A shaft is made from a tube, the ratio of the inside diameter to the outside diameter is 0.6. The material must not experience a shear stress greater than 500KPa. The shaft must transmit 1.5MW of mechanical power at 1500 revolution per minute. Calculate the shaft diameter

Answer:

In engineering and materials science, a stress–strain curve for a material gives the relationship between stress and strain. It is obtained by gradually applying load to a test coupon and measuring the deformation, from which the stress and strain can be determined (see tensile testing).

Explanation:

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Explanation:

Nested piezometers indicate an upward flow if the elevation of the top of the water in the piezometer tube that penetrates the aquifer to the deeper point is greater than the elevation of the water in the shallower tube

Answer:

13.4 mm

Explanation:

Given data :

Load amplitude ( F )  = 22,000 N

factor of safety ( N )= 2.0

Take ( Fatigue limit stress amplitude for this alloy ) б = 310 MPa

calculate the minimum allowable bar diameter to ensure that fatigue failure will not occur

minimum allowable bar diameter = 13.4 * 10^-3 m ≈ 13.4 mm

attached below is a detailed solution

Answer:

t2 = 256 hours

Explanation:

Given data:

Carbon concentration ( C ) at 1.2mm from surface, C = 0.38 wt%

Duration( t ) of heat treatment for 0.38wt% at 1.2mm =  9-hr

Estimate the time (in h) necessary to achieve the same concentration at a 6.4 mm

Assuming same concentration of 0.38wt%  we will apply Fick's second law for constant surface concentration

attached below is the remaining part of the solution

x1 = 1.2 mm

x2 = 6.4 mm

t1 = 9-hr

t2 = ?

t2 = [tex](\frac{6.4}{1.2} )^{2} * 9[/tex]  = 256 hours

True is the answer.....

Answer:

The estimated evaporation is 51340 cubic meters.

Explanation:

Let suppose that precipitation is very small in comparison with depth of the Clear Lake. The monthly change in the volume of the lake ([tex]\Delta V[/tex]), in cubic meters, is estimated by the following formula:

[tex]\Delta V = A\cdot \Delta z + (\dot V_{in}-\dot V_{out})\cdot \Delta t -V_{evap}[/tex] (1)

Where:

[tex]A[/tex] - Surface area of the lake, in square meters.

[tex]\Delta z[/tex] - Water precipitation, in meters.

[tex]\dot V_{in}[/tex] - Average water inflow, in cubic meters per second.

[tex]\dot V_{out}[/tex] - Average water outflow, in cubic meters per second.

[tex]\Delta t[/tex] - Monthly time, in seconds.

[tex]V_{evap}[/tex] - Evaporation, in cubic meters.

If we know that [tex]A = 700000\,m^{2}[/tex], [tex]\Delta z = 0.0762\,m[/tex], [tex]\dot V_{in} = 1.5\,\frac{m^{3}}{s}[/tex], [tex]\dot V_{out} = 1.25\,\frac{m^{3}}{s}[/tex], [tex]\Delta t = 2.592\times 10^{6}\,s[/tex] and [tex]\Delta V = 650000\,m^{3}[/tex], then the estimated evaporation is:

[tex]650000 = 701340-V_{evap}[/tex] (2)

[tex]V_{evap} = 51340\,m^{3}[/tex]

The estimated evaporation is 51340 cubic meters.

Explanation:

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Answer: My favorite color is Red favorite letter C favorite fruit watermelon.

Answer:

0.08704 W

Explanation:

converting the mm to m (1000mm = 1m)

cross-sectional area of the fins, Ac = (0.003) (0.0004) = 0.0000012‬m^2

The wetted perimeter of the cross-section, P = 2 (0.003 + 0.0004) = 0.0068‬m

Thickness of solid in direction of heat flow, B^2 = (heat transient coefficient, h) (The wetted perimeter of the cross-section, P) ÷ (Thermal conductivity, k) (cross-sectional area of the fins, Ac)

B^2 = (8 W/m2K)(0.0068‬m) ÷ (175 W/mK)(0.0000012‬m^2)

=259.0476m^-2

B= square root of the result

B = 16.09m^-1

we now look for:

The Coordinate, x = B, multiplied by Length, L

x = (16.09m^-1) (0.04m) = 0.6436‬

 finding the side area of a fin =  P multiplied by Length, L

= 0.0068‬m X 0.04m = 0.000272m^2

Neglecting inefficiency, assuming the fins are all 100% efficient, the power they would dissipate =

h, Heat-transfer coefficient (PL) (temperature of at the base - temperature at the ambient air)

= (8) (0.000272m^2)(340 K- 300k)

= 0.08704 W

Answer:

(a) Precipitation hardening - 1, 2, 4

(b) Dispersion strengthening - 1, 3, 5

Explanation:

The correct options for each are shown as follows:

Precipitation hardening

From the first statement; Dislocation movement is limited by precipitated particles. This resulted in an expansion in hardness and rigidity. Precipitates particles are separated out from the framework after heat treatment.

