Answer:
Industrial Engineers study various types of math including calculus, numerical analysis, statistics, linear algebra, numerical methods, operations research, etc. We do not necessarily use these in our day-to-day activities, but they help to build an analytical mindset that many employers value.
Answer:
105.70 mm
Explanation:
Poisson’s ratio, v is the ratio of lateral strain to axial strain.
E=2G(1+v) where E is Young’s modulus, v is poisson’s ratio and G is shear modulus
Since G is given as 25.4GPa, E is 65.5GPa, we substitute into our equation to obtain poisson’s ratio

Original length 

Where
is final diameter,
is original diameter,
is final length and
is original length.


Therefore, the original length is 105.70 mm
Answer:
i) 0.610 m or 610 mm
ii) 0.4 m or 400 mm
Explanation:
The pressure difference between the pipes is
a) Air
Pa + πha +Ha = Pb + πhb +Hb
Pa - Pb = π(hb-ha) + Hb-Ha
Relative density of air = 1.2754 kg /m3
Pa - Pb = 1.2754 * 0.4 + (0.3-0.2) = 0.610 m or 610 mm
b) paraffin of relative density of 0.75
Pa - Pb = π(hb-ha) + Hb-Ha
Pa - Pb = 0.75 * 0.4 + (0.3-0.2) = 0.4 m or 400 mm
Answer:

Explanation:
given data
Load P = 35 kN
Width of bar W = 50.8 mm
Breadth of bar B = 25 mm
Ratio of crack length to width α = a/W = 0.2
solution
we get here KI for a rectangular bar that is express as
................................1
here Y is the geometrical function
so
Y =
Y =
Y =
Y = 0.9878
so put here value in equation 1

= 5210.45 × 10³
= 5.21 MPa 
Answer:
Heat losses by convection, Qconv = 90W
Heat losses by radiation, Qrad = 5.814W
Explanation:
Heat transfer is defined as the transfer of heat from the heat surface to the object that needs to be heated. There are three types which are:
1. Radiation
2. Conduction
3. Convection
Convection is defined as the transfer of heat through the actual movement of the molecules.
Qconv = hA(Temp.final - Temp.surr)
Where h = 6.4KW/m2K
A, area of a square = L2
= (0.25)2
= 0.0625m2
Temp.final = 250°C
Temp.surr = 25°C
Q = 64 * 0.0625 * (250 - 25)
= 90W
Radiation is a heat transfer method that does not rely upon the contact between the initial heat source and the object to be heated, it can be called thermal radiation.
Qrad = E*S*(Temp.final4 - Temp.surr4)
Where E = emissivity of the surface
S = boltzmann constant
= 5.6703 x 10-8 W/m2K4
Qrad = 5.6703 x 10-8 * 0.42 * 0.0625 * ((250)4 - (25)4)
= 5.814 W