Convert 5m/h to yds/week

Water is a great erode. Man-made levees must be tough. They are often built of rock and concrete. The correct option is A.

Man-made levees are designed to withstand the erosive power of water and prevent flooding in areas near bodies of water.

As such, they are constructed using strong and durable materials, such as rock and concrete, that can resist the forces of water and prevent erosion.

These materials provide a solid and stable foundation for the levees, which is important for ensuring their effectiveness in protecting against flooding.

Packed mud and dirt or sandbags and gravel may not provide sufficient strength and durability to withstand the erosive power of water and may not be suitable for building effective levees.

Thus, the correct option is A.

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Answer:

it is weak walls

Explanation:

Answer:

Between 90 and 110 degrees

Explanation:

Ideally, the angle should not be less than 90° so that your elbows could be parallel to the workstation. This will in turn mean that the shoulders will be relaxed and it will make it a lot less stressful to keep the wrists in a straight manner.

Looking at the options, there is no exact figure of 90°, however, it could be between 90° to 110° because the difference is not going to be much in terms of making the shoulders relaxed.

Answer:

90 degrees i did the test and i also looked it up to make sure i was right.

Explanation:

Answer:

784.090909mph

Explanation:

1ft/s=0.681818 mph

1150ft/s=0.681818 x 1150=784.090909

Answer:

the saturated density should be

Explanation:

Answer:

a) Ef = 0.755

b) length of specimen( Lf )= 72.26mm

  diameter at fracture = 9.598 mm

c) max load ( Fmax ) = 52223.24 N

d) Ft = 51874.67 N

Explanation:

a) Determine the true strain at maximum load and true strain at fracture

True strain at maximum load

Df = 9.598 mm

True strain at fracture

Ef = 0.755

b) determine the length of specimen at maximum load and diameter at fracture

Length of specimen at max load

Lf = 72.26 mm

Diameter at fracture

= 9.598 mm

c) Determine max load force

Fmax = 52223.24 N

d) Determine Load ( F ) on the specimen when a true strain et = 0.25 is applied during tension test

F = 51874.67 N

attached below is a detailed solution of the question above

Answer:

b

Explanation:

Trust me. I am not the smartest though...

Answer:

Hello your question is incomplete attached below is the missing part

a) p1 = 454.83 kPa,  p2 = 283.359 Kpa , p3 = 536.423 kpa , p4 = 860.959kPa

b) W12 = 3.4 kJ, W23 = -3.5875 KJ, W34 = -4.0735 KJ, W41 = 3.5875 KJ

c) 5

Explanation:

Given data:

mass of air ( m ) = 1/10 kg

adiabatic index ( k ) = 1.4

temperature for isothermal expansion = 250K

rate of heat transfer ( Q12 ) = 3.4 KJ

temperature for Isothermal compression ( T4 ) = 300k

final volume ( V4 ) = 0.01m ^3

a)  Calculate the pressure, in Kpa, at each of the four principal states

from an ideal gas equation

P4V4 = mRT4 ( input values above )

hence P4 = 860.959kPa

attached below is the detailed solution

b) Calculate work done for each processes

attached below is the detailed solution

C) Calculate the coefficient of performance

attached below is detailed solution

Answer:

Uh- dude if you think I'm gonna download that, think twice...

Answer:

The answer is "slightly before the disc brakes".

Explanation:

This dose appropriate secondary that perhaps the brake pads get hydrostatic fluid slightly even before brake discs on vehicles of rear trumpet brake and front disc brakes.

These same brakes are far more powerful and efficient than brake pads. It's why the brakes of disc-brakes could be found fading as much as on drum breaks.

Answer:

B

Explanation:

Answer:

[tex]1.18\ \text{kPa}[/tex]

Explanation:

g = Acceleration due to gravity = [tex]9.81\ \text{m/s}^2[/tex]

h = Height of reading = 12 cm

[tex]\rho[/tex] = Density of water = [tex]1000\ \text{kg/m}^3[/tex]

Pressure due to height difference is given by

[tex]P=\rho gh\\\Rightarrow P=1000\times 9.81\times 12\times 10^{-2}\\\Rightarrow P=1177.2\ \text{Pa}=1.1772\ \text{kPa}\approx 1.18\ \text{kPa}[/tex]

The pressure differential is [tex]1.18\ \text{kPa}[/tex].

