The largest tensile force that can be applied to the cables given a rod with diameter 1.5 is 2013.15lb
<h3>The static equilibrium is given as:</h3>
F = P (Normal force)
Formula for moment at section
M = P(4 + 1.5/2)
= 4.75p
Solve for the cross sectional area
Area = 
d = 1.5

= 1.767 inches²
<h3>Solve for inertia</h3>

= 0.2485inches⁴
Solve for the tensile force from here

30x10³ = 
30000 = 14.902 p
divide through by 14.902
2013.15 = P
The largest tensile force that can be applied to the cables given a rod with diameter 1.5 is 2013.15lb
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Organic material
I think it is the correct answer please forgive me if I am incorrect
Answer:
a). 139498.24 kg
b). 281.85 ohm
c). 10.2 ohm
Explanation:
Given :
Diameter, d = 22 m
Linear strain,
= 3%
= 0.03
Young's modulus, E = 30 GPa
Gauge factor, k = 6.9
Gauge resistance, R = 340 Ω
a). Maximum truck weight
σ = Eε
σ = 


= 342119.44 N
For the four sensors,
Maximum weight = 4 x P
= 4 x 342119.44
= 1368477.76 N
Therefore, weight in kg is 
m = 139498.24 kg
b). Change in resistance

, since 

Ω
For 4 resistance of the sensors,
Ω
c). 
If linear strain,
, where k = 1


Ω
The future perfect tense forms are made by putting ‘will / shall + have’ before the past participle from the verb. these sentences can be changed into the passive if the active verb has an object
i hope this helps :D thanks
Answer:
A) 3.868x10^26 W
B) 8176.678 K
C) 3.544x10^-7 m
D) 393.187 K
Explanation:
Detailed explanation and calculation is shown in the image below.