Answer:
total weight of aggregate = 5627528 lbs = 2814 tons
Explanation:
we get here volume of space to be filled with aggregate that is
volume = 2000 × 48 × 0.5
volume = 48000 ft³
now filling space with aggregate of the density that is
density = 0.95 × 119.7
density = 113.72 lb/ft³
and dry weight of this aggregate is
dry weight = 48000 × 113.72
dry weight = 5458320 lbs
we consider here percent moisture is by weigh
so weight of moisture in aggregate will be
weight of moisture = 0.031 × 5458320
weight of moisture = 169208 lbs
so here total weight of aggregate is
total weight of aggregate = 5458320 + 169208
total weight of aggregate = 5627528 lbs = 2814 tons
Answer:
A pet
Explanation:
Latin time I checked animals aren't made by people? I honestly don't know if this helps but I'm technically not wrong.
Answer:
the percent increase in the velocity of air is 25.65%
Explanation:
Hello!
The first thing we must consider to solve this problem is the continuity equation that states that the amount of mass flow that enters a system is the same as what should come out.
m1=m2
Now remember that mass flow is given by the product of density, cross-sectional area and velocity
(α1)(V1)(A1)=(α2)(V2)(A2)
where
α=density
V=velocity
A=area
Now we can assume that the input and output areas are equal
(α1)(V1)=(α2)(V2)
Now we can use the equation that defines the percentage of increase, in this case for speed
Now we use the equation obtained in the previous step, and replace values
the percent increase in the velocity of air is 25.65%
Answer:
0.71 lbf
Explanation:
Use ideal gas law:
PV = nRT
where P is absolute pressure,
V is volume,
n is number of moles,
R is universal gas constant,
and T is absolute temperature.
The absolute pressure is the sum of the atmospheric pressure and the gauge pressure.
P = 32 lbf/in² + 14.7 lbf/in²
P = 46.7 lbf/in²
Absolute temperature is in Kelvin or Rankine:
T = 75 + 459.67 R
T = 534.67 R
Given V = 3.0 ft³, and R = 10.731 ft³ psi / R / lb-mol:
PV = nRT
(46.7 lbf/in²) (3.0 ft³) = n (10.731 ft³ psi / R / lb-mol) (534.67 R)
n = 0.02442 lb-mol
The molar mass of air is 29 lbm/lb-mol, so the mass is:
m = (0.02442 lb-mol) (29 lbm/lb-mol)
m = 0.708 lbm
The weight of 1 lbm is lbf.
W = 0.708 lbf
Rounded to two significant figures, the weight of the air is 0.71 lbf.