The aging process occurs in the second statement; because it speaks volumes on how heated solutions are treated with alloys above raised elevated temperature. As such when aging increases, there exists a decrease in the hardness of the alloy.

Also, for the third option for precipitation hardening; This cycle includes the application of heat the alloy (amalgam) to a raised temperature, maintaining such temperature for an extended period of time. This temperature relies upon alloying components. e.g. Heating of steel underneath eutectic temperature. Subsequent to heating, the alloy is extinguished and immersed in water.

Dispersion strengthening

Here: The effect of hearting is not significant to the hardness of alloys hardening by the method in statement 3.

In statement 5: The process only involves the dispersion of particles and not the application of heat.

Answer:

Distribution (or its more sophisticated counterpart, supply chain management) can add value to goods and services by making them more easily and conveniently available to consumers. ... This means that you need good wholesalers and good transportation systems to get your products to the retailers.

Explanation:

Distribution (or its more sophisticated counterpart, supply chain management) can add value to goods and services by making them more easily and conveniently available to consumers. ... This means that you need good wholesalers and good transportation systems to get your products to the retailers.

Answer:

Distribution (or its more sophisticated counterpart, supply chain management) can add value to goods and services by making them more easily and conveniently available to consumers.

Explanation:

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Answer:

Both technicians.

Explanation:

An oscilloscope can be defined as an electronic instrument used for observing, measuring, analyzing, and displaying the waveform of an electric signal.

The output of an oscilloscope is a graph of instantaneous velocity with respect to time.

Generally, an oscilloscope can be used to check ignition system operating voltages. Also, it can be used to check output signals from computer system sensors.

Therefore, both Technician A and Technician B are right.

Answer:

3.52×10⁶ atoms of Si and 7.05×10⁶ atoms of O

Explanation:

It is all about unit conversions.

The area of our MOS is 1 mm². So, we know that the thickness is 160 nm. This data can give us the volume. We convert nm to mm.

160 nm . 1×10⁻⁶ mm /1nm = 1.6×10⁻⁴ mm

By the way, now we can determine the volume of MOS, in order to work with density.

1.6×10⁻⁴ mm . 1 mm² = 1.6×10⁻⁴ mm³

But density is mg/m³, so we convert mm³ to m³

1.6×10⁻⁴ mm³ . 1×10⁻⁹ m³/mm³ = 1.6×10⁻¹³ m³

Now, we apply density to determine the mass of MOS

Density = mass /volume → Density . volume = mass

1.6×10⁻¹³ m³ . 2.20mg/m³ = 3.52×10⁻¹³ mg

To make more easier the calculate, we convert mg to g.

3.52×10⁻¹³ mg . 1g /1000mg = 3.52×10⁻¹⁶ g

To count the atoms, we determine molar mass of SiO₂ → 60.08 g/mol

We need to know moles of Si and O₂ in the MOS

Firstly, we determine amount of MOS: 3.52×10⁻¹⁶ g / 60.08 g/mol = 5.86×10⁻¹⁸ moles

1 mol of SiO₂ has 1 mol of Si and 2 mol of O so:

5.86×10⁻¹⁸ mol of SiO₂ may have:

(5.86×10⁻¹⁸ . 1) /1 = 5.86×10⁻¹⁸ moles of Si

(5.86×10⁻¹⁸ .2) /1 =  1.17×10⁻¹⁷ moles of O₂

Let's count the atoms (1 mol of anything contain NA particles)

5.86×10⁻¹⁸ mol of Si . 6.02×10²³ atoms/ mol = 3.52×10⁶ atoms of Si

1.17×10⁻¹⁷ mol of O₂  . 6.02×10²³ atoms/ mol = 7.05×10⁶ atoms of O

The number of Si and O atoms present per mm² of the oxide layer are respectively; 3.52 × 10⁶ atoms of Si and 7.05×10⁶ atoms of O

We are given;

Area of MOS device; A = 1 mm²

Thickness of layer; t = 160 nm = 1.6 × 10⁻⁴ mm

Formula for volume is;

