Avertical solid steel post of diameter d = 21 cm and length l = 2.70 m is required to support a load of mass m = 8300 kg . you can ignore the weight of the post. take free fall acceleration to be g=9.8m/s2. a. what is the stress in the post?
b. what is the strain in the post?
c. what is the change in the post's length when the load is applied?

Explanation:

Cross sectional area of post  = 3.14 x(10.5)² x 10⁻⁴ = 346.18 x 10⁻⁴ m²

Weight = mg = 8300 x 9.8 = 8.134 x 10⁴ N

Stress = weight / area

= 8.134 x 10⁴ / (346.18 x 10⁻⁴)

23.49 x 10⁵ N / m²

Strain = stress / modulus of elasticity of steel

modulus of elasticity = 190 x 10⁹ Pa

strain =  Stress \ young modulus

=

= 12.36 x 10⁻⁶

strain = change in length / original length

change in length = original length x strain

= 2.7 x 12.36 x 10⁻⁶

= 33.372 x 10⁻³ mm

.033 mm.

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

does anyone know or what?