A thin metal disk of mass m=2.00 x 10^-3 kg and radius R=2.20cm is attached at its center to a long fiber. When the disk is turn
ed from the relaxed state through a small angle theta, the torque exerted by the fiber on the disk is proportional to theta. Find an expression for the torsional constant k in terms of the moment of inertia I of the disk and the angular frequency w of small, free oscillations. The disk, when twisted and released, oscillates with a period T of 1.00 s. Find the torsional constant k of the fiber.
1 answer:
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Answer:
a) Q = 251.758 kJ/mol
b) creep rate is
Explanation:
we know Arrhenius expression is given as
where
Q is activation energy
C is pre- exponential constant
At 700 degree C creep rate is% per hr
At 800 degree C creep rate is% per hr
activation energy for creep is =
solving for Q we get
Q = 251.758 kJ/mol
b) creep rate at 500 degree C
we know