Answer:
L = 0.475 m = 475 mm = 18.7 inches
Explanation:
A cylindrical specimen of a nickel alloy having an elastic modulus of 207 GPa and an original diameter of 10.2 mm (0.40 in.) will experience only elastic deformation when a tensile load of 8900 N (2000 lb ) is applied. Compute the maximum length of the specimen before deformation if the maximum allowable elongation is 0.25 mm (0.010 in).
E = 207 GPa = 207*10⁹ Pa
D = 10.2 mm = 0.0102 m
P = 8900 N
ΔL = 0.25 mm = 2.5*10⁻⁴ m
L = ?
We can use the Equation of the Hooke's Law
ΔL = P*L / (A*E) ⇒ L = ΔL*A*E / P
⇒ L = (2.5*10⁻⁴ m)*(π*(0.0102 m)²*0.25)*(207*10⁹ Pa) / (8900 N)
⇒ L = 0.475 m = 475 mm = 18.7 inches
Answer:
137.2 in pounds and in Newton's it's 588.399
Answer:
Dilute
Explanation:
A dilute is a substance <u>mixed</u> with water to make it thinner or weaker.
Answer:
5 J
Explanation:
F = spring force required to stretch the spring = 10 N
x = stretch in the spring = 1 m
k = spring constant of the spring
Spring force is given as
F = k x
Inserting the values
10 = k (1)
k = 10 N/m
W = work done in stretching the spring
Work done is given as
W = (0.5)k x²
Inserting the values
W = (0.5) (10) (1)²
W = 5 J