Answer:
stress = 16.9 MPa
Explanation:
The stress in the cable can be calculated as:
Where F is the force and A is the area. So, the area can be calculated as:
Where r is the radius. Since the radius is half the diameter, the radius is 4.0 mm and the area will be equal to:
Then, replacing the force F by 850 N, and A by 50.24 mm², we get that the stress is equal to:
Therefore, the answer is 16.9 MPa
Answer:
Plot ln K vs 1/T
(a) -0.5004; (b) 0.002 539 K⁻¹; (c) -197.1 K⁻¹; (d) 1.64 kJ/mol
Explanation:
This is an example of the Arrhenius equation:
Thus, if we plot ln k vs 1/T, we should get a straight line with slope = -Eₐ/R and a y-intercept = lnA
Data:
Calculations:
(a) Rise
Δy = y₂ - y₁ = -0.9545 - (-0.4541) = -0.9545 + 0.4541 = -0.5004
(b) Run
Δx = x₂ - x₁ = 0.004 444 - 0.001 905 = 0.002 539 K⁻¹
(c) Slope
Δy/Δx = -0.5004/0.002 539 K⁻¹ = -197.1 K⁻¹
(d) Activation energy
Slope = -Eₐ/R
Eₐ = -R × slope = -8.314 J·K⁻¹mol⁻¹ × (-197.1 K⁻¹) = 1638 J/mol = 1.64 kJ/mol
