Answer:
a) 254.6 GPa
b) 140.86 GPa
Explanation:
a) Considering the expression of rule of mixtures for upper-bound and calculating the modulus of elasticity for upper bound;
Ec(u) = EmVm + EpVp
To calculate the volume fraction of matrix, 0.63 is substituted for Vp in the equation below,
Vm + Vp = 1
Vm = 1 - 0.63
Vm = 0.37
In the first equation,
Where
Em = 68 GPa, Ep = 380 GPa, Vm = 0.37 and Vp = 0.63,
The modulus of elasticity upper-bound is,
Ec(u) = EmVm + EpVp
Ec(u) = (68 x 0.37) + (380 x 0.63)
Ec(u) = 254.6 GPa.
b) Considering the express of rule of mixtures for lower bound;
Ec(l) = (EmEp)/(VmEp + VpEm)
Substituting values into the equation,
Ec(l) = (68 x 380)/(0.37 x 380) + (0.63 x 68)
Ec(l) = 25840/183.44
Ec(l) = 140.86 GPa
Answer:
Cc= 12.7 lb.sec/ft
Explanation:
Given that
m = 22 lb
g= 32 ft/s²

x= 4.5 in
1 in = 0.083 ft
x= 0.375 ft
Spring constant ,K

K= 58.66 lb/ft
The damper coefficient for critically damped system


Cc= 12.7 lb.sec/ft
Umm the Water cycle sorry I’m trying
Answer:
The amount of cargo the plane can carry is 8707.89 N
Explanation:
The surface area of the wings facing the air = 30×6×2 × sin(2.5) = 15.7 m²
The speed of the plane 550 km/h = 152.78 m/s
The volume of air cut through per second = 15.7 × 152.78 = 2399.07 m³
The mass of air = Volume × Density = 2399.07 × 0.37 = 887.65 kg
Weight of air = Mass × Acceleration due to gravity = 887.65 × 9.81 = 8707.89 N
Given that the plane is already airborne, the additional cargo the plane can carry is given by the available lift force of the plane.
The amount of cargo the plane can carry = 8707.89 N