Answer:
0.304 L of Freon is needed
Explanation:
Q = mCT
Q is quantity of energy that must be removed = 47 BTU = 47×1055.06 = 49587.82 J
C is specific heat of Freon = 74 J/mol.K = 74 J/mol.K × 1 mol/120 g = 0.617 J/g.K
T is temperature in the area of Mars = 189 K
m = Q/CT = 49587.82/(0.617×189) = 452.23 g = 452.24/1000 = 0.45223 kg
Density of Freon = specific gravity of Freon × density of water = 1.49 × 1000 kg/m^3 = 1490 kg/m^3
Volume of Freon = mass/density = 0.45223/1490 = 0.000304 m^3 = 0.000304×1000 = 0.304 L
It would have environmental and societal impacts
Answer:
Explanation:
cross sectional area A = 1.9 x 2.6 x 10⁻⁶ m²
= 4.94 x 10⁻⁶ m²
stress = 42 x 9.8 / 4.94 x 10⁻⁶
= 83.32 x 10⁶ N/m²
strain = .002902 / 2.7
= 1.075 x 10⁻³
Young's modulus = stress / strain
= 83.32 x 10⁶ / 1.075 x 10⁻³
= 77.5 x 10⁹ N/m²
Well you could always do what Hispanics do and repeat the question 2-3 times while yelling till they finally answer
