Answer:
3A
Explanation:
Using Ohms law U=I×R solve for I by I=U/R
For powder compaction using a single-action punch, derive an expression for the distribution of axial pressure within a die of r
ectangular cross-section. (b) Reduce the equation you derived in part (a) for the special case of a die with a square cross-section.
1 answer:
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Answer:
A) attractive force between ions = 5.09x10^-19 N of attractive force
B) there will be no repulsion since both ions have opposite charges.
Explanation:
Detailed explanation and calculation is shown in the image below.
Answer:
hello your question is incomplete attached below is the complete question
answer :
Slopes : B = 180 mm , C = 373 mm
Deflection: B = 0.0514 rad , C = 0.077 rad
Explanation:
Given data :
I = 500(10^6) mm^4
E = 70 GPa
The M / EI diagram is attached below
<u><em>Deflection angle at B</em></u>
∅B = ∅BA = [ 150 (6) + 1/2 (300)*6 ] / EI
= 1800 / ( 500 * 70 ) = 0.0514 rad
<u><em>slope at B </em></u>
ΔB = ΔBA = [ 150(6)*3 + 1/2 (300)*6*4 ] / EI
= 6300 / ( 500 * 70 ) = 0.18 m = 180 mm
<u><em>Deflection angle at C </em></u>
∅C = ∅CA = [ 1800 + 300*3 ] / EI
= 2700 / ( 500 * 70 )
= 2700 / 35000 = 0.077 rad
<u><em>Slope at C </em></u>
ΔC = [ 150 * 6 * 6 + 1/2 (800)*6*7 + 300(3) *1.5 ]
= 13050 / 35000 = 373 mm
