Answer:
(A) Because the angle of twist of a material is often used to predict its shear toughness
Explanation:
In engineering, torsion is the solicitation that occurs when a moment is applied on the longitudinal axis of a construction element or mechanical prism, such as axes or, in general, elements where one dimension predominates over the other two, although it is possible to find it in diverse situations.
The torsion is characterized geometrically because any curve parallel to the axis of the piece is no longer contained in the plane initially formed by the two curves. Instead, a curve parallel to the axis is twisted around it.
The general study of torsion is complicated because under that type of solicitation the cross section of a piece in general is characterized by two phenomena:
1- Tangential tensions appear parallel to the cross section.
2- When the previous tensions are not properly distributed, which always happens unless the section has circular symmetry, sectional warps appear that make the deformed cross sections not flat.
Answer:
D=41.48 ft
Explanation:
Given that
y=0.5 x²
Vx= 2 t
We know that
At t= 0 ,x=0
At t= 3 s
![x=[t^2\left\right ]_0^3](https://tex.z-dn.net/?f=x%3D%5Bt%5E2%5Cleft%5Cright%20%5D_0%5E3)
x= 9 ft
When x= 9 ft then
y= 0.5 x 9² ft
y= 40.5 ft
So distance from origin is
x= 9 ft ,y= 40.5 ft
D=41.48 ft
Vx= 2 t
At t= 3 s , x= 9 ft
y=0.5 x²
y=0.5 x²
Given that
Answer:
2.77mpa
Explanation:
compressive strength = 20 MPa. We are to find the estimated flexure strength
We calculate the estimated flexural strength R as
R = 0.62√fc
Where fc is the compressive strength and it is in Mpa
When we substitute 20 for gc
Flexure strength is
0.62x√20
= 0.62x4.472
= 2.77Mpa
The estimated flexure strength is therefore 2.77Mpa
It costs 300
Perimeter = 2(L+B)
2(10+5)
2(15) = 30
10 — 1metre
X — 30metres
30metres = 300
Hope that helps :)
