Isothermal Work = PVln(v₂/v₁)
PV = nRT = 2 mole * 8.314 J/ (k.mol) * 330 k = 5487.24 J
Isothermal Work = PVln(v₂/v₁) v₂ = ? v₁ = 19L,
1.7 kJ = (5487.24)In(v₂/19)
1700 = (5487.24)In(v₂/19)
In(v₂/19) = (1700/5487.24) = 0.3098
In(v₂/19) = 0.3098
(v₂/19) =

v₂ = 19*

v₂ = 25.8999
v₂ ≈ 26 L Option b.
Solution:
initial sphere mvr = final sphere mvr + Iω
where I = mL²/3 = 2.3g * (2m)² / 3 = 3.07 kg·m²
0.25kg * (12.5 + 9.5)m/s * (4/5)2m = 3.07 kg·m² * ω
where: ω = 2.87 rad/s
So for the rod, initial E = KE = ½Iω² = ½ * 3.07kg·m² * (2.87rad/s)²
E = 12.64 J becomes PE = mgh, so
12.64 J = 2.3 kg * 9.8m/s² * h
h = 0.29 m
h = L(1 - cosΘ) → where here L is the distance to the CM
0.03m = 1m(1 - cosΘ) = 1m - 1m*cosΘ
Θ = arccos((1-0.29)/1) = 44.77 º
Answer:
about: 110.14 million mi
Explanation:
the distance to Mars from Earth is 140 million miles (225 million kilometers).But, distance to Mars from Earth is constantly changing.
Hope that was helpful.Thank you!!!
Answer:
A downward sloped line means the object is returning to the starting point.