I think because if you’ve already turned it in they might as well grade asap instead of waiting
Answer:
1.887 m
Explanation:
(15 *pi)/180
= 0.2618 rad
Polar moment
= Pi*d⁴/32
= (22/7*20⁴)/32
= 15707.96
Torque on shaft
= ((22/7)*20³*110)/16
= 172857.14
= 172.8nm
Shear modulus
G = 79.3
L = Gjθ/T
= 79.3x10⁹x(1.571*10^-8)x0.2618/172.8
= 1.887 m
The length of the bar is therefore 1.887 meters
Answer:
Gravitational force (pulled downward by the Earth)
Normal force (pushed upward by the ground)
Applied force (pushed by the person)
Friction force (pulled opposite the direction of motion by the roughness of the ground)
Answer:
B. operator error or misuse
Explanation:
A product is a failure if it is not able to achieve the anticipated life cycle as expected by the organization.
In such a case, there is a withdrawal of the product from the market as a result of its ultimate failure of a product to achieve profitability.
Four common causes of product failure are poor design, poor construction, poorly communicated operating instructions, and <u>operator error or misuse</u>
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