Answer:
Rate of corrosion = 24.95 mpy
Rate of corrosion = 0.63 mm/yr
Explanation:
given data
steel sheet area = 150 in²
weight loss = 485 g
density of steel = 7.9 g/cm³
time taken = 1 year
to find out
rate of corrosion in (a) mpy and (b) mm/yr
solution
we get here the rate of corrosion that is express as
rate of corrosion = (k × W) ÷ (D × A × T) ..................1
here k is constant and w is total weight lost and t is time taken for loss and A is surface area and D is density of steel
so put her value in equation 1 we get
Rate of corrosion = 
Rate of corrosion = 24.95 mpy
and
Rate of corrosion = 
Rate of corrosion = 0.63 mm/yr
Answer:
The required mechanical work is required to reduce each day by 1.05×10^8 Joules.
Explanation:
Coefficient of Performance (COP) = Q/W
Q is thermal energy absorbed by the air conditioner
W is mechanical work done
Q = 3.9×10^8 J
COP of old air conditioner = 2.3
W = Q/COP = 3.9×10^8/2.3 = 1.70×10^8 J
COP of new air conditioner = 6
W = Q/COP = 3.9×10^8/6 = 6.5×10^7 J
Reduction in mechanical work = (1.7×10^8) - (6.5×10^7) = 1.05×10^8 J
Answer
Assuming
At 10000 m height temperature T = -55 C = 218 K
At 1000 m height temperature T = 0 C = 273 K

R = 287 J/kg K



V₂ = V₁ ×1.1222
V₁ = 0.5 × C₁ = 0.5 × 295 = 147.5 m/s
V₂ = 1.1222 × 147.5 = 165.49 m/s
so, the jetliner need to increase speed by ( V₂ -V₁ )
= 165.49 - 147.5
= 17.5 m/s
Answer:
Overall project duration
Explanation:
Scheduling can best be defined as the process used to determine a overall project duration.
Answer:
a₁= 1.98 m/s² : magnitud of the normal acceleration
a₂=0.75 m/s² : magnitud of the tangential acceleration
Explanation:
Formulas for uniformly accelerated circular motion
a₁=ω²*r : normal acceleration Formula (1)
a₂=α*r: normal acceleration Formula (2)
ωf²=ω₀²+2*α*θ Formula (3)
ω : angular velocity
α : angular acceleration
r : radius
ωf= final angular velocity
ω₀ : initial angular velocity
θ : angular position theta
r : radius
Data
r =0.4 m
ω₀= 1 rad/s
α=0.3 *θ , θ= 2π
α=0.3 *2π= 0,6π rad/s²
Magnitudes of the normal and tangential components of acceleration of a point P on the rim of the disk when theta has completed one revolution.
We calculate ωf with formula 3:
ωf²= 1² + 2*0.6π*2π =1+2.4π ²= 24.687
ωf=
=4.97 rad/s
a₁=ω²*r = 4.97²*0.4 = 1.98 m/s²
a₂=α*r = 0,6π * 0.4 = 0.75 m/s²