Question 3 (5 points)
1 answer:
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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
Answer:
For - 5.556 lb/s
For - 7.4047 lb/s
Solution:
As per the question:
System Load = 96000 Btuh
Temperature, T =
Temperature rise, T' =
Now,
The system load is taken to be at constant pressure, then:
Specific heat of air,
Now, for a rise of in temeprature:
Now, for :