For a bronze alloy, the stress at which plastic deformation begins is 267 MPa and the modulus of elasticity is 115 GPa. (a) What is the maximum load (in N) that may be applied to a specimen having a cross-sectional area of 206 mm2 without plastic deformation?

The maximum load that may be applied without plastic deformation should be less than 55,002 N.

Explanation:

The load that may be applied at which plastic deformation begins is obtained by multiplying the stress of the bronze alloy by its cross-sectional area.

Stress = 267 MPa = 267×10^6 = 2.67×10^8 Pa = 2.67×10^8 N/m^2

Cross-sectional area = 206 mm^2 = 206 mm^2 × (1 m/1000 mm)^2 = 2.06×10^-4 m^2

Load at which plastic deformation begins = 2.67×10^8 × 2.06×10^-4 = 55,002 N

Maximum load that may be applied without plastic deformation should be less than 55,002 N.

see earlier answer

the old price per ounce is c/4.

the new price per ounce is c/3.6.

the ratio of the old price to the new one is

(c/4)/(c/3.6) = (c/c)×(3.6/4) = 0.9 . . = 9/10 . . "nine tenths" in either form