‏What is the potential energy in joules of a 12 kg ( mass ) at 25 m above a datum plane ?

In this exercise we have to calculate the formula that will be able to determine the length of the cantilevered, like this:

[tex]\sigma_{max}C=\frac{M_{max}C}{I}[/tex]

So to determinated the maximum tensile and compreensive stress due to bending we can describe the formula as:

[tex]\sigma_b = \frac{MC}{I}[/tex]

Where,

So making this formula for the max, we have:

[tex]\sigma_c=\frac{MC}{I} \\\sigma_T=-\sigma_c=-\frac{MC}{I}\\\sigma_{max}=M_{max}\\[/tex]

With all this information we can put the formula as:

[tex]\sigma_{max}C=\frac{M_{max}C}{I}[/tex]

See more about stress in the beam at brainly.com/question/23637191

Answer:

A.) 0 > K > 9.6

B.) K = 9.6

C.) w = +/- 2 sqrt (3)

Explanation:

G(s)= K(s+4)/s(s+1.2)(s+2)

For a closed loop stability, we can analyse by using Routh - Horwitz analysis.

To make the pole completely imaginary, K must be equal to 9.6 Because for oscillations. Whereas, one pair of pole must lie at the imaginary axis.

Please find the attached files for the solution

Answer:

59°C

Explanation:

Given that, Cc = McCp,c = 5000 /3600 × 4178 = 5803.2(W/K)

and Ch = MhCp,h = 10000 / 3600 × 4188 = 11634.3(W/K)

Therefore the minimum and maximum heat capacities are:

Cmin = Cc = 5803.2(W/K)

Cmax = Ch = 11634.3(W/K)

The capacity ratio is:

Cr = Cmin / Cmax = 0.499 = 0.5

The maximum possible heat transfer rate is:

Qmax = Cmin (Th,i - Tc,i) = 5803.2 (80 - 20) = 348192(W)

And the number of transfer units is: NTU = UA / Cmin = 11600 / 5803.2 = 1.99

Given that from the appropriate graph in the handouts we can read  = 0.7. So the actual heat transfer rate is: Qact = Qmax = 0.7 × 348192 = 243734.4(W)

Hence, the outlet hot temperature is: Th,o = Th,i - Qact / Ch = 59°C

Answer:

true

Explanation:

true

Answer:

I believe this is true

Explanation:

If your looking at something and you look at something else everything is still in perfect view and clear, in focus.

hope this helps :)

Answer: 164.2253 MPa

Explanation:

First we find the half internal crack which is  = length of surface crack /2

so α = 8.6 /2 = 4.3mm ( 4.3×10⁻³m )

Now we find the dimensionless parameter using the critical stress crack propagation equation

∝ = K / Y√πα

stress level ∝ = 112Mpa

fracture toughness K = 26Mpa

dimensionless parameter Y = ?

SO working the formula

Y = K / ∝√πα

Y = 26 / 112 (√π × 4.3× 10⁻³)

Y = 1.9973

We are asked to find stress level for internal crack length of  4m

so half internal crack is  = length of surface crack /2

4/2 = 2mm ( 2 × 10⁻³)

from the previous formula critical stress crack propagation equation

∝ = K / Y√πα

∝ = 26 / 1.9973 √(π × 2 × 10⁻³)

∝ = 164.2253 Mpa

Answer:

[tex]\t(x)_{max} =\dfrac{p\times L}{2\times \pi}[/tex]

Explanation:

Given that

Shear modulus= G

Sectional area = A

Torsional load,

[tex]t(x) = p sin( \frac{2\pi}{ L} x)[/tex]

For the maximum value of internal torque

[tex]\dfrac{dt(x)}{dx}=0[/tex]

Therefore

[tex]\dfrac{dt(x)}{dx} = p cos( \frac{2\pi}{ L} x)\times \dfrac{2\pi}{L}\\ p cos( \frac{2\pi}{ L} x)\times \dfrac{2\pi}{L}=0\\cos( \frac{2\pi}{ L} x)=0\\ \dfrac{2\pi}{ L} x=\dfrac{\pi}{2}\\\\x=\dfrac{L}{4}[/tex]

Thus the maximum internal torque will be at x= 0.25 L

[tex]t(x)_{max} = \int_{0}^{0.25L}p sin( \frac{2\pi}{ L} x)dx\\t(x)_{max} =\left [p\times \dfrac{-cos( \frac{2\pi}{ L} x)}{\frac{2\pi}{ L}} \right ]_0^{0.25L}\\t(x)_{max} =\dfrac{p\times L}{2\times \pi}[/tex]

Hey,

Who plays a role in the financial activities of a company?

