Answer:
The Euler buckling load of a 160-cm-long column will be 1.33 times the Euler buckling load of an equivalent 120-cm-long column.
Explanation:
160 - 120 = 40
120 = 100
40 = X
40 x 100 / 120 = X
4000 / 120 = X
33.333 = X
120 = 100
160 = X
160 x 100 /120 = X
16000 / 120 = X
133.333 = X
Answer:
s= 20.4 m
Explanation:
First lets write down equations for each ball:
s=so+vo*t+1/2a_c*t^2
for ball A:
s_a=30+5*t+1/2*9.81*t^2
for ball B:
s_b=20*t-1/2*9.81*t^2
to find time deeded to pass we just put that
s_a = s_b
30+5*t-4.91*t^2=20*t-4.9*t^2
t=2 s
now we just have to put that time in any of those equations an get distance from the ground:
s = 30 + 5*2 -1/2*9.81 *2^2
s= 20.4 m
Answer:
y ≈ 2.5
Explanation:
Given data:
bottom width is 3 m
side slope is 1:2
discharge is 10 m^3/s
slope is 0.004
manning roughness coefficient is 0.015
manning equation is written as
where R is hydraulic radius
S = bed slope
P is perimeter 
![Q = (2+2y) y) \times 1/0.015 [\frac{(3+2y) y}{(3+2\sqrt{5} y)}]^{2/3} 0.004^{1/2}](https://tex.z-dn.net/?f=Q%20%3D%20%282%2B2y%29%20y%29%20%5Ctimes%201%2F0.015%20%5B%5Cfrac%7B%283%2B2y%29%20y%7D%7B%283%2B2%5Csqrt%7B5%7D%20y%29%7D%5D%5E%7B2%2F3%7D%200.004%5E%7B1%2F2%7D)
solving for y![100 =(2+2y) y) \times (1/0.015) [\frac{(3+2y) y}{(3+2\sqrt{5} y)}]^{2/3} \times 0.004^{1/2}](https://tex.z-dn.net/?f=100%20%3D%282%2B2y%29%20y%29%20%5Ctimes%20%281%2F0.015%29%20%5B%5Cfrac%7B%283%2B2y%29%20y%7D%7B%283%2B2%5Csqrt%7B5%7D%20y%29%7D%5D%5E%7B2%2F3%7D%20%5Ctimes%200.004%5E%7B1%2F2%7D)
solving for y value by using iteration method ,we get
y ≈ 2.5
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:
18 kJ
Explanation:
Given:
Initial volume of air = 0.05 m³
Initial pressure = 60 kPa
Final volume = 0.2 m³
Final pressure = 180 kPa
Now,
the Work done by air will be calculated as:
Work Done = Average pressure × Change in volume
thus,
Average pressure = 
= 120 kPa
and,
Change in volume = Final volume - Initial Volume = 0.2 - 0.05 = 0.15 m³
Therefore,
the work done = 120 × 0.15 = 18 kJ
