Answer: 33 mm
Explanation:
Given
Diameter of the tank, d = 9 m, so that, radius = d/2 = 9/2 = 4.5 m
Internal pressure of gas, P(i) = 1.5 MPa
Yield strength of steel, P(y) = 340 MPa
Factor of safety = 0.3
Allowable stress = 340 * 0.3 = 102 MPa
σ = pr / 2t, where
σ = allowable stress
p = internal pressure
r = radius of the tank
t = minimum wall thickness
t = pr / 2σ
t = 1.5*10^6 * 4.5 / 2 * 102*10^6
t = 0.033 m
t = 33 mm
The minimum thickness of the wall required is therefore, 33 mm
340 ms
I got it right and I hope you do as well
I think d because they are both increasing
With the increase in the temperature of the star, the brightness of the stars will also increase.
<u>Explanation:</u>
The brightness and surface temperature of stars ordinarily increment with age. A star stays close to its underlying situation on the fundamental arrangement until a lot of hydrogen in the center has been devoured, at that point starts to advance into a progressively brilliant star.
The brightness of a star relies upon its structure and how far it is from the planet. Space experts characterize star brilliance as far as clear extent — how splendid the star shows up from Earth — and outright greatness — how brilliant the star shows up at a standard separation
Answer:
Electric field by charged disk is given as
E = (Charge Density/2u0)*[1 - (z/sqrt(z^2 - R^2))]
R = 9.54cm = 0.0954m, z = 1.01m, Charge density = 4.07 x 10^-6C/m2, e0 = 8.85 x 10^-12F/m.
Substituting all the values in to equation,
E = (2.299 x 10^5) x (8.931 x 10^-3)
E = 2.053 x 10^3N/C
Explanation:
