Math, asked by tahira9731, 13 hours ago

A mild steel rod of 12mm diameter was tested for tensile strength with the gauge of 60mm following were the observation. Final length=78mm, final diameter=7mm, yeild load=34KN, ultimate load= 61KN. Find yield stress, ultimate stress, percentage reduction, percentage elongation

Answers

Answered by hk4087935
0

Step-by-step explanation:

Question 3.

Give the steps you will use to separate the variables and then solve the equation:

(a) 3n – 2 = 46

(b) 5m + 7 = 17

(c) 20p3=40

(d) 3p10=6

Solution:

(a) 3n – 2 = 46

⇒ 3n- 2 + 2 = 46+ 2 (adding 2 to both sides)

⇒ 3n = 48 Question 3.

Give the steps you will use to separate the variables and then solve the equation:

(a) 3n – 2 = 46

(b) 5m + 7 = 17

(c) 20p3=40

(d) 3p10=6

Solution:

(a) 3n – 2 = 46

⇒ 3n- 2 + 2 = 46+ 2 (adding 2 to both sides)

⇒ 3n = 48

⇒ 3n + 3 = 48 ÷ 3

⇒ 3n + 3 = 48 ÷ 3

Answered by ramashishyadav054
0

Answer:

A mild steel rod whose diameter 12 mm and guage length are

60mm atensile test is done on the rod and following data are

recorded final length 78mm, yield load 3.40 ton . Ultimate load

6.10 ton calculate

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