Answer:
The shortest distance d to the edge of the plate is 66.67 mm
Concepts and reason
Moment of a force:
Moment of a force refers to the propensity of the force to cause rotation on the body it acts upon. The magnitude of the moment can be determined from the product of force’s magnitude and the perpendicular distance to the force.
Moment(M) = Force(F)×distance(d)
Moment of inertia ( I )
It is the product of area and the square of the moment arm for a section about a reference. It is also called as second moment of inertia.
First prepare the free body diagram of sectioned plate and apply moment equilibrium condition and also obtain area and moment of inertia of rectangular cross section. Finally, calculate the shortest distance using the formula of compressive stress (σ) in combination of axial and bending stress
Solution and Explanation:
[Find the given attachments]
Answer:
The external pressure is p = -21.9 psf or p = -8.85 psf
Explanation:
Given :
Velocity of wind, v = 120 mi / hr
(wind direction factor)
= topographical factor (for flat terrain)
= velocity pressure at height h
But for height h = 30 ft, 
= 0.98 (from table)
= 36.16
Now, 
, so 
(from table)
where, p = external pressure
G = 0.85 = gust factor (for typical rigid building)
(internal pressure co efficient)
Therefore putting the values,
p = -21.9 psf or p = -8.85 psf
The sample of 81 students was selected with a mean score of 90, this illustrates an example of a right tailed one sample z test.
<h3>How to illustrate the sample?</h3>
From the information given, the teacher claims that the mean score of students in his class is greater than 82 with a standard deviation of 20. In this case, the right tailed one sample z test is used.
The z value will be:
= (90 - 82)/20/✓81
= 3.6
Since 3.6 > 1.645, the null hypothesis will be rejected as there's enough evidence to support the teacher's claim.
When an online medicine shop claims that the mean delivery time for medicines is less than 120 minutes with a standard deviation of 30 minutes, the left tailed one sample test is used.
The z value will be:
= (100 - 120)/(30/✓49)
= -4.66
The null hypothesis is rejected as there is enough evidence to support the claim of the medicine shop.
Learn more about sampling on:
brainly.com/question/17831271
#SPJ1
Answer:
Zero 1 = -1
Zero 2 = -3
Pole 1 = 0
Pole 2 = -2
Pole 3 = -4
Pole 4 = -6
Gain = 4
Explanation:
For any given transfer function, the general form is given as
T.F = k [N(s)] ÷ [D(s)]
where k = gain of the transfer function
N(s) is the numerator polynomial of the transfer function whose roots are the zeros of the transfer function.
D(s) is the denominator polynomial of the transfer function whose roots are the poles of the transfer function.
k [N(s)] = 4s² + 16s + 12 = 4[s² + 4s + 3]
it is evident that
Gain = k = 4
N(s) = (s² + 4s + 3) = (s² + s + 3s + 3)
= s(s + 1) + 3 (s + 1) = (s + 1)(s + 3)
The zeros are -1 and -3
D(s) = s⁴ + 12s³ + 44s² + 48s
= s(s³ + 12s² + 44s + 48)
= s(s + 2)(s + 4)(s + 6)
The roots are then, 0, -2, -4 and -6.
Hope this Helps!!!
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.
