The user needs to complete the entire disk surface first before starting another surface.
Explanation:
When you are using a multiple disk storage system to write the data the disk automatically writes the disk based on the algorithm for better efficiency and availability of the disk space.
hence when you are recording a data on a multiple disk storage system, it is recommended to fill the complete disk surface initially before you start the another surface to record the data.
Answer:
Check explanation
Explanation:
Two stacks can make use of one array by utilizing various stack pointers that begins from different ends of an array. Looking at the array A[1...
], the first stack will drive elements that starts from position 1 as well as to move its' pointer to
.
The Second stack will begin at the
position and motion its' pointer to 1. The best likely divide is to offer each stack a half of an array. whenever any of two stacks transverse the half-point, an overflow can happen but for that overall number of elements, it must be
MOHR-COULOMB FAILURE CRITERIA:
In 1900, MOHR-COULOMB states Theory of Rupture in Materials which defines as “A material fails due to because of a critical combination of normal and shear stress, not from maximum normal or shear stress”. Failure Envelope is approached by a linear relationship.
If you can not understand the below symbols see the attachment below
f f ()
Where: f = Shear Stress on Failure Plane
´= Normal Stress on Failure Plane
See the graph in the attachment
For calculating the shear stress, when Normal stress, cohesion and angle of internal friction are given. Use this formula: shear stress = f c tan
Where,
• f is Shear Stress on Failure Plane
• c is Cohesion
• is Normal Total Stress on Failure Plane
• is Friction Angle
Answer:
Selective Repeat protocols
Explanation:
It is better to make use of the selective repeat protocol here. From what we have here, there is a high error rate on this channel.
If we had implemented Go back N protocol, the whole N packets would be retransmitted. Much bandwidth would be needed here.
But we are told that bandwidth is limited. So if packet get lost when we implement selective protocol, we would only need less bandwidth since we would retransmit only this packet.