with a digital system, if you have measured incorrectly and use too low of a kvp for adequate penetration, what do you need to d
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Answer:
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Explanation:
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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:
The shear plane angle and shear strain are 28.21° and 2.155 respectively.
Explanation:
(a)
Orthogonal cutting is the cutting process in which cutting direction or cutting velocity is perpendicular to the cutting edge of the part surface.
Given:
Rake angle is 12°.
Chip thickness before cut is 0.32 mm.
Chip thickness is 0.65 mm.
Calculation:
Step1
Chip reduction ratio is calculated as follows:
r = 0.4923
Step2
Shear angle is calculated as follows:
Here, is shear plane angle, r is chip reduction ratio and
is rake angle.
Substitute all the values in the above equation as follows:
Thus, the shear plane angle is 28.21°.
(b)
Step3
Shears train is calculated as follows:
.
Thus, the shear strain rate is 2.155.