For laminar flow over a flat plate the local heat transfer coefficient varies as hx 5 Cx20.5, where x is measured from the leading edge of the plate and C is a constant. Determine the ratio of the average convection heat transfer coefficient over the entire plate of length L to the local convection heat transfer coefficient at the end of the plate (x 5 L).

Answer:

a) 4.286 m² b) $ 55.44/mo

Explanation:

If we assume that the sun is behaving as an isotropic radiator, the power density that is arriving to the house, is constant and equal to the quoted solar irradiance.

If the energy conversion capability of the solar panels were 100%, the roof area needed to supply the power required, would be simply the quotient between the power required and the solar irradiance, as follows:

A = P / SI = 10 A* 120 V / 800 W/m² = 1200 W / 800 W/m²= 1,5 m²

As the solar panels are only 35% efficient in converting the solar energy to useful electrical energy, we will need more roof area, according to this expression:

Ae = At / 0.35 = 1,5 / 0.35 = 4.286 m²

b) If we can get 1200 W during 7 hs/day, the energy supplied by the solar panels will be the product of the power times the time, as follows:

E= 1200 W* 7 hs = 8.4 Kwh

If the cost per Kwh, is $0.22, assuming 7 hs. of use in average during a month (assumed to be of 30 days), we can have savings as follows:

Cost = 0.22($/Kwh)* 8.4 (Kwh/day)*30 (days/mo) = $ 55.44

Copper cables are used to communicate binary data by altering the voltage between two ranges.

A pair of Twisted copper cables send data through a network by transmitting pulses of electricity that represent binary data.

On this note, to make sure cables are transmitting information in a way that can be understood by the recipient, they follow the Ethernet standards. This twisted pair cables are commonly known as Ethernet cables.

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Answer / Explanation:

The divergence in my complex undulating movement assumed a sinusoidal wave movement that it began to form harmonic melodies even though the standing wave form seem static.

Never knew my exponential increment were causing intrinsic impedance causing a constant phase shift  allowing me cross my boundary conditions.

I now therefore concur that the curl in every angle, gradient and wave pattern i assume are as a result of constant propagation of or the bid to exceed my boundary condition.

Now i feel caged in this electromagnetic field force while i dance i phasor form. Alas let it be known that every boundary met was polarized by me.

Answer:

The viscous shear stress is 968 Pa (or can be written as 968 [tex]N/m^2[/tex]) and the drag force is 968 N.

Explanation:

Viscous shear stress

For a parallel flow of a Newtonian fluid, the shear stress is proportional to the gradient of the velocity,

[tex]\tau_{yx}= \mu \cfrac{du}{dy}[/tex]

Considering a large square plate moving on a thin film of thickness h, the velocity profile is

[tex]u(y) = U\cfrac{y}{h}[/tex]

Thus its derivative will be just

[tex]\cfrac{du}{dy}= \cfrac U h[/tex]

So replacing on the viscous shear stress formula we get

[tex]\tau_{yx}= \mu \cfrac {U}h[/tex]

We can then replace the given information

[tex]\tau_{yx}= 0.968 \times \cfrac{kg}{m\times  s}\times \cfrac{0.1 \cfrac ms}{0.0001 \, m}[/tex]

Evaluating we get

[tex]\boxed{\tau_{yx}=968\, Pa}[/tex]

The viscous shear stress is 968 Pascals or [tex]N/m^2[/tex].

Drag force

Using the given equation for the drag force we have

[tex]D = \tau_{yx} A[/tex]

And since we have a large square plate of sides L we can write the area as

[tex]A = L^2[/tex]

So the drag force is

[tex]D = \tau_{yx} L^2[/tex]

Replacing values

[tex]D = 968\, Pa \times (1\, m)^2[/tex]

We get

[tex]\boxed{D= 968 \,N}[/tex]

The drag force is 968 N.

Elongation in a bar is

[tex]\epsilon = \frac{PL}{\pi r^2 E}[/tex]

Where,

P = Applied force

L = Lengthof the specified rod

A = Cross-sectional area [tex](\pi r^2)[/tex]

E = Modulus of Elasticity

Performing a quick analysis we can realize that the larger the radius, the lower the elongation percentage. The radius is inversely proportional to the percentage of elongation. For this reason in a 2in bar the change will be GREATER than that of an 8in bar.