O D. Everyone at the company, including managers and employees

Answer:

Everyone at the company, including managers and employees

Explanation:

Answer:

(a) The magnitude of force is 116.6 lb, as exerted by the rod CD

(b) The reaction at A is (-72.7j-38.1k) lb and at B it is (37.5j) lb.

Explanation:

Step by step working is shown in the images attached herewith.

For this given system, the coordinates are the following:

A(0, 0, 0)

B(26, 0, 0)

And the value of angle alpha is 20.95°

Hope that answers the question, have a great day!

Answer:

hello your question is incomplete here is the complete question

. “ABC construction Inc.” company becomes the lowest in the bed process to get a $21 million construction project for “Northern Inc.”. Now “ABC construction Inc.” planning to make a formal contract agreement with the “Northern Inc.”. What are the main elements of this agreement to consider it as a legal contract?

Answer : elements of the agreement

Explanation:

Offer : an offer is the beginning element for any valid agreement to be started or reached between two or more bodies. ABC construction would have to make an offer first for the agreement to be valid

Acceptance: This is part where by the company "Northern Inc" after receiving the offer from ABC construction Inc would have to consent to the approval of the offer made.

capacity : This the element of the agreement that helps to ensure that both parties have the legal and financial backings to embark on the contract agreement .

certainty : This element ensures that both parties understands the terms and conditions attached to the agreement and this to ensure that there are no bogus conditions

Consideration : This is a very vital element because the both parties have to give something in return while going into a valid agreement

Intention to create legal relation : Legal relations are applied to contract agreements whereby both parties want the contract agreement to b legally enforced and this is important in order to prevent contract breach by any party involved in the agreement

The red and green lights are sidelights that are positioned on the port side (red) (left as facing the bow) and starboard (green) (right as facing the bow) side of the boat. Various white lights are required depending on the size of the boat, but generally, a white masthead light and stern light are required. See the US Coast Guard site in the link below for more specific information.

Hope this helps

Answer:

max shear = R = V = 15 kN

Explanation:

given:

load = 10 kn/m

span = 3m

max shear = R = V = wL / 2

max shear = R = V = (10 * 3) / 2

max shear = R = V = 15 kN

Explanation:

1. increase This due to increase in the pore volume.

2.increase . This is due to the fact that more liquid phase will be present at the firing.

3. Increase. This increase is because of the fact that clay on cooling forms glass.Thus, gaining more strength as the liquid phase formed fills in pore volume.

4. Increase, Rate of corrosion depends upon the surface area exposed.Since, upon vitrification surface area would increase, therefore corrosion increases.

5. Increase , glass has higher thermal conductivity than the pores it fills.

Answer:

the required stress level  at which fracture will occur for a critical internal crack length of 3.0 mm is 189.66 MPa

Explanation:

From the given information; the objective is to compute the stress level at which fracture will occur for a critical internal crack length of 3.0 mm.

The Critical Stress for a maximum internal crack can be expressed by the formula:

[tex]\sigma_c = \dfrac{K_{lc}}{Y \sqrt{\pi a}}[/tex]

[tex]Y= \dfrac{K_{lc}}{\sigma_c \sqrt{\pi a}}[/tex]

where;

[tex]\sigma_c[/tex] = critical stress required for initiating crack propagation

[tex]K_{lc}[/tex] = plain stress fracture toughness = 26 Mpa

Y = dimensionless parameter

a = length of the internal crack

given that ;

the maximum internal crack length is 8.6 mm

half length of the internal  crack will be 8.6 mm/2 = 4.3mm

half length of the internal  crack a = 4.3 × 10⁻³ m

From :

[tex]Y= \dfrac{K_{lc}}{\sigma_c \sqrt{\pi a}}[/tex]

[tex]Y= \dfrac{26}{112 \times \sqrt{\pi \times 4.3 \times 10 ^{-3}}}[/tex]

[tex]Y= \dfrac{26}{112 \times0.1162275716}[/tex]

[tex]Y= \dfrac{26}{13.01748802}[/tex]

[tex]Y=1.99731315[/tex]

[tex]Y \approx 1.997[/tex]

For this same component and alloy, we are to also compute the stress level at which fracture will occur for a critical internal crack length of 3.0 mm.

when the length of the internal crack a = 3mm

half  length of the internal  crack will be 3.0 mm / 2 = 1.5 mm

half length of the internal  crack a =1.5 × 10⁻³ m

From;

[tex]\sigma_c = \dfrac{K_{lc}}{Y \sqrt{\pi a}}[/tex]

[tex]\sigma_c = \dfrac{26}{1.997 \sqrt{\pi \times 1.5 \times 10^{-3}}}[/tex]

[tex]\sigma_c = \dfrac{26}{0.1370877444}[/tex]

[tex]\sigma_c =189.6595506[/tex]

[tex]\sigma_c =[/tex] 189.66 MPa

Thus; the required stress level  at which fracture will occur for a critical internal crack length of 3.0 mm is 189.66 MPa

Answer:

Engineering drawing abbreviations and symbols are used to communicate and detail the characteristics of an engineering drawing.

