(a) The stress in the post is 1,568,000 N/m²
(b) The strain in the post is 7.61 x 10⁻⁶
(c) The change in the post’s length when the load is applied is 1.9 x 10⁻⁵ m.
<h3>Area of the steel post</h3>
A = πd²/4
where;
d is the diameter
A = π(0.25²)/4 = 0.05 m²
<h3>Stress on the steel post</h3>
σ = F/A
σ = mg/A
where;
- m is mass supported by the steel
- g is acceleration due to gravity
- A is the area of the steel post
σ = (8000 x 9.8)/(0.05)
σ = 1,568,000 N/m²
<h3>Strain of the post</h3>
E = stress / strain
where;
- E is Young's modulus of steel = 206 Gpa
strain = stress/E
strain = (1,568,000) / (206 x 10⁹)
strain = 7.61 x 10⁻⁶
<h3>Change in length of the steel post</h3>
strain = ΔL/L
where;
- ΔL is change in length
- L is original length
ΔL = 7.61 x 10⁻⁶ x 2.5
ΔL = 1.9 x 10⁻⁵ m
Learn more about Young's modulus of steel here: brainly.com/question/14772333
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String the·o·ry
noun
a cosmological theory based on the existence of cosmic strings.
Answer:
We are given x= bt +ct²
So
A. bxt= m
Because m/s*s= m
So b= m/s and c= m/s²
B.
x= bt-ct²
So at x=0 t=0
x=0 t= 2
We have
bt = ct² so t = b/c at x= 0
So b-2ct= 0
B. To find velocity we use
dx / dt = b - 2 Ct
C. At rest wen V= 0
We have t= b/2c
D. To find acceleration we use
dv / dt = - 2C
Answer:
If the wishing well was in a vacuum, then s=ut + 0.5 a t^2 (s=distance, ... wishing well if you drop a coin into it and hear the splash 10 seconds
Explanation:
Answer:
0.6983 m/s
Explanation:
k = spring constant of the spring = 0.4 N/m
L₀ = Initial length = 11 cm = 0.11 m
L = Final length = 27 cm = 0.27 m
x = stretch in the spring = L - L₀ = 0.27 - 0.11 = 0.16 m
m = mass of the mass attached = 0.021 kg
v = speed of the mass
Using conservation of energy
Kinetic energy of mass = Spring potential energy
(0.5) m v² = (0.5) k x²
m v² = k x²
(0.021) v² = (0.4) (0.16)²
v = 0.6983 m/s