Answer:

[tex]P_m_i_n= 356.9 KPa[/tex]

Explanation:

Coefficient of performance (COP) of refrigeration cycle is given by:

[tex]COP=\frac{1}{\frac{T_H}{T_L}-1 }[/tex]

We are given:

[tex]T_H=1.2T_L[/tex]

[tex]COP=\frac{1}{1.2-1 }[/tex]

COP= 5

We can also write Coefficient of performance (COP) of refrigeration cycle  as:

[tex]COP_R=\frac{Q_L}{W_i_n}[/tex]

Amount of heat absorbed by low temperature reservoir can be found as:

[tex]Q_L=COP_R * W_i_n[/tex]

[tex]Q_L=5 * 22 KJ[/tex]

[tex]Q_L=110 KJ[/tex]

According to first law of thermodynamics amount of heat rejected by hot reservoir is given by:

[tex]Q_H=Q_L + W_i_n[/tex]

[tex]Q_H=110 KJ + 22 KJ[/tex]

[tex]Q_H=132 KJ[/tex]

We are given the mass of 0.96 kg. So,

[tex]q_H=\frac{Q_H}{m}[/tex]

[tex]q_H=\frac{132 KJ}{0.96Kg}[/tex]

[tex]q_H=137.5 KJ/Kg[/tex]

Since it is a saturated liquid-vapour mixture [tex]q_H=h_f_g[/tex].

[tex]q_H=h_f_g=137.5 KJ/Kg[/tex]

From Refrigerant 134-a tables [tex]T_H[/tex] at [tex]h_f_g=137.5 KJ/Kg[/tex] is 61.3 C. (We calculated this by interpolation)

Converting [tex]T_H[/tex] from Celsius to Kelvin:

[tex]61.3^{o} C+273 = 334.3^{o} K[/tex]

[tex]T_H= 334.3^{o} K[/tex]

We are given:

[tex]T_H=1.2T_L[/tex]

[tex]T_L=\frac{T_H}{1.2}[/tex]

[tex]T_L=\frac{334.3}{1.2}[/tex]

[tex]T_L=278.58^{o} K[/tex]

Converting [tex]T_L[/tex] from Kelvin to Celsius:

[tex]278.58^{o} K-273 = 5.58^{o} C [/tex]

[tex]T_L= 5.58^{o} C [/tex]

From Refrigerant 134-a tables [tex]P_m_i_n[/tex] at [tex]T_L=5.58^{o} C[/tex] is 356.9 KPa. (We calculated this by interpolation).

[tex]P_m_i_n= 356.9 KPa[/tex]

The minimum pressure in the Carnot refrigeration cycle is 356.9 kPa.

Using,

Coefficient of performance (COP) of the refrigeration cycle. The COP is a measure of how efficient the refrigeration cycle is.

It is given by the formula:

COP = 1 / ([tex]T_H[/tex] /[tex]T_L[/tex] - 1),

where

[tex]T_H[/tex] is the maximum absolute temperature and [tex]T_L[/tex] is the minimum absolute temperature.

We are told that [tex]T_H[/tex] is 1.2 times [tex]T_L[/tex].

COP = 1 / (1.2 - 1).

Simplifying

COP = 5.

We can also express the COP as the ratio of the heat absorbed by the low-temperature reservoir to the net work input.

Using this relationship,

The heat absorbed ([tex]Q_L[/tex]) is equal to 5 times the net work input, which is 5 * 22 kJ

= 110 kJ.

According to the first law of thermodynamics, the heat rejected by the hot reservoir ([tex]Q_H[/tex]) is equal to the sum of the heat absorbed ([tex]Q_L[/tex]) and the net work input (22 kJ).

So,

[tex]Q_H[/tex] = 110 kJ + 22 kJ

= 132 kJ.

Since we are given the mass of the refrigerant (0.96 kg), we can find the heat transfer per unit mass ([tex]q_H[/tex]) using QH/m.

[tex]q_{H}[/tex]= 132 kJ / 0.96 kg

= 137.5 kJ/kg.

Since the refrigerant is in a saturated liquid-vapor mixture state, [tex]q_H[/tex] represents the enthalpy change from liquid to vapor.

By looking up the tables for refrigerant-134a, we can find the corresponding temperature ([tex]T_H[/tex]) at [tex]q_H[/tex] = 137.5 kJ/kg.

After conversion to Kelvin, [tex]T_H[/tex] is found to be 334.3 K.

