The answer is: [A]: " 2 × 3 × 5 × 7 " .
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210 = 70 * 3 ;
70 * 3 = 35 * 2 * 3 ;
35* 2* 3 = 7 * 5 * 2 * 3 ; which corresponds to "Answer choice: [A]." .
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Answer:
P_max = 9.032 KN
Step-by-step explanation:
Given:
- Bar width and each side of bracket w = 70 mm
- Bar thickness and each side of bracket t = 20 mm
- Pin diameter d = 10 mm
- Average allowable bearing stress of (Bar and Bracket) T = 120 MPa
- Average allowable shear stress of pin S = 115 MPa
Find:
The maximum force P that the structure can support.
Solution:
- Bearing Stress in bar:
T = P / A
P = T*A
P = (120) * (0.07*0.02)
P = 168 KN
- Shear stress in pin:
S = P / A
P = S*A
P = (115)*pi*(0.01)^2 / 4
P = 9.032 KN
- Bearing Stress in each bracket:
T = P / 2*A
P = T*A*2
P = 2*(120) * (0.07*0.02)
P = 336 KN
- The maximum force P that this structure can support:
P_max = min (168 , 9.032 , 336)
P_max = 9.032 KN
The answer is A,(b + 3)(b - 2)
Im Not Sure But Im Thinking 7/10 Wrote As in A Fraction .
Factoring the expression would be 6a(2a^2+4a-1)
