Refer to the diagram shown below.
Still-water speed = 9.5 m/s
River speed = 3.75 m/s down stream.
The velocity of the swimmer relative to the bank is the vector sum of his still-water speed and the speed of the river.
The velocity relative to the bank is
V = √(9.5² + 3.75²) = 10.21 m/s
The downstream angle is
θ = tan⁻¹ 3.75/9.5 = 21.5°
Answer: 10.2 m/s at 21.5° downstream.
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The periodic table is in increasing atomic mass. Hope this helped.
Explanation:
Given data:
d = 30 mm = 0.03 m
L = 1m
S
= 70 Mpa
Δd = -0.0001d
Axial force = ?
validity of elastic deformation assumption.
Solution:
O'₂ = Δd/d = (-0.0001d)/d = -0.0001
For copper,
v = 0.326 E = 119×10³ Mpa
O'₁ = O'₂/v = (-0.0001)/0.326 = 306×10⁶
∵δ = F.L/E.A and σ = F/A so,
σ = δ.E/L = O'₁ .E = (306×10⁻⁶).(119×10³) = 36.5 MPa
F = σ . A = (36.5 × 10⁻⁶) . (π/4 × (0.03)²) = 25800 KN
S
= 70 MPa > σ = 36.5 MPa
∵ elastic deformation assumption is valid.
so the answer is
F = 25800 K N and S
> σ