Using the relationship [tex]T_H[/tex] = 1.2[tex]T_L[/tex],

Dividing [tex]T_H[/tex] by 1.2,  

[tex]T_L[/tex] = 334.3 K / 1.2 =

278.58 K.

Converting this back to Celsius, we get [tex]T_L[/tex] = 5.58 °C.

Finally, by looking up the tables for refrigerant-134a, we can find the minimum pressure ([tex]P_{min}[/tex]) at [tex]T_L[/tex] = 5.58 °C.

The value is determined to be 356.9 kPa.

So, the minimum pressure in the cycle is 356.9 kPa.

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Answer:

Saponification is a process in which soap is formed from mixtures of  sodium or potassium salts of fatty acids. These fatty acids are reacted high temperature of At 80°C-100°C with alkali to extract salt. These alkali can be sodium hydroxide or potassium hydroxide.

Soap has both polar (ionic) and non polar molecules due to which it has characteristics of both hydrophilic substance (having tendency to mix with water) and hydrophobic substance (have tendency to mix with oils) and due to this nature it can act as an emulsifier.

An emulsifier has tendency to diffuse one liquid into another   liquid which is incapable of mixing with homogeneous liquid like water.

Cleansing action takes place due to presence of ionic and non-polar properties at same time, in combination with solubility principles. The ionic end of soap molecule is the salt end. It is hydrophilic (water soluble) in nature. The non-polar end cotains long hydrocarbon chains and is hydrophobic (water repellent).

When immiscible liquids like grease or oil mixed with soap water, non polar end (hydrophobic end) absorbs the dirt which means the soap will form the micelles and trap the dirt in it. As micelles is soluble in water it will remove the dirt with it.

Answer:

it is not possible to place the wires in the condui

Explanation:

given data

total area = 2.04 square inches

wires total area = 0.93 square inches

maximum fill conduit =  40%

to find out

Can it is possible place wire in conduit conduit

solution

we know maximum fill is 40%

so here first we get total area of conduit that will be

total area of conduit = 40% × 2.04

total area of conduit = 0.816 square inches

but this area is less than required area of wire that is 0.93 square inches

so we can say it is not possible to place the wires in the conduit

Conduit fill is a term used in electrical engineering to describe the maximum amount of space that can be occupied by wires in a conduit. In this case, the percentage of fill exceeds the allowed limit, indicating that the wires cannot be accommodated within the given conduit diameter.

Conduit fill is a term used in electrical engineering to describe the maximum amount of space that can be occupied by wires in a conduit. It is important to ensure that the conduit is not overfilled, as this can cause overheating and other safety issues.

In this case, the conduit has a total area of 2.04 square inches and the wires occupy 0.93 square inches. To determine if the maximum fill is exceeded, we can calculate the percentage of fill:

Percentage of fill = (occupied area / total area) * 100

Plugging in the values given:

Calculating the percentage of fill:

Percentage of fill = (0.93 / 2.04) * 100 = 45.59%

Since the calculated percentage of fill (45.59%) is greater than the maximum fill allowed (40%), the 1½" conduit cannot be filled with the given wires within the allowed limit.

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Answer: The answer would depend on the experiment but in most cases zero load is not used to ensure that there is no overflow of current.

Explanation: The internal resistance of any typical wire is minute compared to the resistive components that you are likely to utilize in any given experiment. Using I=V/R (Ohms law) , if the resistance is zero then the current that will flow through the wire would be much too high, causing the wire to burn out. The level of currents that the wire can tolerate are provided by the manufacturers. If you exceed those limits, than the wire will surely get permanently damaged. If you have no load at all than this will be known as a short circuit and might damage the source as well. If you have other components in the circuit than the overflow of current might in effect damage them too. This could also damage the bread board if you are using one.

Answer:

Q1. Why is the percentage elongation in 2 inches greater than in 8 inches?

Q2. Describe the Type of Information which may obtained from a character of fracture

Q3. Explain the difference, on the basis of test result between the ultimate strength and The true stress at fracture, cite your specific value to enhance your explanation.

Q4. What is the purpose of using the spherically seated compression plate?

Q5. Why is the experiment started with an initial load rather than starting at zero load?

Following are the answers to each of the questions

Answer 1.

The larger the radius of an elastic material the lower the elongation because the radius is inversely proportional to the percentage of elongation. A wire with a lower radius has a higher elongation than a material with higher radius  

∈=PL/AE  

Where P is the force applied, L is the original length of the material, A is the cross-sectional area which is πr^2, E is the Elastic Modulus.  

Answer 2.  

