The load is 17156 N.
<u>Explanation:</u>
First compute the flexural strength from:
σ = FL / π
= 3000 
(40 10^-3) / π (5 10^-3)^3
σ = 305 10^6 N / m^2.
We can now determine the load using:
F = 2σd^3 / 3L
= 2(305 10^6) (15 10^-3)^3 / 3(40 10^-3)
F = 17156 N.
Answer:
8.85 Ω
Explanation:
Resistance of a wire is:
R = ρL/A
where ρ is resistivity of the material,
L is the length of the wire,
and A is the cross sectional area.
For a round wire, A = πr² = ¼πd².
For aluminum, ρ is 2.65×10⁻⁸ Ωm, or 8.69×10⁻⁸ Ωft.
Given L = 500 ft and d = 0.03 in = 0.0025 ft:
R = (8.69×10⁻⁸ Ωft) (500 ft) / (¼π (0.0025 ft)²)
R = 8.85 Ω
Answer:
C.) Weight and distance I believe
Explanation:
