Answer: 8.33333333 or 6.1989778
Explanation:
Ummm did you try to add or subtract and multiply or divide that can get your answer
Answer:
%Reduction in area = 73.41%
%Reduction in elongation = 42.20%
Explanation:
Given
Original diameter = 12.8 mm
Gauge length = 50.80mm
Diameter at the point of fracture = 6.60 mm (0.260 in.)
Fractured gauge length = 72.14 mm.
%Reduction in Area is given as:
((do/2)² - (d1/2)²)/(do/2)²
Calculating percent reduction in area
do = 12.8mm, d1 = 6.6mm
So,
%RA = ((12.8/2)² - 6.6/2)²)/(12.8/2)²
%RA = 0.734130859375
%RA = 73.41%
Calculating percent reduction in elongation
%Reduction in elongation is given as:
((do) - (d1))/(d1)
do = 72.14mm, d1 = 50.80mm
So,
%RA = ((72.24) - (50.80))/(50.80)
%RA = 0.422047244094488
%RA = 42.20%
Answer:
Expressions are made up of terms.
A term is a product of factors.
Coefficient is the numerical factor in the term
Before moving to terms like monomials, binomials, and polynomials, like and unlike terms are discussed.
When terms have the same algebraic factors, they are like terms.
When terms have different algebraic factors, they are unlike terms.
Explanation:
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Answer:
the heat transfer from the pipe will decrease when the insulation is taken off for r₂< 
where;
r₂ = outer radius
= critical radius
Explanation:
Note that the critical radius of insulation depends on the thermal conductivity of the insulation k and the external convection heat transfer coefficient h .
The rate of heat transfer from the cylinder increases with the addition of insulation for outer radius less than critical radius (r₂< ) 0, and reaches a maximum when r₂ = , and starts to decrease for r₂< . Thus, insulating the pipe may actually increase the rate of heat transfer from the pipe instead of decreasing it when r₂< .