According to the question,

The formula will be:

→ [tex]P = \rho g h[/tex]

By putting the values, we get

      [tex]= 1000\times 9.8\times 12\times 10^{-2}[/tex]

      [tex]= 1177.2 \ Pa[/tex]

      [tex]= 1.18 \ kPa[/tex]

Thus the answer above is correct.

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Answer:

-3 dB point

Explanation:

A filter is a circuit deigned to pass or amplify certain frequencies while also preventing other frequencies from passing. There are different types of filter depending on the frequency range such as the low pass filter, high pass filter, bandstop filter, bandpass filter.

A low pass filter is a filter that allows the passage of frequencies below the cut off frequency and blocks frequencies above the cut off frequency. The cutoff frequency of a low pass filter is the frequency at the -3dB point.

Answer:

team building exercises

Explanation:

Answer:   its not team building excercize

Explanation:

Answer:

The armature voltage required to run the motor at 1500 RPM is 143.75 V

The armature voltage required to run the motor at 100 RPM is 9.58 V

Explanation:

Given;

angular speed of the DC motor, ω = 1200 RPM

source voltage, V = 115 V

The armature voltage required to run the motor at 1500 RPM;

[tex]V_a = 115V \ \times \ \frac{1500 \ RPM}{1200 \ RPM} \\\\V_a = 143.75 \ V[/tex]

The armature voltage required to run the motor at 100 RPM;

[tex]V_a = 115V \ \times \ \frac{100 \ RPM}{1200 \ RPM} \\\\V_a = 9.58 \ V[/tex]

Answer:pp

Explanation:

Answer:

a pneumatic or electric power

Explanation:

The primary energy source for the controller in a typical control system is either "a pneumatic or electric power."

This is because a typical control system has majorly four elements which include the following:

1. Sensor: this calculates the controlled variable

2. Controller: this receives and process inputs from the sensor to the controlled device as output

3. Controlled device: this tweak the controlled variable

4. Source of energy: this is the energy used to power the control system. It could be a pneumatic or electric power

Answer: required tensile stress is 0.889 MPa

Explanation:

Given that;

tensile load is oriented along the [1 1 1] direction

shear stress is 0.242 MPa along [1 0 1] in the (1 1 -1) plane

first we determine

λ which is Angle between  [1 1 1]  and  [1 0 1]

so

cosλ = [ 1(1) +  1(0) + 1(1) ] / [ √(1² + 1² + 1²) √(1² + 0² + 1²)]

= 2 / √3√2 =  2/√6

Next, we determine ∅ which is angle between [1 1 1]  and  [1 1 -1]

so,

cos∅ = [ 1(1) +  1(1) + 1(-1) ] / [ √(1² + 1² + 1²) √(1² + 1² + (-1)²)]

cos∅ = [ 2-1] / [√3√3 ]

cos∅ = 1/3

Now, we know that;

σ = T_stress/cosλcosθ

so we substitute

σ = 0.242 / ( 2/√6 × 1/3 )

σ = 0.242 / 0.2721

σ = 0.889 MPa

Therefore, required tensile stress is 0.889 MPa

No amount of metal removal will restore a cracked surface, rather, there is a need to completely change the metal drum.

This means a from of cylindal metal container that is used for shipping or storage of liquids.

When its surface is cracked, then, no amount of metal removal will restore a cracked surface, rather, there is a need to completely change the metal drum.

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Answer:Animation can occur before an action. Prepares the audience for the action. ... A pose or action should clearly communicate to the audience the attitude, mood, reaction or idea of the character. The effective use of long, medium or close up shots, as well as camera angles helps tell the story.

Explanation:May i have brainlist plz only if u wanna give me brainlist though have an nice day and stay safe.

Answer:

a

Explanation:

a

The correct answer is A. Earning a bachelor's degree in Civil Engineering from a four-year university, completing an internship, and seeking a job at a private firm.

Explanation:

In the U.S. and many countries, the best to start a career is to enroll in a formal educational program at a university or college. This helps students learn concepts, theories, methods, etc. they need for their profession. Moreover, a degree such as a bachelor's degree is required by employers. In this context, the first step for Connor is to earn a bachelor's degree in Civil Engineering.

