A force of 6.0 N is applied horizontally to a 3.0 kg crate initially at rest on a horizontal frictionless surface. After the cra
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Answer: The magnitude of impulse imparted to the ball by the golf club is 2.2 N seconds
Explanation:
Force applied on the golf ball =
Mass of the ball = 0.05 kg
Velocity with which ball is accelerating = 44 m/s
Time period over which forece applied = t
Newton seconds
The magnitude of impulse imparted to the ball by the golf club is 2.2 N seconds
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.
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.