Cyclic stresses a fatigue test was conducted in which the mean stress was 46.2 mpa

a) 265.2 MPa

b)-172.8 MPa

c) -0.6541

d) 438 MPa

Explanation:

we know that

..............(1)

.................(2)

solving equ (1) and equ (2) we get

= 265.2 MPa

= -172.8 MPa

stress ratio =

= = -0.6515

now stress range = -

= 265.2 -(-172.8) = 438 MPa

Explanation:

Here,we've been given that,

We've to find the acceleration of the object. By using the third equation of motion,

→ v² - u² = 2as

→ (420)² - (0)² = 2 × a × 145

→ 176400 - 0 = 290a

→ 176400 = 290a

→ 176400 ÷ 290 = a

→ 608.275862 m/s² = a