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
Answer:
endothermic
Explanation:
An endothermic is any process with an increase in the enthalpy H (or internal energy U) of the system. In such a process, a closed system usually absorbs thermal energy from its surroundings, which is heat transfer into the system.
Explanation:
<h3>The beam balance is a device used</h3><h3> for the determination of the mass of </h3><h3>a body under gravitation. It consists </h3><h3>of a beam supported at the centre </h3><h3>by an agate knife edge resting on a</h3><h3> support moving inside a vertical </h3><h3>pillar. The beam carries a light</h3><h3>pointer which moves over a scale.</h3>
Answer:
Explanation:
Electric field at the surface of the the lead 208 = KQ/ R²
where K = 8.99 × 10⁹ Nm² /C²
Q ( total charge inside the nucleus) and e is the charge of a proton = Ne = 82 × 1.6 × 10⁻¹⁹ C = 1.312 × 10⁻¹⁷ C
V of the lead = 208 v of a proton assuming they both are sphere
4/3 πR³ =208( 4/3 πr³) where R is the radius of the sphere and r is the radius of the proton
R³ = 208 r³
R = ∛( 208 r³) = 5.92r
replace r with 1.20 x 10-15 m
R = 5.92 ×1.20 x 10-15 m = 7.11 × 10⁻¹⁵ m
E = ( 8.99 × 10⁹ Nm² /C² × 1.312 × 10⁻¹⁷ C ) / (7.11 × 10⁻¹⁵ m)² = 0.233 × 10²² N/C = 2.33 × 10²¹ N/C
Answer:
0.002925 m
Explanation:
Lt = LO(1 +α Δt ) here Lt is total length Lo is original length α is coefficient of linear expansion and Δt is change in temperature
<h2>for aluminium</h2>
α=25×10^-6
Lt = 5(1+25×10^-6×(70-20))
Lt = 5 (1+25×10^-6×50)
Lt = 5 ( 1+0.00125)
Lt = 5×1.00125
Lt =5.00625 m
<h2>for nickel </h2>
α=13.3×10^-6
Lt =5(1+13.3×10^-6×50)
Lt = 5(1+0.000665)
Lt =5.003325 m
hence difference in length =5.00625-5.003325
= 0.002925 m
