being cooled by river water that is entering at 20 °C because this temperature (20 °C) of the external cooling water is less than the saturation temperature of steam which is which is 45.81 °C, and heated by a boiler; as a result of this condition, coupled with the assumption that the turbine, pump, and interconnecting tube are adiabatic, and the condenser exchanges its heat with the external cooling river water, it possible to maintain a pressure of 10 kPa.
Answer:
C₁₀ = 6.3 KN
Explanation:
The catalog rating of a bearing can be found by using the following formula:
C₁₀ = F [Ln/L₀n₀]^1/3
where,
C₁₀ = Catalog Rating = ?
F = Design Load = 2.75 KN
L = Design Life = 1800 rev/min
n = No. of Hours Desired = 10000 h
L₀ = Rating Life = 500 rev/min
n₀ = No. of Hours Rated = 3000 h
Therefore,
C₁₀ = [2.75 KN][(1800 rev/min)(10000 h)/(500 rev/min)(3000 h)]^1/3
C₁₀ = (2.75 KN)(2.289)
<u>C₁₀ = 6.3 KN</u>
Answer:
Option D
160 kHz
Explanation:
Since we must use at least one synchronization bit, total message signal is 15+1=16
The minimum sampling frequency, fs=2fm=2(5)=10 kHz
Bandwith, BW required is given by
BW=Nfs=16(10)=160 kHz
