Answer:
8 for dual-op-amp package, and 14 for quad-op-amp
Explanation;
This is because every op-amp has 2 input terminal 4 pns
So one output terminal that is 2 pins which are required for power
and the same for a minumum number of pins required by quad op amp which is 14
Answer:
theoretical fracture strength = 16919.98 MPa
Explanation:
given data
Length (L) = 0.28 mm = 0.28 × 10⁻³ m
radius of curvature (r) = 0.002 mm = 0.002 × 10⁻³ m
Stress (s₀) = 1430 MPa = 1430 × 10⁶ Pa
solution
we get here theoretical fracture strength s that is express as
theoretical fracture strength = 
.............................1
put here value and we get
theoretical fracture strength = 
theoretical fracture strength = 
theoretical fracture strength = 16919.98 MPa
Answer:
The compressive stress of aplying a force of 708 kN in a 81 mm diamter cylindrical component is 0.137 kN/mm^2 or 137465051 Pa (= 137.5 MPa)
Explanation:
The compressive stress in a cylindrical component can be calculated aby dividing the compressive force F to the cross sectional area A:
fc= F/A
If the stress is wanted in Pascals (Pa), F and A must be in Newtons and square meters respectively.
For acylindrical component the cross sectional area A is:
A=πR^
If the diameter of the component is 81 mm, the radius is the half:
R=81mm /2 = 40.5 mm
Then A result:
A= 3.14 * (40.5 mm)^2 = 5150.4 mm^2
In square meters:
A= 3.14 * (0.0405 m)^2 = 0.005150 m^2
Replacing 708 kN to the force:
fc= 708 kN / 5150.4 mm^2 = 0.137 kN/mm^2
Using the force in Newtons:
F= 70800 N
Finally the compressive stress in Pa is:
fc= 708000 / 0.005150 m^2 = 137465051 Pa = 137 MPa
