Answer:
radius = 9.1 × 
m
Explanation:
given data
applied load = 5560 N
flexural strength = 105 MPa
separation between the support = 45 mm
solution
we apply here minimum radius formula that is
radius = ![\sqrt[3]{\frac{FL}{\sigma \pi}}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B%5Cfrac%7BFL%7D%7B%5Csigma%20%5Cpi%7D%7D)
.................1
here F is applied load and is length
put here value and we get
radius = ![\sqrt[3]{\frac{5560\times 45\times 10^{-3}}{105 \times 10^6 \pi}}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B%5Cfrac%7B5560%5Ctimes%2045%5Ctimes%2010%5E%7B-3%7D%7D%7B105%20%5Ctimes%2010%5E6%20%5Cpi%7D%7D)
solve it we get
radius = 9.1 × m
analyzing building materials???????????? but i can try i think it is analyzing the materials of the building
Answer:
The distance measure from the wall = 36ft
Explanation:
Given Data:
w = 10
g =32.2ft/s²
x = 2
Using the principle of work and energy,
T₁ +∑U₁-₂ = T₂
0 + 1/2kx² -wh = 1/2 w/g V²
Substituting, we have
0 + 1/2 * 100 * 2² - (10 * 3) = 1/2 * (10/32.2)V²
170 = 0.15528V²
V² = 170/0.15528
V² = 1094.796
V = √1094.796
V = 33.09 ft/s
But tan ∅ = 3/4
∅ = tan⁻¹3/4
= 36.87°
From uniform acceleration,
S = S₀ + ut + 1/2gt²
It can be written as
S = S₀ + Vsin∅*t + 1/2gt²
Substituting, we have
0 = 3 + 33.09 * sin 36.87 * t -(1/2 * 32.2 *t²)
19.85t - 16.1t² + 3 = 0
16.1t² - 19.85t - 3 = 0
Solving it quadratically, we obtain t = 1.36s
The distance measure from the wall is given by the formula
d = VCos∅*t
Substituting, we have
d = 33.09 * cos 36. 87 * 1.36
d = 36ft
