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3 years ago

If i build thing a and thing a builds thing b did i build thing b

Engineering

2 answers:

lianna [129]3 years ago

8 0

Answer:

No, I don't think so.

Explanation:

It depends on if you programmed thing a to build thing b.

postnew [5]3 years ago

3 0

Answer:

Explanation:

If you build thing "a" and thing "a" builds thing "b" you <u>indirectly</u> build thing "b".

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Estimate the theoretical fracture strength of a brittle material if it is known that fracture occurs by the propagation of an el

baherus [9]

Answer:

The theoretical fracture strength of the brittle material is 11864.5 MPa

Explanation:

Fracture strength is the ability of a material to withstand fracture. It is also known as the breaking stress, it is the stress at which the material fails as a result of fracture. It usually determined from the stress-strain curve after performing a tensile test.

Given that:

Length (L) = 0.15 mm = 0.15 × 10⁻³ m

radius of curvature (r) = 0.002 mm = 0.002 × 10⁻³ m

Stress (s₀) = 1370 MPa = 1370 × 10⁶ Pa

theoretical fracture strength (s) = ?

The theoretical fracture strength is given as:

$s=s_{0} .\sqrt{\frac{L}{r} }$

Substituting values:

$s=1370*10^6 .\sqrt{\frac{0.15*10^{-3}}{0.002*10^{-3}} }\\s=1370*10^6 *8.66=11864.5*10^6\\s=11864.5*10^6$

s = 11864.5 MPa

The theoretical fracture strength of the brittle material is 11864.5 MPa

3 0

3 years ago

Read 2 more answers

Name the point of intersection, where the axis meet.

kakasveta [241]

Answer:

origin

Explanation:

The point of intersection of axis is called origin.

In 2D origin is the intersection point of x-axis and y-axis if we go right to the origin then it is positive x axis, if we go left side of origin then it is negative x- axis

Similarly when we go above the origin then it positive y axis , and if we go bellow the origin then it is negative x axis

In 3D origin is the intersection of x-axis, y-axis and z-axis

NOTE- For defining i take here x axis as horizontal axis and y-axis as vertical axis

4 0

3 years ago

Calculate the thermal efficiency (ηth) for the actual cycle using pump efficiency (ηpump) = 0.85. You’ll need to find the high-p

dangina [55]

Answer:

Thermal efficiency(ηth) = 1 - (TH/Tc), where TH is temperature of hot reservoir and Tc is temperature of cold reservoir.

ηth = 1 - (300/Tc)

We can assume a value for TH to be 400°C

Then,

ηth = 1 - (300/400) = 1 - 0.75 = 0.25

Explanation:

Thermal efficiency, ηth, of an heat engine is define as the ratio of the work it does, W, to the heat input at the high temperature, QH.

The thermal efficiency is expressed mathematically as:

ηth= QH/W

Thermal efficiency(ηth) = 1 - (TH/Tc).

In other words, the fraction of heat that is converted to work is the thermal efficiency. That is, the measure of performance of engines that uses heat energy in their operation. It has no dimension.

Examples of such engines are:

Steam turbine;

Any internal combustion engine;

A refrigerator.

In operation of a refrigeration or heat pumps, thermal efficiency indicates the extent to which the energy added by work is converted to net heat output. Since it is dimensionless number, we must always express W, QH, and QC in the same units

8 0

3 years ago

Read 2 more answers

A HSS152.4 × 101.6 × 6.4 structural steel [E = 200 GPa] section (see Appendix B for crosssectional properties) is used as a colu

Temka [501]

Answer:

(a) 126.66 kN (b) 31.665 kN (c) 258.49 kN (d) 506.64 kN

Explanation:

Solution

Given

A HSS152.4 × 101.6 × 6.4 structural steel is used as a column

Actual length of the column , L= 6 m

The elasticity modules, E = 200 GPa

The factor of safety with respect to failing buckling . F.S =2

Geometric properties of structural steel shapes for, A HSS152.4 × 101.6 × 6.4

the moment of inertial about y axis Iy =4 .62 * 10^ 6 mm ^4

For

(a) If the end condition is pinned - pinned

The effective length factor, K =1

The critical buckling load , Pcr = π²EI/(KL)²

Pcr = π² * 200 * 10 ^3 * 4.62 * 10 ^6/( 1* 6* 10 ^3)

= 253319.85N

= 253.32N

The maximum safe load , Pallow = 253.32 /2 = 126.66kN

hence, the maximum safe for the column is 126.66kN

(b)If the end condition is fixed free-free

the effective length factor, K= 2

The critical buckling load , Pcr = π²EI/(KL)²

Pcr = π² * 200 * 10 ^3 * 4.62 * 10 ^6/( 2 * 6 * 10 ^3)²

= 63329.96N

=63.33kN

The maximum safe load, P allow = Pcr/F.S

P allow = 63.33/2

31.665 kN

Therefore the maximum safe for the column is 31.665 kN

The effective length factor K =0.7

The critical buckling load , Pcr = π²EI/(KL)²

Pcr = π² * 200 * 10 ^3 * 4.62 * 10 ^6/( 0.7 * 6 * 10 ^3)²

=516979.2 8N

=516.98 kN

The maximum safe load, P allow = Pcr/F.S

P allow = 516.98 kN/2

=258.49 kN

Therefore the maximum safe for the column is 258.49 kN

(d) If the end condition is fixed -fixed

The effective factor, K =0.5

The critical buckling load , Pcr = π²EI/(KL)²

Pcr = π² * 200 * 10 ^3 * 4.62 * 10 ^6/( 0.5 * 6 * 10 ^3)²

=1013279.4 N

=1013.28 kN

The maximum safe load, P allow = Pcr/F.S

P allow = 1013.28 / 2

= 506.64 kN

P allow = 506.64 kN

Therefore the maximum safe for the column is 506.64 kN

8 0

4 years ago

Since the application can be the prospective employers first impression, it is important to a. always answer questions truthfull

loris [4]

Answer:

Definitely D

Explanation:

You should do all of those things in an application.

7 0

4 years ago

Read 2 more answers