Besides this, an internship is recommended after earning a degree because this is the way students can gain real-life work experience, which is considered positive by employers. This means the next step should be an internship.

Finally, Connor can seek a job to design bridges and other buildings because after the degree and internship he will have the experience and knowledge required by employers and by the job.

Answer:

a) [tex]E_{m}[/tex] = 133.75 Gpa

b) Fnet = 560 N

c) thermal expansion of the composite material = 14.31 [tex]10^{-6 }[/tex] / °C

Explanation:

Solution:

a) Elastic Modulus of the composite:

Area of steel wire = [tex]\frac{\pi }{4}[/tex] x ([tex]0.001^{2}[/tex]) = 0.8 x [tex]10^{-6}[/tex] [tex]m^{2}[/tex]

Area of Copper wire = [tex]\frac{\pi }{4}[/tex] x ([tex]0.002^{2}[/tex]) - 0.8 x [tex]10^{-6}[/tex] [tex]m^{2}[/tex]

Area of Copper wire = 2.4 x [tex]10^{-6}[/tex] [tex]m^{2}[/tex]

Young's Modulus of Composite mixture:

[tex]E_{m}[/tex] = [tex]F_{st}[/tex][tex]E_{st}[/tex] +  [tex]F_{Cu}[/tex][tex]E_{Cu}[/tex]     Equation 1

here,

[tex]F_{st}[/tex] = Stress in Steel

[tex]F_{Cu}[/tex] = Stress in Copper.

We know that,

F = P/A

F is inversely proportional to Area, so if area is large, stress will less and vice versa. So, Take

Ratio for area of steel = [tex]\frac{0.8. 10^{-6} }{(0.8 + 2.4) .10^{-6} }[/tex]

Ratio for area of steel = [tex]\frac{0.8}{3.2 }[/tex] = 0.25

Similarly, for Copper,

Ratio for area of copper = [tex]\frac{2.4. 10^{-6} }{(0.8 + 2.4) .10^{-6} }[/tex]

Ratio for area of copper = [tex]\frac{2.4 }{3.2}[/tex] = 0.75

Put these values in equation 1:

[tex]E_{m}[/tex] = [tex]F_{st}[/tex][tex]E_{st}[/tex] +  [tex]F_{Cu}[/tex][tex]E_{Cu}[/tex]    

[tex]E_{m}[/tex] = (0.25) [tex]E_{st}[/tex] +  (0.75)[tex]E_{Cu}[/tex]

We are given that,

  [tex]E_{st}[/tex] = 205 Gpa

[tex]E_{Cu}[/tex]  = 110 Gpa

So,

[tex]E_{m}[/tex] = (0.25) (205 Gpa) +  (0.75) (110 GPa)

[tex]E_{m}[/tex] = 51.25GPa + 82.5 Gpa

Hence, the Elastic Modulus of the composite will be:

[tex]E_{m}[/tex] = 133.75 Gpa

b) maximum force:

Fnet = Fst + Fcu

We know that F = (Yield Stress x Area)

F = fst x Ast + fcu x Acu

And we are given that,

Yield stress of Steel = 280 Mpa

Yield stress of Copper = 140 Mpa

And,

Ast = 0.8 x [tex]10^{-6}[/tex] [tex]m^{2}[/tex]

Acu = 2.4 x [tex]10^{-6}[/tex] [tex]m^{2}[/tex]

Just plugging in the values, we get:

F = (280 Mpa) (0.8 x [tex]10^{-6}[/tex] [tex]m^{2}[/tex]) + (140 Mpa) (2.4 x [tex]10^{-6}[/tex] [tex]m^{2}[/tex])

F = 224 + 336

Fnet = 560 N    ( because Mpa = [tex]10^{6}[/tex] N/[tex]m^{2}[/tex])

So, it means the composite will carry the maximum force of 560N

c) Coefficient of Thermal Expansion:

Strain on both material is same upon loading so,

(ΔL/L)st = (ΔL/L)cu

by thermal expansion equation:

([tex]\alpha .[/tex]ΔT  + [tex]\frac{F}{A}[/tex][tex]. \frac{1}{Est}[/tex]) = [tex]\alpha .[/tex]ΔT  + [tex]\frac{F}{A}[/tex][tex]. \frac{1}{Ecu}[/tex])

Where [tex]\alpha[/tex] = Coefficient of Thermal expansion

Here, fst = -fcu = F

and ΔT = 1°

So,

Plugging in the values, we get.