There are many abbreviations common to the vocabulary of people who work with engineering drawings in the manufacture and inspection of parts and assemblies.

Technical standards exist to provide glossaries of abbreviations, acronyms, and symbols that may be found on engineering drawings. Many corporations have such standards, which define some terms and symbols specific to them; on the national and international level, like BS8110 or Eurocode 2 as an example.

Explanation:

Answer:

Maneuvering speed.

Explanation:

The speed above which an airplane will experience structural damage when a load is applied, instead of stalling, is called the maneuvering speed and varies with weight.

In aeronautical engineering, the maneuvering speed (Va) of an aircraft such as an aeroplane, helicopter, or jet is an airspeed limitation which is mainly selected by an aircraft designer.

Generally, at speeds higher or greater than the manoeuvring speed, aircraft pilots are advised not to attempt a full deflection of any flight control surface because it's capable of resulting in a damage to the structure of an aircraft.

If you're a pilot, to find the maneuvering speed of an aircraft, you should look at the flight manual of the aircraft or on the cockpit placard in the aircraft. The maneuvering speed of an aircraft is a calibrated speed and should not be exceeded by any pilot.

Answer:

Hello the table which is part of the question is missing and below are the table values

For a 5V belt the available diameters are : 5.5, 5.8, 5.9, 6.2, 6.3, 6.6, 12.5, 13.9, 15.5, 16.1, 18.5, 20.1

Answers:

belt size = 140 in with diameter of 20.1n

actual speed of belt = 288.49 in/s

actual center distance = 49.345 in

Explanation:

Given data :

Electric motor (driver sheave) speed (w1) = 950 rpm

Driven sheave speed (w2) = 250 rpm

pick D1 ( diameter of driver sheave)  = 5.8 in  ( from table )

To select an appropriate belt size we apply the equation for the velocity ratio to get the diameter first

VR = [tex]\frac{w1}{w2}[/tex] = 950 / 250

also since the speed of  belt would be constant then ;

Vb = w1r1 = w2r2 ------- equation 1

r = d/2

substituting the value of r into equation 1

equation 2 becomes : [tex]\frac{w1}{w2} = \frac{d2}{d1}[/tex]    = VR

Appropriate belt size ( d2) can be calculated as

d2 = [tex]\frac{w1d1}{w2}[/tex] = [tex]\frac{950 * 5.8}{250}[/tex] = 22.04

From the given table the appropriate belt size would be : 20.1 because it is the closest to the calculated value

next we have to determine the belt length /size

[tex]L = 2C + \frac{\pi }{2} ( d1+d2) + \frac{(d2-d1)^2}{4C}[/tex]

inputting  all the values into the above equation including the value of C as calculated below

L ≈ 140 in

Calculating the center distance

we use this equation to get the ideal center distance

[tex]d2< C_{ideal} < 3( d1 +d2)[/tex]

22.04 < c < 3 ( 5.8 + 20.1 )

22.04 < c < 77.7

the center distance is between 22.04 and 77.7  but taking an average value

ideal center distance would be ≈ 48 in

To calculate the actual center distance we use

[tex]C = \frac{B+\sqrt{B^2 - 32(d2-d1)^2} }{16}[/tex] -------- equation 3

B = [tex]4L -2\pi (d2 + d1 )[/tex]

inputting all the values into (B)

B = 140(4) - 2[tex]\pi[/tex]( 20.01 + 5.8 )

B ≈ 399.15 in

inputting all the values gotten Back to equation 3 to get the actual center distance

C = 49.345 in ( actual center distance )

Calculating the actual belt speed

w1 = 950 rpm = 99.48 rad/s

belt speed ( Vb) = w1r1 = w1 * [tex]\frac{d1}{2}[/tex]

                           = 99.48 * 5.8 / 2 = 288.49 in/s

Answer:

The normal stress is 0.7821 MPa

Explanation:

The external diameter D = 160 mm

The thickness t = 3.8 mm = 3.8 x 10^-3 m

gauge pressure P = 78 kPa = 78 x 10^3 Pa

The maximum shear stress τmax = ?

The external radius of the shell from the external surface R = D/2 = 160/2 = 80 mm

The internal radius of the shell r = R - t

==> 80 - 3.8 = 76.2 mm

Therefore the internal diameter d = 2r = 2 x 76.2 = 152.4 mm

==> d = 152.4 x 10^-3 m

The normal stress σ = [tex]\frac{Pd}{4t}[/tex] = [tex]\frac{78*10^{3}*152.4*10^{-3} }{4*3.8*10^{-3} }[/tex] = 782052.63 Pa

==>  σ = 0.7821 MPa

Answer:

A to C = 6.4 km

Explanation:

A to B = 4 km

B to C = 5 km

A to C =  using pythagorean theorem

a² + b² = c²

a = A to B = 4

b = B to C = 5

c = A to C

c² = 4² + 5²

c = 6.4 km (A to C)

According to the scenario, the distance between town C from town A is found to be 6.40 Km.