A fracture is the separation of material into two or more pieces usually caused by ultimate load or stress. Fracture strength is the stress required for a material to fail.  

TYPES OF FRACTURE  

Brittle fracture: brittle materials are materials that are liable to break easily under load for example cast iron. In brittle fracture no plastic deformation will occur before fracture i.e there are no signs of distress like cracks to notify that a certain material is about to fail.  

In brittle fracture cracks spread rapidly with little or no plastic deformation. These cracks continue to grow once it is initiated  

Ductile fracture: they are material that are capable of being molded in to any shape easily. In ductile fracture a plastic deformation occurs like cracks or rough surface sign will be visible before it attains it ultimate failure state.  

In ductile material, the cracks move slowly and in a gradual process together with plastic deformation. Cracks in ductile fracture will not grow unless there is increase in stress applied on the material  

Answer 3.  

The divergence in the values of true stress and engineering stress occurs only at large loads and displacements; or typically when the specimen is undergoing plastic deformation. That is because most materials have a elastic strain limit close to 0.2%. Note that the values of true stresses and strains are similar to their engineering counterparts below 0.2%.  

If a material that strain hardens is tested in tension, the true stress-true strain curve will keep increasing till the specimen fails as the cross section area keeps decreasing continuously (even during necking). However, the slope of engineering stress-strain curve becomes positive and negative before and after necking respectively . This is because necking reduces the material's ability to harden and hence take larger loads. Hence the load starts dropping but remember that one still measures the stress as the load divided by the original cross section area. If the instantaneous cross section area were taken into account, as was done for true stress, the load drop is compensated by the acute reduction in cross section area.  

In compression, the necking instability does not occur and this difference is not stark. the true stress-strain curve and engineering stress strain curve under compression more-or-less converge (unless specimen 'barrelling' is very prominent).  

Answer 4

They are designed to be placed at the center at a point where loading is exerted in an electromechanical or hydraulic universal test machine. It provides a hardened surface when performing complex compression test in which uniform stress distribution is critical. They can be used to test the variety of materials like concrete, metals, wood and composite.  

Answer 5

From ohms law, if the resistance is zero then the current that will flow through the experiment will be too high causing the wire to burn, since the level of current that the wire can tolerate has been produced by the manufacturer and if it exceed this limit, the wire will damage permanently. If there is no load at all this will be known as short circuit and might damage the source as well.

Explanation:

1. Elongation is the increase in length of a material after it has been stressed within the gauge length e.g. elastic materials like wires, spring and other materials that obeys Hooke's law.  

2. In brittle fracture no plastic deformation will occur before fracture i.e there are no signs of distress like cracks to notify that a certain material is about to fail.  

In brittle fracture cracks spread rapidly with little or no plastic deformation. These cracks continue to grow once it is initiated .

4. Spherically seated compression plate are mounted to the cross head electromechanical or hydraulic universal test machine.

5. The internal resistance of any typical wire is low compared to the resistance of the component that are likely to be used for the experiment.

Answer:

Aggressive behavior

Explanation:

Alcohol consumption tends to cause more Aggressive behavior.

The consumption of alcohol plays a more role in our culture but drinking of too much alcohol can cause drowsiness, vomiting, Upset stomach, slurred speech, heart damage, infertile, numbness lung infections, and many more. Also too much alcohol can cause violence, anger and so on in the society.    

Answer:

Tractive force or traction

Explanation:

The main purpose of the tractive forces is to improve the ability to transform the engine's energy into the vehicle's movement. There are several systems that have different qualities and uses. Here we explain how they work and what they are for.

In a traction vehicle with one of its axles, its ability to transmit engine power to the ground is limited for two reasons:

-  At least one of the wheels must have adhesion, and this as long as it has a self-locking differential. Otherwise, simply with one lacking grip, we can no longer move forward.

- If there are two wheels that must distribute the power, it will always be easier to saturate the traction capacity of the tire than if we divide the force by four. The example is very simple: if we try to drag an object on the ground by pulling a rubber, it will stretch more than if we pull 4 identical tires, although the force we make is the same.

Contrary to what one might think, all-wheel drive vehicles are nothing recent. What happens is that it did take time to reduce the weight and size, as well as to increase the resistance of the homokinetic joints (they are articulations on the axles to allow the wheel to go up and down with the suspension or turn with the steering) to to adapt these systems to cars.

Answer:

C. Decline the invitation and explain to your manager that to do otherwise is inappropriate for a registered professional engineer.

