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Julli [10]
3 years ago
5

The engineer must place a catch basin at the north west corner of the yard of a subdivision lot. The catchment area will be calc

ulated with a portion of the pervious yard area and the area of the whole impervious roof area.
A. True
B. False
Engineering
1 answer:
castortr0y [4]3 years ago
5 0

Answer:

A = true.

Explanation:

The statement made in the Question or problem above is TRUE/correct. The term known as " catch basin" is an important and essential part of the drainage system that is should be built in a community(whether housing community, industrial community or commercial community).

The catch basin is very important because it helps in making sure that they act as sieve to get all the dirt such as leaves from water running on the surface(runoffs).

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A Gaussian random voltage X volts is input to a half-wave rectifier and the output voltage is Y = Xu (X) Volts were u (x) is the
adelina 88 [10]

Answer:

Please look at attachment carefully.

7 0
4 years ago
If the bolt head and the supporting bracket are made of the same material having a failure shear stress of 'Tra;i = 120 MPa, det
Nina [5.8K]

Answer:

P=361.91 KN

Explanation:

given data:

brackets and head of the screw are made of material with T_fail=120 Mpa

safety factor is F.S=2.5

maximum value of force P=??

<em>solution:</em>

to find the shear stress

                            T_allow=T_fail/F.S

                                         =120 Mpa/2.5

                                         =48 Mpa

we know that,

                               V=P

<u>Area for shear head:</u>

                              A(head)=π×d×t

                                           =π×0.04×0.075

                                           =0.003×πm^2

<u>Area for plate:</u>

                               A(plate)=π×d×t  

                                            =π×0.08×0.03

                                            =0.0024×πm^2

now we have to find shear stress for both head and plate

<u>For head:</u>

                                   T_allow=V/A(head)

                                    48 Mpa=P/0.003×π                 ..(V=P)

                                             P =48 Mpa×0.003×π

                                                =452.16 KN

<u>For plate:</u>

                                   T_allow=V/A(plate)

                                    48 Mpa=P/0.0024×π                 ..(V=P)

                                             P =48 Mpa×0.0024×π

                                                =361.91 KN

the boundary load is obtained as the minimum value of force P for all three cases. so the solution is

                                                P=361.91 KN

note:

find the attached pic

7 0
4 years ago
Students are expected to respond to one of the two questions described below. Students should provide examples to clarify their
Flura [38]

Answer:

can u explain what we are answering?

Explanation:

7 0
4 years ago
0-0 what is Boyle’s Law
Dima020 [189]
A law stating that the pressure of a given mass of an ideal gas is inversely proportional to its volume at a constant temperature.

Hope it helps
8 0
3 years ago
A sedimentation basin in a water treatment plant has a length = 48 m, width = 12 m, and depth = 3 m. The flow rate = 4 m 3 /s; p
Slav-nsk [51]

Answer:

The minimum particle diameter that is removed at 85% is 1.474 * 10 ^⁻4 meters.

Solution

Given:

Length = 48 m

Width = 12 m

Depth = 3m

Flow rate = 4 m 3 /s

Water density = 10 3 kg/m 3

Dynamic viscosity = 1.30710 -3 N.sec/m

Now,

At the minimum particular diameter it is stated as follows:

The Reynolds number= 0.1

Thus,

0.1 =ρVTD/μ

VT = Dp² ( ρp- ρ) g/ 10μ²

Where

gn = The case/issue of sedimentation

VT = Terminal velocity

So,

0.1 = Dp³ ( ρp- ρ) g/ 10μ²

This becomes,

0.1 = 1000 * dp³ (1100-1000) g 0.1/ 10 *(1.307 * 10 ^⁻3)²

= 3.074 * 10 ^⁻6 = dp³ (.g01 * 10^6)

dp³=3.1343 * 10 ^⁻12

Dp minimum= 1.474 * 10 ^⁻4 meters.

8 0
3 years ago
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