Answer:
The minimum volume requirement for the granite stones is 1543.64 cm³
Explanation:
1 granite stone weighs 10 denarium
100 granted stones will weigh 1000 denarium
1 denarium = 3.396g
1000 denarium = 3396g.
But we're told that 20% of material is lost during the making of these stones.
This means the mass calculated represents 80% of the original mass requirement, m.
80% of m = 3396
m = 3396/0.8 = 4425 g
This mass represents the minimum mass requirement for making the stones.
To now obtain the corresponding minimum volume requirement
Density = mass/volume
Volume = mass/density = 4425/2.75 = 1543.64 cm³
Hope this helps!!!
Answer:
it is f all of the above
Explanation:
let me know if im right
im not positive if im right but i should be right
Answer:
a)W=12.62 kJ/mol
b)W=12.59 kJ/mol
Explanation:
At T = 100 °C the second and third virial coefficients are
B = -242.5 cm^3 mol^-1
C = 25200 cm^6 mo1^-2
Now according isothermal work of one mole methyl gas is
W=-
a=
b=
from virial equation

And

a=
b=
Now calculate V1 and V2 at given condition

Substitute given values
= 1 x 10^5 , T = 373.15 and given values of coefficients we get

Solve for V1 by iterative or alternative cubic equation solver we get

Similarly solve for state 2 at P2 = 50 bar we get

Now

a=241.33
b=30780
After performing integration we get work done on the system is
W=12.62 kJ/mol
(b) for Z = 1 + B' P +C' P^2 = PV/RT by performing differential we get
dV=RT(-1/p^2+0+C')dP
Hence work done on the system is

a=
b=
by substituting given limit and P = 1 bar , P2 = 50 bar and T = 373 K we get work
W=12.59 kJ/mol
The work by differ between a and b because the conversion of constant of virial coefficients are valid only for infinite series
Answer:
471 days
Explanation:
Capacity of Carvins Cove water reservoir = 3.2 billion gallons i.e. 3.2 x 10˄9 gallons
As,
1 gallon = 0.133 cubic feet (cf)
Therefore,
Capacity of Carvins Cove water reservoir in cf = 3.2 x 10˄9 x 0.133
= 4.28 x 10˄8
Applying Mass balance i.e
Accumulation = Mass In - Mass out (Eq. 01)
Here
Mass In = 0.5 cfs
Mass out = 11 cfs
Putting values in (Eq. 01)
Accumulation = 0.5 - 11
= - 10.5 cfs
Negative accumulation shows that reservoir is depleting i.e. at a rate of 10.5 cubic feet per second.
Converting depletion of reservoir in cubic feet per hour = 10.5 x 3600
= 37,800
Converting depletion of reservoir in cubic feet per day = 37, 800 x 24
= 907,200
i.e. 907,200 cubic feet volume is being depleted in days = 1 day
1 cubic feet volume is being depleted in days = 1/907,200 day
4.28 x 10˄8 cubic feet volume will deplete in days = (4.28 x 10˄8) x 1/907,200
= 471 Days.
Hence in case of continuous drought reservoir will last for 471 days before dry-up.
Answer:d
Explanation:
Given
Temperature
Also 
R=287 J/kg
Flow will be In-compressible when Mach no.<0.32
Mach no.
(a)
Mach no.
Mach no.=0.63
(b)
Mach no.
Mach no.=0.31
(c)
Mach no.
Mach no.=1.27
(d)
Mach no.
Mach no.=0.127
From above results it is clear that for Flow at velocity 200 km/h ,it will be incompressible.