Explanation:NCEES has 3 major rules with some sub sections which helps to uphold the professional conducts of it's members.

The rule that supports this professional conduct is rule II. LICENSEE’S OBLIGATION TO EMPLOYER AND CLIENTS subsection(d)

Subsection d states that a licensee shall not reveal any information about a client,contractor or his employer to a another party except it is required by Law.

Answer:

False

Explanation:

The Engineer of Record for all projects shall be a registered Professional Engineer (P.E.)

Some Detailed Responsibilities of Engineer of Record

The Engineer of record shall attend

All these responsibilties require his physical presence which cannot be achieved through electronic communication.

Answer:

Explanation:

we have to make three functions namely

1 Housekeeping

2 Details

3 end of Job

for variable int

for character string

The hierarchy chart.

As per the chart of the hierarchy, the plan of logic is for making a program that is needed for doing hometown banks and the program determines the checking for an account fee. This includes the account balance and the number of accounts overdrawn. The fee is measured in percentage and each time the account was overdrawn.

Thus the answer is functions include housekeeping, details, and end of job module.

The chart follows the three main parts as housekeeping, details, and job module where the housekeeping is making several deletions, followed by the details and end at the stop.

The details have a outbalance prompt then input balance and then return. The job module has output overdraft then input fee and then output fee and return.

Find out more information about the hierarchy chart.

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Answer:

The problems faced in designing a roadway based on a single design vary. The challenges are described in the explanation section below. The first part of the question "If you wanted to create a vehicle that requires a very low tractive effort in order to overcome resistance forces, what parameter would you change and how would you change it" is also answered in italics below

Explanation:

When you are designing the roadway geometry you commonly use a design vehicle that gives you the worst scenario viable in that road, which is translated in terms of speed, meaning that after all of the topography problems, localisation of essential regions that can´t be modified (urban regions or national parks for example), ambiental effect and economic issues are handled, the horizontal and vertical geometry goes to depend of the speed value that is going to regulate camber, radius of curvature, acceleration and deceleration longitudes and others.

Now, in terms of infrastructure or pavement engineering, after handling climate challenges, the worst scenario is about weight, that has to do with the total equivalent load that is going to resist that pavement till first failure.So you see, that the challenges of using a single design vehicle in the geometry and infrastructure design of a roadway has to do with the selection of the worst scenario possible.

On the other hand, knowing the LOS of the roadway, which is crucial for knowing how is going to be transit management, depends on evaluating the level of service in all kinds of scenarios, that means different types of vehicles speed and acceleration configurations. So using a single vehicle, in this case, is not going to give you the best real knowledge of the road.

For the first question, designing a vehicle with a low tractive effort means that, from the civil engineering point of view, in order to enhance that vehicle, vertical geometry is key for solving resistance forces, because you can design a road that brings much more speed and power to the vehicle. Friction is also a parameter that reduces speed in vehicles but is necessary for a safe travel so you have see if it is logic to reduce this parameter in order to enhance the design vehicle behavior.

If you want to bring more tractive effort to the vehicle; handling resistance forces caused by rolling resistance or air resistance, drawbar pull and other should be key for enhancing the vehicle behavior. If you want a more detailed answer in this topic you should ask a mechanical engineering expert

Answer:

Those products are generally called Work in Process WIP

Explanation:

Work in process (WIP), or work in progress (WIP), goods in process, or in-process inventory in a manufacturing industry/company refer to the company's partially finished goods waiting for completion and eventual sale or the value of these items.

These items are either just being produced or require further processing (like purification, separation, packaging or handling) in a queue or a buffer storage.

Answer:

Explanation:

effective delay = delay when no traffic x [tex]\frac{100}{100- network\r usage}[/tex]

effective delay = [tex]84.1 \times \frac{100}{100-39.3}=138.55024711697ms[/tex]

Answer:

what r u on

Explanation:

Answer:

Explanation:

.......................................................................................................................

Answer:

24.87 mm

Explanation:

The area of the bolt is given by

[tex]A_b=\pi r^{2}[/tex]

Since diameter is 16mm, the radius is 16/2= 8 mm= 0.008 m

Area, [tex]A_b=\pi\times 0.008^{2}=0.000201062 m^{2}[/tex]

For safe design of the bolt, we use stress of 201 Mpa

[tex]\sigma_b=\frac {P}{A_b}[/tex] where P is the load and [tex]\sigma_b[/tex] is normal stress.

