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barxatty [35]
3 years ago
15

Who wanna rp?????????????????????????!

Engineering
1 answer:
Nata [24]3 years ago
7 0

:/??????????????????

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A structural element for a new bridge is designed for a constant load of 1000 psi. Its mean resistance is 1200 psi and the proba
andrew11 [14]

Answer:

i) SF = 0.83

ii) 10 psi

iii) 3.16 psi

iv) 0.0083

Explanation:

Constant load = 1000 Psi

mean resistance = 1200 psi

Probability of failure = 1.9 * 10^-3

<u>Determine </u>

<u>i) Safety factor </u>

S.F =  1 / (mean load / load design ) = 1 / ( 1200 / 1000 ) = 0.83

<u>ii) Standard deviation </u>

1.9 X 10^-3 = e  ( -1/2 ( μ / б) ^3  /  (2.5 * 6  )

0.004756 = e^- 20000 / б^3

hence std = 10 psi

<u>iii) Variance </u>

= \sqrt{10} = 3.16 psi

<u>iv) coefficient of variation </u>

Cv = std / mean resistance

    = 10 / 1200 = 0.0083

<u />

4 0
3 years ago
MITM can present in two forms: eavesdropping and manipulation. Discuss the process involved when an attacker is eavesdropping an
Nikitich [7]

Answer / Explanation:

Eavesdropping attack is also sometimes refereed to as sniffing attack. It is simply the process by which an attacker or hacker tries to penetrate very suddenly into an unaware individuals network or server with the intention to steal information transmitted over the network or server through that computer.  

To prevent such attack, there are several mean which include installing network monitoring software to check who else is connected to the network but the most common method of preventing such attack is to encrypt the Hypertext Transfer Protocol (http) and the way to do this is by securing it with a sort of security key.

On installing the security key, the network becomes encrypted and secured such that whatever network transmitted over the network becomes encrypted and unable to read. The protocol then converts to (https).

5 0
3 years ago
The closed tank of a fire engine is partly filled with water, the air space above being under pressure. A 6 cm bore connected to
skelet666 [1.2K]

Answer:

The air pressure in the tank is 53.9 kN/m^{2}

Solution:

As per the question:

Discharge rate, Q = 20 litres/ sec = 0.02\ m^{3}/s

(Since, 1 litre = 10^{-3} m^{3})

Diameter of the bore, d = 6 cm = 0.06 m

Head loss due to friction, H_{loss} = 45 cm = 0.45\ m

Height, h_{roof} = 2.5\ m

Now,

The velocity in the bore is given by:

v = \frac{Q}{\pi (\frac{d}{2})^{2}}

v = \frac{0.02}{\pi (\frac{0.06}{2})^{2}} = 7.07\ m/s

Now, using Bernoulli's eqn:

\frac{P}{\rho g} + \frac{v^{2}}{2g} + h = k                  (1)

The velocity head is given by:

\frac{v_{roof}^{2}}{2g} = \frac{7.07^{2}}{2\times 9.8} = 2.553

Now, by using energy conservation on the surface of water on the roof and that in the tank :

\frac{P_{tank}}{\rho g} + \frac{v_{tank}^{2}}{2g} + h_{tank} = \frac{P_{roof}}{\rho g} + \frac{v_{tank}^{2}}{2g} + h_{roof} + H_{loss}

\frac{P_{tank}}{\rho g} + 0 + 0 = \0 + 2.553 + 2.5 + 0.45

P_{tank} = 5.5\times \rho \times g

P_{tank} = 5.5\times 1\times 9.8 = 53.9\ kN/m^{2}

4 0
3 years ago
How much does the transportation sector contribute to current greenhouse gas emissions? Where do most of these emissions come fr
Mekhanik [1.2K]

Answer:

29 percent

Explanation:

The transportation sector is the highest contributing factor to greenhouse gas emissions in the world. Vehicles such as cars, motorcycles, airplanes, ships etc all emit these gases which causes global warming.

Electric cars is known not to little or zero direct emission and known to be safer. It however emits nitrogen oxide which is fog like in a very minute quantity.

6 0
3 years ago
2. What is the charge, expressed in micro coulombs on two equally and similarly charges spheres placed in air with their centres
9966 [12]

Answer:

q = 0.1086 micro Coulombs

Explanation:

By Coulombs law, we have;

F =k \times  \dfrac{ q_1 \times q_2}{r^2}

Where;

F = The electric force = 120 mgm

q₁ and q₂ = Charge

r = The separating distance = 30 cm = 0.3 m

k = 8.9876×10⁹ kg·m³/(s²·C²)

Where, q₁ and q₂, we have;

F =k \times  \dfrac{ q^2}{r^2}

Whereby the force is the force of 120 milligram mass, we have;

0.00012 × 9.81 = 000011772 N

q = \sqrt{ \dfrac{ F\times r^2}{k}}

Substituting the values, we have;

q = \sqrt{ \dfrac{ 000011772 \times (0.3)^2}{8.9876 \times 10^9}} = 1.086 \times 10 ^{-7} \ Coulombs

q = 0.1086 micro Coulombs.

5 0
3 years ago
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