The background that this question depends on is known as the direction-based question. These types of questions completely depend on the distance of moving bodies like cars, persons, or any other objects as well with respect to the initial position.

According to the question,

The distance between town A to town B = 4 km.

The distance between town B to town C = 5 km.

Now, according to the Pythagoras theorem, the distance between town C to town A is as follows:

[tex]AC^2[/tex] = [tex]AB^2 +BC^2[/tex].

[tex]AC^2[/tex] = [tex]4^2+5^2[/tex]

[tex]AC^2[/tex] = 16 + 25 = 41.

AC = √41 = 6.40 km.

Therefore, the distance between town C from town A is found to be 6.40 Km.

To learn more about Pythagoras' theorem, refer to the link:

https://brainly.com/question/343682

#SPJ5

Answer:

(a) The value of the resistance is 3.431 Ω

(b) The value of the inductance is 0.0264 H

Explanation:

Given;

frequency of the source, f = 60 Hz

rms voltage, V-rms = 220 V

real power, Pr = 1500 W

apparent power, Pa = 4600 VA

(a). Determine the value of the resistance

[tex]P_r = I_{rms}^2R[/tex]

where;

R is resistance

[tex]I_{rms} = \frac{Apparent \ Power}{V_{rms}} \\\\I_{rms} = \frac{P_a}{V_{rms}}\\\\I_{rms}= \frac{4600}{220} \\\\I_{rms}= 20.91 \ A[/tex]

Resistance is calculated as;

[tex]R = \frac{P_r}{I_{rms}^2} \\\\R = \frac{1500}{(20.91)^2} \\\\R = 3.431 \ ohms[/tex]

(b). Determine the value of the inductance.

[tex]Q_L = I_{rms}^2 X_L[/tex]

where;

[tex]Q_L[/tex] is reactive power

[tex]X_L[/tex] is inductive reactance

[tex]Apparent \ power = \sqrt{Q_L^2 + P_r^2} \\\\P_a^2 = Q_L^2 + P_r^2\\\\Q_L^2 = P_a^2 - P_r^2\\\\Q_L^2 = 4600^2 - 1500^2\\\\Q_L^2 = 18910000\\\\Q_L = \sqrt{18910000}\\\\Q_L = 4348.56 \ VA[/tex]

inductive reactance is calculated as;

[tex]X_L = \frac{Q_L}{I_{rms}^2} \\\\X_L = \frac{4348.56}{(20.91)^2} \\\\X_L = 9.95 \ ohms[/tex]

inductance is calculated as;

[tex]X_L = \omega L\\\\X_L = 2\pi f L\\\\L = \frac{X_L}{2\pi f} \\\\L = \frac{9.95}{2\pi *60} \\\\L = 0.0264 \ H\\\\L = 26.4 \ mH[/tex]

Answer:

at  t = 45 s :  

To = 61.7⁰c,  T1 = 55.6⁰c, T2 = 49.5⁰c, T3 = 34.8⁰C

Explanation:

Wall thickness = 0.12 m

thermal diffusivity = 1.5 * 10^-6 m^2/s

Δt ( time increment ) = 300 s

Δ x   = 0.03 m ( dividing wall thickness into 4 parts assuming the system to be one dimensional )

using the explicit finite-difference technique

Detailed solution is attached below  

Answer:

a) cross sectional area of the core = 0.0187 m²

b) The secondary voltage on no-load = 413 V

c) The primary currency and power factor on no load is 1.21 A and 0.168 lagging respectively.

Explanation:

See attached solution.

Answer:

no need for that

Explanation:

they are not the same at all

Answer:

From the calculation, we can see that the invention's COP of 2.67 does not exceed the maximum theoretical COP of 14.65. Hence his claim is valid and could be possible.

Explanation:

Heat generated Q = 200 kW

power input W = 75 kW

Temperature of heated region [tex]T_{h}[/tex] = 293 K

Temperature of heat source [tex]T_{c}[/tex] = 273 K

For this engine,

coefficient of performance COP = Q/W  = 200/75 = 2.67

The maximum theoretical COP obtainable for a heat pump is given as

COP = [tex]\frac{T_{h} }{T_{h} - T_{c} }[/tex] =  [tex]\frac{293 }{293 - 273 }[/tex] = 14.65

From the calculation, we can see that the invention's COP of 2.67 does not exceed the maximum theoretical COP of 14.65. Hence his claim is valid and could be possible.