Answer:
it is not possible to place the wires in the condui
Explanation:
given data
total area = 2.04 square inches
wires total area = 0.93 square inches
maximum fill conduit = 40%
to find out
Can it is possible place wire in conduit conduit
solution
we know maximum fill is 40%
so here first we get total area of conduit that will be
total area of conduit = 40% × 2.04
total area of conduit = 0.816 square inches
but this area is less than required area of wire that is 0.93 square inches
so we can say it is not possible to place the wires in the conduit
Answer:
Following are the proving to this question:
Explanation:
using the energy equation for entry and exit value
:
![\to \frac{p_o}{y} +\frac{V^{2}_{o}}{2g}+Z_0 = \frac{p_1}{y} +\frac{V^{2}_{1}}{2g}+Z_1+ f \frac{l}{D}\frac{V^{2}}{2g}](https://tex.z-dn.net/?f=%5Cto%20%5Cfrac%7Bp_o%7D%7By%7D%20%2B%5Cfrac%7BV%5E%7B2%7D_%7Bo%7D%7D%7B2g%7D%2BZ_0%20%20%3D%20%5Cfrac%7Bp_1%7D%7By%7D%20%2B%5Cfrac%7BV%5E%7B2%7D_%7B1%7D%7D%7B2g%7D%2BZ_1%2B%20f%20%5Cfrac%7Bl%7D%7BD%7D%5Cfrac%7BV%5E%7B2%7D%7D%7B2g%7D)
where
![= (\frac{1}{(2f (\frac{l}{D} ))^{\frac{1}{4}}})^4\ V^{2}_{1}\\\\](https://tex.z-dn.net/?f=%3D%20%28%5Cfrac%7B1%7D%7B%282f%20%28%5Cfrac%7Bl%7D%7BD%7D%20%29%29%5E%7B%5Cfrac%7B1%7D%7B4%7D%7D%7D%29%5E4%5C%20%20V%5E%7B2%7D_%7B1%7D%5C%5C%5C%5C)
![= \frac{1}{(2f (\frac{l}{D}) )} \ V^{2}_{1}\\](https://tex.z-dn.net/?f=%3D%20%5Cfrac%7B1%7D%7B%282f%20%28%5Cfrac%7Bl%7D%7BD%7D%29%20%20%29%7D%20%5C%20%20V%5E%7B2%7D_%7B1%7D%5C%5C)
![\to \frac{p_o}{y} +\frac{V^{2}_{o}}{2g}+Z_0 =\frac{p_1}{y} +\frac{V^{2}_{1}}{2g}+Z_1+ f \frac{l}{D}\frac{V^{2}}{2g} \\\\](https://tex.z-dn.net/?f=%5Cto%20%5Cfrac%7Bp_o%7D%7By%7D%20%2B%5Cfrac%7BV%5E%7B2%7D_%7Bo%7D%7D%7B2g%7D%2BZ_0%20%20%3D%5Cfrac%7Bp_1%7D%7By%7D%20%2B%5Cfrac%7BV%5E%7B2%7D_%7B1%7D%7D%7B2g%7D%2BZ_1%2B%20f%20%5Cfrac%7Bl%7D%7BD%7D%5Cfrac%7BV%5E%7B2%7D%7D%7B2g%7D%20%5C%5C%5C%5C)
![\to 0+0+Z_0 = 0 +\frac{V^{2}_{1} }{2g} +Z_1+ f \frac{l}{D} \frac{\frac{1}{(2f(\frac{l}{D}))}\ V^{2}_{1}}{2g} \\\\\to Z_0 -Z_1 = +\frac{V^{2}_{1}}{2g} \ (1+f\frac{l}{D}\frac{1}{(2f(\frac{l}{D}) )} ) \\\\\to H= \frac{V^{2}_{1}}{2g} (\frac{3}{2}) \\\\\to \frac{V^{2}_{1}}{2g} = H(\frac{3}{2})](https://tex.z-dn.net/?f=%5Cto%200%2B0%2BZ_0%20%3D%200%20%20%2B%5Cfrac%7BV%5E%7B2%7D_%7B1%7D%20%7D%7B2g%7D%20%2BZ_1%2B%20f%20%5Cfrac%7Bl%7D%7BD%7D%20%5Cfrac%7B%5Cfrac%7B1%7D%7B%282f%28%5Cfrac%7Bl%7D%7BD%7D%29%29%7D%5C%20V%5E%7B2%7D_%7B1%7D%7D%7B2g%7D%20%20%20%5C%5C%5C%5C%5Cto%20Z_0%20-Z_1%20%3D%20%2B%5Cfrac%7BV%5E%7B2%7D_%7B1%7D%7D%7B2g%7D%20%5C%20%281%2Bf%5Cfrac%7Bl%7D%7BD%7D%5Cfrac%7B1%7D%7B%282f%28%5Cfrac%7Bl%7D%7BD%7D%29%20%29%7D%20%29%20%20%5C%5C%5C%5C%5Cto%20H%3D%20%5Cfrac%7BV%5E%7B2%7D_%7B1%7D%7D%7B2g%7D%20%28%5Cfrac%7B3%7D%7B2%7D%29%20%5C%5C%5C%5C%5Cto%20%20%5Cfrac%7BV%5E%7B2%7D_%7B1%7D%7D%7B2g%7D%20%3D%20H%28%5Cfrac%7B3%7D%7B2%7D%29)
L.H.S = R.H.S
Answer:
a) A suspended floor is a ground floor with a void underneath the structure. The floor can be formed in various ways, using timber joists, precast concrete panels, block and beam system or cast in-situ with reinforced concrete. However, the floor structure is supported by external and internal walls.
