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:
13.95
Explanation:
Given :
Vector A polar coordinates = ( 7, 70° )
Vector B polar coordinates = ( 4, 130° )
To find A . B we will
A ( r , ∅ ) = ( 7, 70 )
A = rcos∅ + rsin∅
therefore ; A = 2.394i + 6.57j
B ( r , ∅ ) = ( 4, 130° )
B = rcos∅ + rsin∅
therefore ; B = -2.57i + 3.06j
Hence ; A .B
( 2.394 i + 6.57j ) . ( -2.57 + 3.06j ) = 13.95
Answer:
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Answer: 5.36×10-3kg/h
Where 10-3 is 10 exponential 3 or 10 raised to the power of -3.
Explanation:using the formula
M =JAt = -DAt×Dc/Dx
Where D is change in the respective variables. Insulting the values we get,
=5.1 × 10-8 × 0.13 × 3600 × 2.9 × 0.31 / 4×10-3.
=5.36×10-3kg/h