Answer:
When the lift is greater than the weight, the aircraft gains altitude. ... Drag must be overcome for the aircraft to move, and movement is essential to obtain lift. To overcome drag and move the aircraft forward, another force is essential. This force is thrust.
Explanation:
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Answer:
Efficent at low speeds.
Explanation:
Since water is 1,000 times more dense than air, electricity can generated from tides much more efficiently at slower speeds than wind turbines can.
Answer:
a)σ₁ = 265.2 MPa
b)σ₂ = -172.8 MPa
c)
d)Range = 438 MPa
Explanation:
Given that
Mean stress ,σm= 46.2 MPa
Stress amplitude ,σa= 219 MPa
Lets take
Maximum stress level = σ₁
Minimum stress level =σ₂
The mean stress given as
2 x 46.2 = σ₁ + σ₂
σ₁ + σ₂ = 92.4 MPa --------1
The amplitude stress given as
2 x 219 = σ₁ - σ₂
σ₁ - σ₂ = 438 MPa --------2
By adding the above equation
2 σ₁ = 530.4
σ₁ = 265.2 MPa
-σ₂ = 438 -265.2 MPa
σ₂ = -172.8 MPa
Stress ratio
Range = 265.2 MPa - ( -172.8 MPa)
Range = 438 MPa
Answer:
minimum flow rate provided by pump is 0.02513 m^3/s
Explanation:
Given data:
Exit velocity of nozzle = 20m/s
Exit diameter = 40 mm
We know that flow rate Q is given as
where A is Area
minimum flow rate provided by pump is 0.02513 m^3/s
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.
