If a steel containing 1.88 wt%C is cooled relatively slowly to room temperature, what is the expected weight fraction of pearlite in the as-cooled microstructure? Answer Format: X. XX Unit: unitless (weight fraction)

The answer is %pearlite = 0.06%

Explanation:

according to the exercise we have that the percentage is 1.88% C, therefore, the percentage of perlite is equal to:

%pearlite = (B*C)/(A*C) = (2-1.88)/(2-0) = 0.06%

The percentage of cementite is equal to:

%cementite = (1.88-0)/(2-0) = 0.94%

harry byrd declared organized resistance by southern states would prevent racial integration.

initial speed when it is dropped is 150 ft/s

now at the same moment the acceleration due to gravity is perpendicular to its velocity

so as we know that

now we have centripetal acceleration a = 32 ft/s^2 as its perpendicular to velocity

while tangential acceleration at the same moment will be zero

so

now radius of curvature is given as

so the radius of curvature just after it is dropped is 703.125 ft