V = Area * thickness

V = 1.6 × 10⁻⁴ mm × 1 mm² = 1.6 × 10⁻⁴ mm³

Converting volume to m³ gives;

V = 1.6 × 10⁻¹³ m³

Now, to get the mass of MOS, we will use the formula;

Mass = Density * volume

we are given density = 2.20mg/m³

Thus;

Mass = 1.6 × 10⁻¹³ m³ × 2.20mg/m³

Mass = 3.52 × 10⁻¹³ mg = 3.52×10⁻¹⁶ g

From periodic table, the molar mass of SiO₂ = 60.08 g/mol

Then;

Number of moles of MOS = (3.52 × 10⁻¹⁶ g)/60.08 g/mol

Number of moles of MOS = 5.86 × 10⁻¹⁸ moles

Now, 1 mol of SiO₂  is formed from 1 mol of Si and 2 mol of O. Thus;

5.86×10⁻¹⁸ mol of SiO₂ will have;

5.86×10⁻¹⁸ moles of Si

and (5.86×10⁻¹⁸ .2) =  11.72 × 10⁻¹⁸ moles of O₂

From Avogadro's number that 1 mol equals 6.02×10²³ atoms , we can say that;

5.86 × 10⁻¹⁸ mol of Si × 6.02 × 10²³ atoms/mol = 3.52 × 10⁶ atoms of Si

11.72 × 10⁻¹⁸ mol of O₂ × 6.02 × 10²³ atoms/mol = 7.05×10⁶ atoms of O

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Answer:

r = 1.922 mm

Explanation:

We are given;

Yield stress; σ = 250 MPa = 250 N/mm²

Force; F = 29 KN = 29000 N

Now, formula for yield stress is;

σ = F/A

A = F/σ

Where A is area = πr²

Thus;

r² = 2900/250π

r² = 3.6924

r = √3.6924

r = 1.922 mm

Answer:

True, That is correct. Soil removed from an excavation site is indeed called spoil.

Spoil definition: The waste material (such as soil) brought up during the course of an excavation.

Hope I helped! Sorry if not. Have a wonderful week and follow me for more help! Remember your worth and love yourself! Adíos! ;D

Yes, that is true. It is true that spoil refers to soil excavated from an excavation site.

Thus,  The debris that is dug up during an excavation, such as soil. Depending on the location and the type of excavation, the composition of the spoil material can change.

It could consist of dirt, gravel, boulders, debris, or other substances that were removed or moved during the excavation process.

Typically, the spoil is put aside and used for a variety of tasks, including backfilling, grading, or disposal, as necessary for the project and permitted by local laws.

Thus, Yes, that is true. It is true that spoil refers to soil excavated from an excavation site.

Learn more about Soil, refer to the link:

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Answer:

a

Explanation:

Answer:

When a switch is in the "off" position the circuit is open. Electric charges cannot flow when a switch is in the off position.

Explanation:

A switch that can open or close an electric circuit can be used to stop the flow of current.

A switch can be defined as an electrical component (device) that is typically designed and developed for interrupting the flow of current or electrons in an electric circuit.

This ultimately implies that, a switch that can open or close an electric circuit can be used to stop the flow of current.

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Answer:

1.933 KN-M

Explanation:

Determine the largest permissible bending moment when the composite bar is bent  horizontally

Given data :

modulus of elasticity of steel = 200 GPa

modulus of elasticity of aluminum = 75 GPa

Allowable stress for steel = 220 MPa

Allowable stress for Aluminum = 100 MPa

a = 10 mm

First step

determine moment of resistance when steel reaches its max permissible stress

next : determine moment of resistance when Aluminum reaches its max permissible stress

Finally Largest permissible bending moment of the composite Bar = 1.933 KN-M

attached below is a detailed solution

Answer:

Electrical engineering is an engineering discipline concerned with the study, design and application of equipment, devices and systems which use electricity, electronics, and electromagnetism.

Explanation:

Answer:

Explanation:

This is Answer....

Answer: hello your question lacks the required diagram attached below is the diagram

answer :  29528.1  N/m^2

Explanation:

Given data :

dimensions of tank :

Length = 5-m

Width = 4-m

Depth = 2.5-m

acceleration of tank = 2m/s^2

Determine the maximum gage pressure in the tank

Pa ( pressure at point A )  = s*g*h1

    = 10^3 * 9.81 * 3.01

    = 29528.1  N/m^2

attached below is the remaining part of the solution

Where are your answer options?

Answer:

Implement a hazard communication program

Explanation: i took the quiz

Answer:

Gas turbines in Helicopters require lesser space.

