Answer:
A charge q1=7.0mc is located at the origin and a second charge q2=-5.0mc is located on the x axis, 0.3m the origin find the electric field at the point p which he's coordinates (0,0.40)m
Wait what does Jessica want ?
Answer:
Because with this test you can determine complex material parameters like Young’s modulus, yield strength, ultimate strength and elongation at break. This is important because it provides us with the factor of safety that needs to be built-in the products using these materials.
Explanation:
This test consists in place the material between to tweezers to subdued the material into a stress-strain test. The figure shows the procedure.
In the x axe the strain and the y axe the stress.
Answer:
(a) 6.91 mm (b) 160 MPa
Explanation:
Solution
Given that:
E = 200 GPa
The rod length = 48 mm
P =P¹ = 6 kN
Recall that,
1 kN = 10^3 N
1 m =10^3 mm
I GPa = 10^9 N/m²
Thus
The rod deformation is stated as follows:
δ = PL/AE-------(1)
σ = P/A----------(2)
Now,
(a) We substitute the values in equation and obtain the following:
48 * 10 ^⁻3 m = (6 * 10³ N) (60 m)/A[ 200 * 10^9 N/m^2]
Thus, we simplify
A = (6 * 10³) (60)/ ( 200 * 10^9) (48 * 10 ^⁻3)m²
A =0.0375 * 10 ^⁻3 m²
A =37.5 mm²
A = π/4 d²
Thus,
d² = 4A /π
After inserting the values we have,
d = √37.5 * 4/3.14 mm
= 6.9116 mm
or d = 6.91 mm
Therefore, the smallest that should be used is 6.91 mm
(B) To determine the corresponding normal stress that is caused by the tensile force, we input the values in equation (2)
Thus,
σ = P/A
σ= 6 * 10 ^ 3 N/ 37. 5 * 10 ^⁻6 m²
σ= 160 MPa
Note: I MPa = 10^6 N/m²
Hence the the corresponding normal stress is σ= 160 MPa
