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iogann1982 [59]

2 years ago

8

Which of these credit building options do you personally think is the easiest method that you can see yourself doing? Explain yo

ur reasoning.

1 answer:

enot [183]2 years ago

7 0

Answer:

Explanation:

Your 3-B Annual Credit Reports & Scores with Enhanced Credit Report Monitoring.

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AveGali [126]

Answer:

b

Explanation:

4 0

3 years ago

Tech A says that a cylinder leakage test is performed on a cylinder with low compression to determine the severity of the leak a

Karolina [17]

Tech A djjdjdndnndndbdbx

4 0

3 years ago

. The flexure strength test was performed on a concrete beam having a cross section of 0.15m by 0.15m and a span of 0.45m. If th

ivann1987 [24]

Answer:

σ =5.39Mpa

Explanation:

step one:

The flexure strength is defined as the tendency with which unreinforced concrete yield to bending forces

Flexural strength test Flexural strength is calculated using the equation:

σ = FL/ (bd^2 )----------1

Where

σ = Flexural strength of concrete in Mpa

F= Failure load (in N).

L= Effective span of the beam

b= Breadth of the beam

step two:

Given data

F=40.45 kN= 40450N

b=0.15m

d=0.15m

L=0.45m

step three:

substituting into the expression we have

σ = 40450*0.45/ (0.15*0.15^2 )

σ =18202.5/ (0.15*0.15^2 )

σ =18202.5/ (0.15*0.0225 )

σ =18202.5/0.003375

σ =5393333.3

σ =5393333.3/1000000

σ =5.39Mpa

Therefore the flexure strength of the concrete is 5.39Mpa

5 0

3 years ago

1. The equilibrium number of vacancies in Ni at 1123 K is 4.7x1022m-3. The atomic weight and density of Ni at 1123 K are 58.69 g

Ahat [919]

Answer:

vacancy formation energy of Ni is 1.400 eV

Explanation:

given data

number of vacancies in Ni = 4.7 x $10^{22}$ $m^{-3}$

atomic weight = 58.69 g/mol

density = 8.8 g/cm³

solution

we get here N that is

N = $\frac{N_A \times \rho}{A}$ ...........1

N = $\frac{6.023\times 10^{23} \times 8.8 \times 10^6}{58.69}$

N = $9.030 \times 10^{28}$

and here no of vacancy will be

Nv = $N \times e^{\frac{-Qv}{kT}}$ .................2

put here value

$4.7 \times 10^{22} = 9.030 \times 10^28 \times e^{\frac{-Qv}{8.62\times 10^{-5}\times 1123}}$

$10^{-7} \times 5.20487 = e^{\frac{-Qv}{0.0968}}$

take ln both side

$ln (10^{-7} \times 5.20487 ) = ln (e^{\frac{-Qv}{0.0968}})$

$-14.468 = \frac{-Qv}{0.0968}$

Qv = 1.400 eV

so vacancy formation energy of Ni is 1.400 eV

3 0

3 years ago

The bar DA is rigid and is originally held in the horizontal position when the weight W is supported from C. The wires are made

Rom4ik [11]

Answer:

W = 112 lb

Explanation:

Given:

- δb = 0.025 in

- E = 29000 ksi (A-36)

- Area A_de = 0.002 in^2

Find:

Compute Weight W attached at C

Solution:

- Use proportion to determine δd:

δd/5 = δb/3

δd = (5/3) * 0.025

δd = 0.0417 in

- Compute εde i.e strain in DE:

εde = δd / Lde

εde = 0.0417 / 3*12

εde = 0.00116

- Compute stress in DE, σde:

σde = E*εde

σde = 29000*0.00116

σde = 33.56 ksi

- Compute the Force F_de:

F_de = σde *A_de

F_de = 33.56*0.002

F_de = 0.0672 kips

- Equilibrium conditions apply:

(M)_a = 0

3*W - 5*F_de = 0

W = (5/3)*F_de

W = (5/3)* 0.0672 = 112 lb

4 0

2 years ago