Explanation:

[1] In terms of Space Requirements:

The gas used in helicopters requires lesser space as compared to Automotive gas turbines. The gas in automobile have higher thermal efficiency.

[2]. In terms of Environmental impact:

The occurrence of environmental solution is very slim when  used in helicopters' engines.

[3]. In terms of power-to-weight ratio:

The vibrations in engines of helicopters make it to have lesser efficiency as compared to automobile.

[4]. In terms of Fuel availability:

Fuel is available. Automobile can make use of gas as fuel.

Answer:

Magnitude of force P = 25715.1517 N

Explanation:

Given - The wires each have a diameter of 12 mm, length of 0.6 m, and are made from 304 stainless steel.

To find - Determine the magnitude of force P so that the rigid beam tilts 0.015∘.

Proof -

Given that,

Diameter = 12 mm = 0.012 m

Length = 0.6 m

[tex]\theta[/tex] = 0.015°

Youngs modulus of elasticity of 34 stainless steel is 193 GPa

Now,

By applying the conditions of equilibrium, we have

∑fₓ = 0, ∑[tex]f_{y}[/tex] = 0, ∑M = 0

If ∑[tex]M_{A}[/tex] = 0

⇒[tex]F_{BC}[/tex]×0.9 - P × 0.6 = 0

⇒[tex]F_{BC}[/tex]×3 - P × 2 = 0

⇒[tex]F_{BC}[/tex] = [tex]\frac{2P}{3}[/tex]

If ∑[tex]M_{B}[/tex] = 0

⇒[tex]F_{AD}[/tex]×0.9 = P × 0.3

⇒[tex]F_{AD}[/tex] ×3 = P

⇒[tex]F_{AD}[/tex] = [tex]\frac{P}{3}[/tex]

Now,

Area, A = [tex]\frac{\pi }{4} X (0.012)^{2}[/tex] = 1.3097 × 10⁻⁴ m²

We know that,

Change in Length , [tex]\delta[/tex] = [tex]\frac{P l}{A E}[/tex]

Now,

[tex]\delta_{AD} = \frac{P(0.6)}{3(1.3097)(10^{-4}) (193)(10^{9} }[/tex] = 9.1626 × 10⁻⁹ P

[tex]\delta_{BC} = \frac{2P(0.6)}{3(1.3097)(10^{-4}) (193)(10^{9} }[/tex] = 1.83253 × 10⁻⁸ P

Given that,

[tex]\theta[/tex] = 0.015°

⇒[tex]\theta[/tex] = 2.618 × 10⁻⁴ rad

So,

[tex]\theta = \frac{\delta_{BC} - \delta_{AD}}{0.9}[/tex]

⇒2.618 × 10⁻⁴ = (  1.83253 × 10⁻⁸ P - 9.1626 × 10⁻⁹ P) / 0.9

⇒P = 25715.1517 N

∴ we get

Magnitude of force P = 25715.1517 N

The magnitude of Force P is; P = 25715.15 N

If we draw a free body diagram of the rigid beam system, then for beam AB we can take moments in the following manner;

Taking moments about point A, we have;

(F_bc * 0.9) - P(0.6) = 0

F_bc = ²/₃P

Taking moments about B gives;

P(0.3) - F_ad * 0.9 = 0

F_ad = ¹/₃P

Normal stress for BC is;

σ_bc = F_bc/A_bc

σ_bc = (²/₃P)/(π * 0.006²)

σ_bc = (²/₃P)/(1.131 × 10⁻⁴) N/m²

σ_ad = (¹/₃P)/(π * 0.006²)

σ_ad = (¹/₃P)/(1.131 × 10⁻⁴) N/m²

We know that;

Elongation is; ΔL = PL/AE = (P/A) * (L/E)

Where E for 304 stainless steel is 193 GPa = 193 × 10⁹ Pa

Thus;

ΔL_bc =  (²/₃P)/(1.131 × 10⁻⁴) * (0.6/(193 × 10⁹))

ΔL_bc = 1.83253P × 10⁻⁸

Likewise;

ΔL_ad = (¹/₃P)/(1.131 × 10⁻⁴) * (0.6/(193 × 10⁹))

ΔL_ad = 9.1626P × 10⁻⁹ m

Converting the beam tilt angle from degrees to radians gives;

θ = 0.015° = 0.00026179939 rads

Using small angle analysis, we can say that;

θ = (ΔL_bc - ΔL_ad)/36

θ = P((1.83253P × 10⁻⁸) - (9.1626P × 10⁻⁹))/36

Solving gives P = 25715.15 N

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