Explanation:
Precision machining is a subtractive process used in cases where material needs to be removed from a raw product to create the finished product. Precision machining can be used to create a wide variety of products, items, and parts for any number of different objects and materials. These parts usually require tight tolerances variation from nominal dimensions and from part to part, which means that there is not much room for error in the production of the piece. Repeatability and well-controlled tolerances are hallmarks of precision machining. Components, parts and finished durable products that are designed to maintain extremely tight tolerance margins and a high degree of durability are essential and common drivers for utilization of precision machining. For example, parts that need to work together as part of a machine may need to always align within a certain margin of 0.01mm to 0.05mm. Precision engineering and machining help to ensure these parts can not only be made precisely but can be produced with this level of accuracy over and over again.
Answer:
The resistance is 24.9 Ω
Explanation:
The resistivity is equal to:
The area is:
A = 60 * 60 = 3600 um² = 0.36x10⁻⁴cm²
If NA is greater, then, the term 1/NA can be neglected, thus the equation:
Where
V = 0.44 V
E = 11.68*8.85x10¹⁴ f/cm
The length is:
L = 10 - 0.335 = 9.665 um
The resistance is:
Answer:
Option B
10,20,24,75,70,18,60,35
Explanation:
The first, second and third iteration of the loop will be as follows
insertion sort iteration 1: 20,24,10,75,70,18,60,35
insertion sort iteration 2:10,20,24,75,70,18,60,35
insertion sort iteration 3: 10,20,24,75,70,18,60,35
Answer:
The diameter is 50mm
Explanation:
The answer is in two stages. At first the torque (or twisting moment) acting on the shaft and needed to transmit the power needs to be calculated. Then the diameter of the shaft can be obtained using another equation that involves the torque obtained above.
T=(P×60)/(2×pi×N)
T is the Torque
P is the the power to be transmitted by the shaft; 40kW or 40×10³W
pi=3.142
N is the speed of the shaft; 250rpm
T=(40×10³×60)/(2×3.142×250)
T=1527.689Nm
Diameter of a shaft can be obtained from the formula
T=(pi × SS ×d³)/16
Where
SS is the allowable shear stress; 70MPa or 70×10⁶Pa
d is the diameter of the shaft
Making d the subject of the formula
d= cubroot[(T×16)/(pi×SS)]
d=cubroot[(1527.689×16)/(3.142×70×10⁶)]
d=0.04808m or 48.1mm approx 50mm