( 10 x [tex]10^{-6}[/tex] x (1) + [tex]\frac{F}{0.8.10^{-6} } . \frac{1}{205 . 10^{9} }[/tex] ) = ( 17 x [tex]10^{-6}[/tex] x (1) + [tex]\frac{-F}{2.4.10^{-6} } . \frac{1}{110 . 10^{9} }[/tex] )

Solving for F, we get:

F = 0.71 N

Here,

fst = F = 0.71 N (Tension on Heating)

fcu = -F = 0.71 N ( Compression on Heating )

So, the combined thermal expansion of the composite material will be:

(ΔL/L)cu = ( 17 x [tex]10^{-6}[/tex] x (1°) + [tex]\frac{-0.71}{2.4.10^{-6} } . \frac{1}{110 . 10^{9} }[/tex] )

(ΔL/L)cu = ( 17 x [tex]10^{-6}[/tex] x (1°) - 2.69 x [tex]10^{-6}[/tex]

combined thermal expansion of the composite material = 14.31 [tex]10^{-6 }[/tex] / °C

Answer:

The answer is below

Explanation:

Case hardening is a form of steel hardening that is applied on mild steel with a high temperature of heat.

It results in material forming a hard surface membrane, while the inner layer is soft.

It is mostly used for universal joints, construction cranes, machine tools, etc.

On the other hand, Through hardening is a form of steel hardening in engineering that involves heat treatment of carbon steel.

It increases the hardness and brittleness of the material.

It is often used for axles, blades, nuts and bolts, nails, etc.

Answer:

2.29mHg

Explanation:

We first do a conversion of gage pressure to mmHg

= 204(10³) / 13595(9.807)

= 204000/133326.165

= 1.53mHg

755 mmhg = 0.755

The pressure is therefore

P = p(atmosphere) + Of

= 1.53 + 0.755

= 2.29 mHg

This is the absolute pressure on this tire. So this answers this question.

The absolute pressure in the tire is equal to 2,285.13 mmHg.

Given the following data:

To calculate the absolute pressure in the tire:

First of all, we would convert the value of gauge pressure in kPa to mmHg.

Conversion:

1 kPa = 7.50062 mmHg

204 kPa = [tex]204 \times 7.50062 = 1530.13\;mmHg[/tex]

Mathematically, absolute pressure is given by this formula:

[tex]P_{abs} = P_{atm} + P_{ga}[/tex]

Where:

Substituting the parameters into the formula, we have;

[tex]P_{abs} = 755 + 1530.13[/tex]

Absolute pressure = 2,285.13 mmHg.

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Answer: the ripple factor is 0.005

Explanation:

Given the data in the question;

we know that expression of ripple factor is;

r = Vr(pp) / Vdc

where Vr(pp) the peak to peak is ripple voltage ( 100mv = 0.1 V)

and Vdc is the dc value of the filter output voltage ( 20 V)

so we substitute our given values;

r = 0.1 / 20

r = 0.005

Therefore; the ripple factor is 0.005

The ripple factor of the given power supply filter is; γ = 0.005

We are given;

Ripple peak to peak voltage; V_r,pp = 100 mV = 0.1 V

DC value of the filter output voltage; V_dc = 20 V

Now, formula for the ripple factor is given as;

γ = V_rms/V_dc

Now, our ripple peak to peak voltage is also known as the rms value of ripple voltage at the output. Thus;

γ = 0.1/20

γ = 0.005

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Answer:

b.

Explanation:

Foresight rod reading is any measurement taken at a given sight to determine a particular elevation. The backsight rod readings are usually used to a point of certain elevation. It is added to the elevation to determine the height of the instrument. These are phenomena are used in differential leveling and are applied to corrections for curvature and refraction.

Answer:

A

Explanation:

It is also a toxic material

Answer:

The correct answer is A. Lead.

Explanation:

I got it right on the Plato test.