Answer:
± 0.003 ft
Explanation:
Since our distance is 10,000 ft and we need to use a full tape measure of 100 ft. We find that 10,000 = 100 × 100.
Let L' = our distance and L = our tape measure
So, L' = 100L
Now by error determination ΔL' = 100ΔL
Now ΔL' = ± 0.30 ft
ΔL = ΔL'/100
= ± 0.30 ft/100
= ± 0.003 ft
So, the maxim error per tape is ± 0.003 ft
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.
