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2 years ago

7

Security __________ are the areas of trust within which users can freely communicate. a. rectangles b. domains c. perimeters d.

layers

1 answer:

babymother [125]2 years ago

8 0

B. Domains

Security domains are a place in which “behaviors, interactions, and outcomes occur and that is defined by a governing security policy” thus providing a safe place for users to freely communicate.

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The present worth of income from an investment that follows an arithmetic gradient is projected to be $475,000. The income in ye

Nikitich [7]

Answer:

G = $37,805.65

Explanation:

I found this on another site:

475,000 = 25,000(P/A,10%,6) + G(P/G,10%,6)

475,000 = 25,000(4.3553) + G(9.6842)

9.6842G = 366,117.50

G = $37,805.65

4 0

3 years ago

10. An engineer is designing a total hip implant. She intends to make the femoral stem out of titanium because it forms a good i

creativ13 [48]

Answer:

Yes. She should be worried about corrosion. The 18-8 stainless exhibits intergranular corrosion due to high (0.08%) carbon content and gross pitting due to low molybdenum content.

Explanation: lol

8 0

4 years ago

For some metal alloy, a true stress of 345 MPa (50040 psi) produces a plastic true strain of 0.02. How much will a specimen of t

saveliy_v [14]

Complete Question

For some metal alloy, a true stress of 345 MPa (50040 psi) produces a plastic true strain of 0.02. How much will a specimen of this material elongate when a true stress of 411 MPa (59610 psi) is applied if the original length is 470 mm (18.50 in.)?Assume a value of 0.22 for the strain-hardening exponent, n.

Answer:

The elongation is $=21.29mm$

Explanation:

In order to gain a good understanding of this solution let define some terms

True Stress

A true stress can be defined as the quotient obtained when instantaneous applied load is divided by instantaneous cross-sectional area of a material it can be denoted as $\sigma_T$ .

True Strain

A true strain can be defined as the value obtained when the natural logarithm quotient of instantaneous gauge length divided by original gauge length of a material is being bend out of shape by a uni-axial force. it can be denoted as $\epsilon_T$ .

The mathematical relation between stress to strain on the plastic region of deformation is

$\sigma _T =K\epsilon^n_T$

Where K is a constant

n is known as the strain hardening exponent

This constant K can be obtained as follows

$K = \frac{\sigma_T}{(\epsilon_T)^n}$

No substituting $345MPa \ for \ \sigma_T, \ 0.02 \ for \ \epsilon_T , \ and \ 0.22 \ for \ n$ from the question we have

$K = \frac{345}{(0.02)^{0.22}}$

$= 815.82MPa$

Making $\epsilon_T$ the subject from the equation above

$\epsilon_T = (\frac{\sigma_T}{K} )^{\frac{1}{n} }$

$Substituting \ 411MPa \ for \ \sigma_T \ 815.82MPa \ for \ K \ and \ 0.22 \ for \ n$

$\epsilon_T = (\frac{411MPa}{815.82MPa} )^{\frac{1}{0.22} }$

$=0.0443$

From the definition we mentioned instantaneous length and this can be obtained mathematically as follows

$l_i = l_o e^{\epsilon_T}$

Where

$l_i$ is the instantaneous length

$l_o$ is the original length

$Substituting \ 470mm \ for \ l_o \ and \ 0.0443 \ for \ \epsilon_T$

$l_i = 470 * e^{0.0443}$

$=491.28mm$

We can also obtain the elongated length mathematically as follows

$Elongated \ Length =l_i - l_o$

$Substituting \ 470mm \ for l_o and \ 491.28 \ for \ l_i$

$Elongated \ Length = 491.28 - 470$

$=21.29mm$

4 0

3 years ago

Determine how much concrete you will need for a slab which is 50 feet by 30 feet wide and 1 foot thick

Levart [38]

Answer:

Amount of concrete need to make slab = 1,500 feet³

Explanation:

Given:

Length of slab = 50 feet

Width of slab = 30 feet

Height of slab = 1 feet

Find:

Amount of concrete need to make slab

Computation;

Amount of concrete need to make slab = Volume of cuboid

Volume of cuboid = (l)(b)(h)

Amount of concrete need to make slab = (50)(30)(1)

Amount of concrete need to make slab = 1,500 feet³

6 0

3 years ago

A circular column is fixed at the base and not supported at the top. If the column needs to be 15ft and hold 10kips, what is the

muminat

Answer:

The required size of column is length = 15 ft and diameter = 4.04 inches

Explanation:

Given;

Length of the column, L = 15 ft

Applied load, P = 10 kips = 10 × 10³ Psi

End condition as fixed at the base and free at the top

thus,

Effective length of the column, $\L_e$ = 2L = 30 ft = 360 inches

now, for aluminium

Elastic modulus, E = 1.0 × 10⁷ Psi

Now, from the Euler's critical load, we have

$P =\frac{\pi^2EI}{L_e^2}$

where, I is the moment of inertia

on substituting the respective values, we get

$10\times10^3 =\frac{\pi^2\times1.0\times10^7\times I}{360^2}$

or

I = 13.13 in⁴

also for circular cross-section

I = $\frac{\pi}{64}\times d^4$

thus,

13.13 = $\frac{\pi}{64}\times d^4$

or

d = 4.04 inches

The required size of column is length = 15 ft and diameter = 4.04 inches

3 0

3 years ago