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Answer:
Explanation:
given
T = 3months = 7.9 × 10⁶s
orbital speed = 88 × 10³m/s
V= 2πr÷T
∴ r = (V×T) ÷ 2π
r = (88km × 7.9 × 10⁶s) ÷ 2π
r = 1.10 × 10⁸km
using kepler's 3rd law
mass of both stars = (seperation diatance)³/(orbital speed)²
M₁ + M₂ = (2r)³/(year)²
= (1.06 × 10²⁵)/(6.2×10¹³)
1.71×10¹²kg
since M₁ = M₂ =1.71×10¹²kg ÷ 2
M₁ = M₂ = 8.55×10¹¹kg
To solve this problem we must apply the concept related to the longitudinal effort and the effort of the hoop. The effort of the hoop is given as
Here,
P = Pressure
d = Diameter
t = Thickness
At the same time the longitudinal stress is given as,
The letters have the same meaning as before.
Then he hoop stress would be,
And the longitudinal stress would be
The Mohr's circle is attached in a image to find the maximum shear stress, which is given as
Therefore the maximum shear stress in the pressure vessel when it is subjected to this pressure is 600Psi
