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
<span>Cobalt-60 is undergoing a radioactivity decay.
The formula of the decay is n=N(1/2)</span>∧(T/t).
<span>Where N </span>⇒ original mass of cobalt
<span> n </span>⇒ remaining mass of cobalt after 3 years
T ⇒ decaying period
t ⇒ half-life of cobalt.
So,
0.675 = 1 × 0.5∧(3/t)
log 0.675 = log 0.5∧(3/t)
3/t = log 0.675 ÷log 0.5
3/t= 0.567
t = 3÷0.567
= 5.290626524
the half-life of Cobalt-60 is 5.29 years.
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the radiogenic heat produced by the radioactive decay of isotopes in the mantle and crust, and the primordial heat left over from the formation of the Earth.
If the cross-section of a wire of fixed length is doubled, the resistance of that wire change into doubled.We know that <span>the total </span>length<span> of the wires will </span>affect<span> the amount of </span>resistance. <span> The longer the wire, the more </span>resistance<span> that there will be so the answer is doubled.</span>
