The cube measures 3.0-ft on all sides and has a density of 3.1 slugs/ft3. How much does it weigh?
1 answer:
Answer:
W = 2695.14 lb
Explanation:
given,
side of cube = 3 ft
density of the cube = 3.1 slugs/ft³
we know,
mass = density x volume
volume = 3³ = 27 ft³
mass = 3.1 x 27
m = 83.7 slugs.
weight calculation
converting mass from slug to pound
weight of 1 slug is equal to 32.2 lb
now,
weight of the cube is equal to
W = 83.7 slugs x 32.2 lb/slug
W = 2695.14 lb
hence, weight is equal to W = 2695.14 lb
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Answer:
a) The minimum acceptable value is 387.5 HV using Vickers hardness test.
b) The minimum acceptable value is 39.4 HRC using Rockwell C hardness test.
Explanation:
To get the tensile strength of a material from its hardness, we multiply it by an empirical constant that depends on things like yield strength, work-hardening, Poisson's ratio and geometrical factors. The incidence of cold-work varies this relationship.
According to DIN 50150 (a conversion table for hardness), the constant for Vickers hardness is ≈ 3.2 (an empirical approximate):
According to DIN 50150, the constant for Rockwell C hardness test is ≈31.5 around this values of tensile strength:
Answer:n=0.973
Explanation:
Given
When True strain
at
When True stress=346.2 MPa
true strain =0.226
We know
where =True stress
=true strain
n=strain hardening exponent
k=constant
Substituting value
Divide 1 & 2 to get
Taking Log both side
n=0.973
Answer:
6.007 x 10^(-14) kg/s
Explanation:
