Answer:
a) 70.29 %
b) 37%
Explanation:
percent reduction can be found from:
PR = 100*(π(do/2)^2-π(df/2)^2)/π(do/2)^2
= 100*(π(11.34/2)^2-π(6.21/2)^2)/π(11.34/2)^2
=70.29 %
percent elongation can be found from:
EL =L_f - Lo/Lo*100
= (73.17 -53.3/53.3)*100
= 37%
Answer:
Yes
Explanation:
Yes it is true that COP of heat pump always greater than 1.But the COP of refrigeration can be greater or less than 1.
We know that
COP of heat pump= 1 + COP of refrigeration
It is clear that COP can not be negative .So from the above expression we can say that COP of heat pump is always greater than one.
Answer:
15,000 psi
Explanation:
The solution / solving is attach below.
Answer:
At the point when the quantity of bit strings is not exactly the quantity of processors, at that point a portion of the processors would stay inert since the scheduler maps just part strings to processors and not client level strings to processors. At the point when the quantity of part strings is actually equivalent to the quantity of processors, at that point it is conceivable that the entirety of the processors may be used all the while. Be that as it may, when a part string obstructs inside the portion (because of a page flaw or while summoning framework calls), the comparing processor would stay inert. When there are more portion strings than processors, a blocked piece string could be swapped out for another bit string that is prepared to execute, in this way expanding the use of the multiprocessor system.When the quantity of part strings is not exactly the quantity of processors, at that point a portion of the processors would stay inert since the scheduler maps just bit strings to processors and not client level strings to processors. At the point when the quantity of bit strings is actually equivalent to the quantity of processors, at that point it is conceivable that the entirety of the processors may be used at the same time. Be that as it may, when a part string hinders inside the piece (because of a page flaw or while summoning framework calls), the relating processor would stay inert. When there are more portion strings than processors, a blocked piece string could be swapped out for another bit string that is prepared to execute, along these lines expanding the usage of the multiprocessor framework.
Answer:
42.50 dB
Explanation:
Determine the minimum voltage gain
amplitude of input signal ( Vi ) = 15 mV
amplitude of output signal ( Vo) = 2 V
Vo = 2 v
therefore ; minimum gain = Vo / Vi = 2 / ( 15 * 10^-3 )
= 133.33
Minimum gain in DB = 20 log ( 133.33 )
= 42.498 ≈ 42.50 dB
