A nylon thread is subjected to a 8.5-N tension force. Knowing that E 5 3.3 GPa and that the length of the thread increases by 1.1%, determine (a) the diameter of the thread, (b) the stress in the thread.

a)diameter of the thread = 0.546mm

b) stress in the thread = 36.3MPa

Explanation:

Hooke's law

strain = 1.1% = 0.011

Stress = strain* Young’s modulus

= 3.3 × 10⁹ × 0.011

= 36.3 × 10⁶Pa

stress = 36.3MPa

stress in the thread = 36.3MPa

b) Now,

stress = Force / area

Force = 8.5N

Stress = 36.3 × 10⁶

Area = Force / stress

=  8.5 /  36.3 × 10⁶

= 0.234 × 10⁻⁶m²

Area =

diameter of the thread = 0.546mm

because it says use -2,0,and 2 for x, we can plug those numbers into the equation. y/4=3/4(-2)+2   y/4= -1 1/2 +2 now to get x by itself, we can multiply each side by 4. y=-6+8 which equals y=2 when x=-2. now you can do the same thing with 0 and 2 as x. you get y=8 when x=0 and y=14 when x=2.

x   |   y

-2 |   2

0   |   8

2   |   14

b.

explanation: