Answer:
The duration of the consolidation process for the same clay is 32 min
Explanation:
for clay 1:
t1=0
H1=thickness=2 cm
for the clay 2:
t2=?
H2=2 cm
The time factor is equal to:
where Cv is the coefficient of consolidation
if Cv is constant, we have:
Clearing t2:
t2=32 min
Answer:The move from hubs (shared networks) to switched networks was a big improvement. Control over collisions, increased throughput, and the additional features offered by switches all provide ample incentive to upgrade infrastructure. But Layer 2 switched topologies are not without their difficulties. Extensive flat topologies can create congested broadcast domains and can involve compromises with security, redundancy, and load balancing. These issues can be mitigated through the use of virtual local area networks, or VLANs. This chapter provides the structure and operation of VLANs as standardized in IEEE 802.1Q. This discussion will include trunking methods used for interconnecting devices on VLANs.
Problem: Big Broadcast Domains
With any single shared media LAN segment, transmissions propagate through the entire segment. As traffic activity increases, more collisions occur and transmitting nodes must back off and wait before attempting the transmission again. While the collision is cleared, other nodes must also wait, further increasing congestion on the LAN segment.
The left side of Figure 4-1 depicts a small network in which PC 2 and PC 4 attempt transmissions at the same time. The frames propagate away from the computers, eventually colliding with each other somewhere in between the two nodes as shown on the right. The increased voltage and power then propagate away from the scene of the collision. Note that the collision does not continue past the switches on either end. These are the boundaries of the collision domain. This is one of the primary reasons for switches replacing hubs. Hubs (and access points) simply do not scale well as network traffic increases.
Answer:
6.65 kPa.
- 73.3 kPa.
Explanation:
Without much ado let's jump right into the solution to the problem given. It is given that the pressure = 50 mmHg and the solution to this question is to write out the value of the pressure in kpa and kPa gauge.
P(a) = 0.05m × 133 kN/m³ = 6.65 kPa.
The P(gauge) =( [ 0.05 × 13.6 × 9810] ÷ 1000 ) - 80 = - 73.3 kPa.
Answer:
a table
Explanation:
because you can saw the table
Answer:
no, it seems that everyone is having the same issue. If you use the app you can still find the answers and see them.
Explanation:
