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ozzi
3 years ago
8

How many squares titles (20cm x 20cm) are needed to coat the sides and base of a pool which is 10m long, 6 meter wide and 3m dee

p?
Engineering
1 answer:
Zanzabum3 years ago
7 0

Answer:

3900 tiles

Explanation:

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An organization is struggling to differentiate threats from normal traffic and access to systems. A security engineer has been a
AnnZ [28]

Answer:

Web application Firewall (WAF)

Explanation:

The Web application Firewall (WAF) will be recommended. This will enormously help any organisation that is having a struggle of differentiating threats from normal traffic and access to systems.

WAF is also going to It aggregate data and provide metrics that will assist in identifying malicious actors. Another vital function of WAF will be to block unwanted web traffic from accessing your site. It will protect against hacks, brute force attacks, DDoS attacks, cross-site scripting, SQL injection, and zero-day exploits.

8 0
4 years ago
Footing: This is a 3 m wide strip footing, located on the ground surface of an isotropic, homogeneous
Rufina [12.5K]

Answer:

3

Explanation:

5 0
3 years ago
Detail a high-end design project and what paper and finishing techniques you would use include your choices for matte or gloss v
DochEvi [55]

Answer   what do you want i know a project but what is it on?

Explanation:

5 0
3 years ago
The natural material in a borrow pit has a mass unit weight of 110.0 pcf and a water content of 6%, and the specific gravity of
enot [183]

Answer:

A. 288,030.91 cy

B. The amount of water that must be removed from the natural material is 483541.04254 gallons of water

Explanation:

The natural material in the barrow properties are;

The mass unit weight, γ = 110.0 pcf

The water content, w = 6%

The specific gravity of the soil solids, G_s = 2.63

The desired dry unit weight, \gamma _d = 122 pcf

The water content, w₁ = 5.5 %

The net section volume, V_T = 245,000 cy = 6,615,000 ft³

A.  \gamma _d = W_s/V_T

∴ W_s = V_T × \gamma _d = 6,615,000 ft³ × 122 lb/ft³ = 807030000 lbs

w = (W_w/W_s) ×  100

∴ W_w = (w/100) × W_s = (6/100) × 807030000 lbs = 48421800 lbs

The weight of solids

W = W_s + W_w = 807030000 lbs + 48421800 lbs = 855451800 lbs

V = W/γ = 855451800 lbs/(110.0 lb/ft.³) = 7776834.54545 ft.³ = 288,030.91 cy

V = 288,030.91 cy

The amount of cubic yards of borrow required = 288,030.91 cy

B. The volume of water in the required soil is found as follows;

W_{w1} = (w₁/100) × W_s = (5.5/100) × 807030000 lbs = 44386650 lbs

The amount of water that must be added =  W_{w1} - W_w = 44386650 lbs - 48421800 lbs = -4,035,150 lbs

Therefore, 4,035,150 lbs of water must be removed

The density of water, ρ = 8.345 lbs/gal

Therefore, V = 4,035,150 lbs/(8.345 lbs/gal) = 483541.04254 gal  of water must be removed from the natural material

7 0
3 years ago
The 240-ft structure is used to provide various support services to launch vehicles prior to liftoff. In a test, a 12-ton weight
alina1380 [7]

Answer:

hello your question lacks the required question attached below is the missing diagram

Forces in GJ = -4.4444 i.e. 4.4444 tons

Forces in IG = 15.382 tons ( T )

Explanation:

Forces in GJ = -4.4444 i.e. 4.4444 tons

Forces in IG = 15.382 tons ( T )

attached below is the detailed solution

3 0
3 years ago
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