Five hundred joules of heat are added to a closed system. The initial internal energy of the system is 87 J, and the final inter
2 answers:
We can solve the problem by using the first law of thermodynamics:
where
is the variation of internal energy of the system
Q is the heat added to the system
W is the work done by the system
In this problem, the variation of internal energy of the system is
While the heat added to the system is
therefore, the work done by the system is
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Answer:
Explanation:
Component of force perpendicular to stick
= F Sin 60°
=√3 / 2 F.
Taking torque about the other end
= √3 / 2 F x 1 Nm
Weight of stick = 60 gm
= 60 x 10⁻³ kg
= 60 x 10⁻³ x 9.8 N
= .588 N
This weight will act from the middle point of stick so torque about the
other end
= .588 x 1 Nm
Balancing these two torques we have
.588 = √3 /2 F
F = 0.679 N