The densities of several materials are given in SI units. Convert these to densities in U.S. customary units (slug/ft3), and als
o compute the specific weights of these materials in U.S. customary units (lb/ft3).The density of ceramic (alumina Al2O3), rho = 3.9 Mg/cm3.
1 answer:
Answer:
a) Density of Lead = 22.0029 slug/ft3
b) Density of Ceramic Alumina = 7567163.99 slug/ft3
c) Density of polyethylene = 1.8626slug/ft3
d) Density of Balsa wood = 0.388slug/ft3
Explanation:
The question is incomplete as other information's are missing, here is the complete question ;
The densities of several materials are given in SI units . Convert these to densities in U.S . Customary units (slug/ft3), and also compute the specific weights of these materials in U.S . Customary units (lb/ft3). (a) Lead (pure), p = 11.34g/cm3. (b) Ceramic (alumina Al2O3), p = 3.90 Mg/cm3. (c) Polyethylene (high density), p = 960 kg/m3. (d) Balsa wood, p = 0.2 Mg/m3.
The specific weight is the weight per unit volume, i.e W = mg
but mass = density x volume
W = rho x v x g = (rho)vg
W/v = rho x g
where W/v = is the specific weight and g = is the acceleration due to gravity depending on the units, but in this case as told, g is in the US Customary units (lb/slug) which is 32.2 from conversion factor, and since we are told to calculate the specific weight of each in lb/ft3.
All you have to do is to multiply each density by the acceleration due to gravity and
repeat the same step for each of the materials.
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Answer:
Explanation:
Given that :
The initial volume of water = 1000.00 mL = 1000000 mm³
The initial temperature of the water = 10° C
The height of the water column h = 1000.00 mm
The final temperature of the water = 70° C
The coefficient of thermal expansion for the glass is ∝ =
The objective is to determine the the depth of the water column
In order to do that we will need to determine the volume of the water.
We obtain the data for physical properties of water at standard sea level atmospheric from pressure tables; So:
At temperature the density of the water is
At temperature the density of the water is
The mass of the water is
Thus; we can say ;
⇒
Thus, the volume of the water after heating to a required temperature of is 1022400 mm³
However; taking an integral look at this process; the volume of the water before heating can be deduced by the relation:
The area of the water before heating is:
The area of the heated water is :
Finally, the depth of the heated hot water is:
Hence the depth of the heated hot water is