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:
Web Browser
Explanation:
Because you dont use a messaging app or presentation software to look up stuff its common knowledge
Answer:
¿Qué aplicación estás usando? podría ser porque te equivocaste un paso
Answer:
MRR = 1.984
Explanation:
Given that
Depth of cut ,d=0.105 in
Diameter D= 1 in
Speed V= 105 sfpm
feed f= 0.015 ipr
Now the metal removal rate given as
MRR= 12 f V d
d= depth of cut
V= Speed
f=Feed
MRR= Metal removal rate
By putting the values
MRR= 12 f V d
MRR = 12 x 0.015 x 105 x 0.105
MRR = 1.984
Therefore answer is -
1.944
