Astronomers determine that a certain square region in interstellar space has an area of approximately 2.4 \times 10^72.4×10 7
1 answer:
You might be interested in
Answer:
T = 273 + (-50) = 273 – 50 = 223 K
R = 188.82 J / kg K for CO2
Density (Martian Atmosphere) = P / RT = 900 / 188.92 x 223 = 900 / 42129.16 = 0.0213 kg /
T = 273 +18 = 291 K, R = 287 J / kg k (for air) P = 101.6 k Pa = 101600 Pa
Density (Earth Atmosphere) = P / RT = 101600 / 287 x 291 = 1.216 kg /
Answer:
Given:
Radius of ball bearing (r) = 1.5 mm = 0.15 cm
Density of iron (ρ) = 7.85 g/cm³
Density of glycerine (σ) = 1.25 g/cm³
Terminal velocity (v) = 2.25 cm/s
Acceleration due to gravity (g) = 980.6 cm/s²
To Find:
Viscosity of glycerine ()
Explanation:
Substituting values of r, ρ, σ, v & g in the equation: