Answer:
Attached is the complete question but the weight of the mailbox and cross bar differs from the given values which are : weight of mail box = 3.2 Ib, weight of the uniform cross member = 10.3 Ib
Answer : moment of inertia = 186.7 Ib - in
Explanation:
Given data
weight of the mailbox = 3.2 Ib
weight of the uniform cross member = 10.3 Ib
The origin is of mailbox and cross member is 0
The perpendicular distance from Y axis of centroid of the mailbox
= 4 + (25/2) = 16.5"
The centroid of the bar =( ( 1 + 25 + 4 + 4 ) / 2 ) - 4 = 13"
therefore The moment of Inertia( Mo) = (3.2 * 16.5) + ( 10.3 * 13)
= 52.8 + 133.9 = 186.7 Ib-in
<h3><u>CSMA/CD Protocol:
</u></h3>
Carrier sensing can transmit the data at anytime only the condition is before sending the data sense carrier if the carrier is free then send the data.
But the problem is the standing at one end of channel, we can’t send the entire carrier. Because of this 2 stations can transmit the data (use the channel) at the same time resulting in collisions.
There are no acknowledgement to detect collisions, It's stations responsibility to detect whether its data is falling into collisions or not.
<u>Example:
</u>
, at time t = 10.00 AM, A starts, 10:59:59 AM B starts at time 11:00 AM collision starts.
12:00 AM A will see collisions
Pocket Size to detect the collision.
CSMA/CD is widely used in Ethernet.
<u>Efficiency of CSMA/CD:</u>
- In the previous example we have seen that in worst case
time require to detect a collision.
- There could be many collisions may happen before a successful completion of transmission of a packet.
We are given number of collisions (contentions slots)=4.
Distance = 1km = 1000m
Answer:
LAOD = 6669.86 N
Explanation:
Given data:
width
thickness 
crack length 2c = 0.5 mm at centre of specimen
stress intensity factor = k will be
we know that
[c =0.5/2 = 2.5*10^{-4}]
K = 0.1724 Mpa m^{1/2} for 1000 load
if
then load will be
LAOD = 6669.86 N