b) Soil exploration consists of determining the profile of the natural soil deposits at the site, taking the soil samples and determining the engineering properties of soils using laboratory tests as well as in-situ testing methods
c) Bulking in sand Occurs When dry sand interacts with the atmospheric moisture. Presence of moisture content forms a thin layer around sand particles. This layer generates the force which makes particles to move aside to each other. This results in the increase of the volume of sand.
d) In a nutshell, bearing capacity is the capacity of soil to support the loads that are applied to the ground above. It depends primarily on the type of soil, its shear strength and its density. It also depends on the depth of embedment of the load – the deeper it is founded, the greater the bearing capacity.
Explanation:
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The modern hydraulic lifts make use of biodegradable fluid to transmit hydraulic power
<em>Question: The options are left out in the question. The details and facts about the modern hydraulic lift are presented here</em>
<em />
Details about the modern hydraulic lifts include;
The development of the modern hydraulic occurred in the Industrial Revolution to perform task done previously by steam powered elevators
The power of the hydraulic lift come from the hydraulic cylinder known as the actuator, which in turn is powered by pressurized hydraulic fluid such as oil
The hydraulic fluid is pushed by a piston rod through which energy is capable of being transferred, such that the applied force is multiplied, to provide more power for lifting
<u>Facts about the modern hydraulic lifts include;</u>
- The dry motor in the modern hydraulic lift is more efficient and consumes 20% less energy
- It comprises of valves that are controlled electronically such that the response is much rapid and the energy consumption is reduced by a further 20%
- The cars used in the modern lift are lighter, as well as the slings, which reduces the power usage by 20%
- It makes use of chemicals which are environmentally friendly as hydraulic fluid
- The flash point of the fluid used is higher, as well as it posses 50% lower compressibility as well elasticity
Learn more here:
brainly.com/question/16942803
Answer:
Total time taken = 0.769 hour
Explanation:
using the velocity method
for sheet flow ;
Tt =
Tt = travel time
n = manning CaH
Pl = 25years
L = how length ( ft )
s = slope
For Location ( 1 )
s = 0.045
L = 1000 ft
n = 0.06 ( from manning's coefficient table )
Tt1 = 0.128 hour
For Location ( 2 )
s = 2.5 %
L= 750
n = 0.13
Tt2 = 0.239 hour
For Location ( 3 )
s = 1.5%
L = 500 ft
n = 0.15
Tt3 = 0.237 hour
For Location (4)
s = 0.5 %
L = 250 ft
n = 0.011
Tt4 = 0.165 hour
hence the Total time taken = Tt1 + Tt2 + Tt3 + Tt4
= 0.128 + 0.239 + 0.237 + 0.165 = 0.769 hour