Answer:
The module is why it’s goin to work
Explanation:
Answer:
(a) 53.94%
(b) 26.61%
Explanation:
Change in area will be given by
where
represent change in area R is radius and subscripts O and n represent original and new respectively.
Substituting 10.55/2 for original radius and 7.16/2 for new radius then

(b)
Similarly, percentage elongation will be found by dividing the change in length by the the original length. In this case, rhe original length was 54.5mm and goes to 69 mm so the change in length is given by subtracting the final length from the original length
Percentage elongation is 
Answer:
2.455 W
Explanation:
The power dissipated in each branch is ...
P = V^2/R
So, the branch powers are ...
branch 1: 18^2/220 ≈ 1.473 W
branch 2: 18^2/330 ≈ 0.982 W
Total power is ...
1.473 W + 0.982 W = 2.455 W
Answer:
- 3/5" = 12'
- 1 3/4" = 35'
- 1 1/4" = 25'
- 9/10" = 18'
- 2 13/20" = 53'
Explanation:
One number is wrong; they all lack units.
The basic ratio is 1" = 20', so you can divide feet by 20 to find inches.
- 3/5" = 12'
- 1 3/4" = 35'
- 1 1/4" = 25'
- 9/10" = 18'
- 2 13/20" = 53'
Perhaps you want decimal inches:
- 0.60" = 12'
- 1.75" = 35'
- 1.25" = 25'
- 0.90" = 18'
- 2.65" = 53'
Answer:
It will not experience fracture when it is exposed to a stress of 1030 MPa.
Explanation:
Given
Klc = 54.8 MPa √m
a = 0.5 mm = 0.5*10⁻³m
Y = 1.0
This problem asks us to determine whether or not the 4340 steel alloy specimen will fracture when exposed to a stress of 1030 MPa, given the values of <em>KIc</em>, <em>Y</em>, and the largest value of <em>a</em> in the material. This requires that we solve for <em>σc</em> from the following equation:
<em>σc = KIc / (Y*√(π*a))</em>
Thus
σc = 54.8 MPa √m / (1.0*√(π*0.5*10⁻³m))
⇒ σc = 1382.67 MPa > 1030 MPa
Therefore, the fracture will not occur because this specimen can handle a stress of 1382.67 MPa before experience fracture.