Answer:
percentage change in volume is 2.60%
water level rise is 4.138 mm
Explanation:
given data
volume of water V = 500 L
temperature T1 = 20°C
temperature T2 = 80°C
vat diameter = 2 m
to find out
percentage change in volume and how much water level rise
solution
we will apply here bulk modulus equation that is ratio of change in pressure to rate of change of volume to change of pressure
and we know that is also in term of change in density also
so
E = 
................1
And 
............2
here ρ is density
and we know ρ for 20°C = 998 kg/m³
and ρ for 80°C = 972 kg/m³
so from equation 2 put all value
dV = 0.0130 m³
so now % change in volume will be
dV % = 
× 100
dV % = 
× 100
dV % = 2.60 %
so percentage change in volume is 2.60%
and
initial volume v1 = 
................3
final volume v2 = 
................4
now from equation 3 and 4 , subtract v1 by v2
v2 - v1 = 
dV = 
put here all value
0.0130 = 
dl = 0.004138 m
so water level rise is 4.138 mm
Answer:
Explanation:
The distance of the chain would be the product of the dislocation density and the volume of the metal.
Dislocation density = 
Volume of the metal = 
The chain would extend
Dislocation density = 
Volume of the metal =
The chain would extend
Answer:
See attached image for diagrams and solution
Answer:
a) P ≥ 22.164 Kips
b) Q = 5.4 Kips
Explanation:
GIven
W = 18 Kips
μ₁ = 0.30
μ₂ = 0.60
a) P = ?
We get F₁ and F₂ as follows:
F₁ = μ₁*W = 0.30*18 Kips = 5.4 Kips
F₂ = μ₂*Nef = 0.6*Nef
Then, we apply
∑Fy = 0 (+↑)
Nef*Cos 12º - F₂*Sin 12º = W
⇒ Nef*Cos 12º - (0.6*Nef)*Sin 12º = 18
⇒ Nef = 21.09 Kips
Wedge moves if
P ≥ F₁ + F₂*Cos 12º + Nef*Sin 12º
⇒ P ≥ 5.4 Kips + 0.6*21.09 Kips*Cos 12º + 21.09 Kips*Sin 12º
⇒ P ≥ 22.164 Kips
b) For the static equilibrium of base plate
Q = F₁ = 5.4 Kips
We can see the pic shown in order to understand the question.
