A thick steel sheet of area 150 in.2 is exposed to air near the ocean. After a one-year period it was found to experience a weig
1 answer:
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
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Answer:
a. which is constant therefore, n = constant
b. The temperature at the end of the process is 109.6°C
c. The work done by the balloon boundaries = 10.81 MJ
The work done on the surrounding atmospheric air = 10.6 MJ
Explanation:
p₁ = 100 kPa
T₁ = 27°C
D₁ = 10 m
v₂ = 1.2 × v₁
p ∝ α·D
α = Constant
v₂ = 1.2 × v₁ = 1.2 × 523.6 = 628.32 m³
Therefore, D₂ = 10.63 m
We check the following relation for a polytropic process;
We have;
n = -1/3
Therefore, the relation, pVⁿ = Constant
b. The temperature T₂ is found as follows;
T₂ = 300.15/0.784 = 382.75 K = 109.6°C
c.
p₂ = 100000/0.941 = 106.265 kPa
The work done by the balloon boundaries = 10.81 MJ
Work done against atmospheric pressure, Pₐ, is given by the relation;
Pₐ × (V₂ - V₁) = 1.01×10⁵×(628.32 - 523.6) = 10576695.3 J
The work done on the surrounding atmospheric air = 10.6 MJ
Answer:
save space = 1.9626 Gibibits
Explanation:
given data
recording song time = 3 minute = 180 seconds
song per sample = 24 bits per sample
frequency = 96 kHz
surround sound system = 5.1
solution
first we get here required space for 24 bits at 96 kHz that is
required space = 24 × 96 × 10³ × 6 × 180
required space = 2.48832 × bits
as 1 Gib bit = bits
so required space = 2.48832 × bits ÷
bits
required space = 2.3174 Gibibits ...............1
and
space required to save record 16 bit at 44.1 kHz
space required = 16 × 44.1 × 10³ × 3 × 180
space required = 0.381024 × bits
space required = 0.381024 × bits ÷
bits
space required = 0.3548 Gibibits ...........2
so
we get here that save space in 16 bit at 44.1 kHz
save space = 2.3174 Gibibit - 0.3548 Gibibits
save space = 1.9626 Gibibits