Making P the subject then

[tex]P=A_b \sigma_b[/tex]

Substituting the figures given and already calculated area of bolt

[tex]P=201\times 10^{6}\times 0.000201062 m^{2}=40413.4479 N[/tex]

The area of spacer is given by

[tex]\pi (r_o^{2}- r_i^{2})[/tex] where r is radius and the subscripts o and I denote inner and outer respectively

The value of 142 Mpa by default becomes the stress on spacer hence

[tex]\sigma_s=\frac {P}{A_s}[/tex] and making [tex]A_s[/tex] the subject then  

[tex]A_s=\frac {P}{\sigma_s}[/tex]

[tex]\pi (r_o^{2}- r_i^{2})=\frac {P}{\sigma_s}[/tex]

Since we already calculated the value of P as 40413.4479 N and inner radius is 8mm= 0.008 m then  

[tex]\pi (r_o^{2}- 0.008^{2})=\frac {40413.4479}{142\times 10^{6}}[/tex]

[tex]r_o^{2}=0.000154592[/tex]

[tex]r=\sqrt{0.000154592}=0.012433 m[/tex]

Therefore, [tex]d=2r=2*0.012433=0.024867 m=24.86697 mm\approx 24.87 mm[/tex]

Answer:

the volume of the oxygen tank is

V ox = 0.844 m³

the volume of the nitrogen tank is

V ni = 2.359 m³

the final pressure is

P = 255. 534 kPa

Explanation:

taking into account that

n=  m/M

where

n= number of moles , m = mass, M = molecular weight

then

n oxigen = m ox / M ox = 6.0 kg/(32 gr/mol) *1000gr/kg = 187.5 moles

n nitrogen = m ni / M ni = 4.0 kg/(28 gr/mol) *1000gr/kg = 142.857 moles

from the ideal gas law

P*V=n*R*T

where P= absolute pressure, V= volume occupied by the gas , R= ideal gas constant= 8.314 J/(mol K) , T= absolute temperature

V=n*R*T/P

replacing values

for the oxygen tank , T ox= 25°C= 298 K , P = 550 kPa= 550000 Pa ,

V ox =n*R*T/P = 187.5 mol* 8.314 J/(mol K)* 298 K/ 550000 Pa = 0.844 m³

V ox = 0.844 m³

for the nitrogen tank, T ni=  298 K ,  P = 150000 Pa

V ni =n*R*T/P = 142.857 mol* 8.314 J/(mol K)* 298 K/ 150000 Pa = 2.359 m³

V ni = 2.359 m³

when the gases mix , they occupy a volume of

V = V ox + V ni = 0.844 m³ + 2.359 m³ = 3.203 m³

and total number of moles of gas of the mixture is

n = n oxigen + n nitrogen = 187.5 moles + 142.857 moles = 330.357 moles

therefore

P*V=n*R*T

P = n*R*T/V

replacing values

P = n*R*T/V = 330.357 mol*8.314 J/(mol K)* 298 K/ 3.203 m³ *1 kPa/1000Pa = 255. 534 kPa

P = 255. 534 kPa

Answer:

appropriate tunnel pressure is 384.64 ATM

Even at this (high) pressure, exact dynamic similarity is not achieved because the specific heat ratio of helium is approximately 1.66 and is not equal to that of air which is approximately 1.40

Explanation:

It is too long to type here.

Kindly check the attached file for the calculation.

Answer:

Consider the following code.

Explanation:

Code:

Unix Terminal> cat loan_calc.py

#!/usr/local/bin/python3

import locale

from decimal import *

def main():

locale.setlocale(locale.LC_ALL, 'en_US')

print('Monthly Payment Calculator')

while True:

print('DATA ENTRY')

loan_amt = input('Loan amount: ')

loan_amt = float(loan_amt)

int_rate = input('Yearly interest rate: ')

int_rate = float(int_rate)

years = input('Years: ')

years = int(years)

mon_rate = int_rate / 12 / 100

months = years * 12

monthly_pay = loan_amt * mon_rate / ( 1 - 1/(1 + mon_rate) ** months)

monthly_pay = Decimal(monthly_pay).quantize(Decimal('.01'), rounding=ROUND_DOWN)

print()

print('FORMATTED RESULT')

print('Loan amount: %30s' %locale.currency(loan_amt))

print('Yearly interest rate: %20.2f' %int_rate + '%')

print('Number of years: %25d' %years)

print('Montly payment: %25s' %locale.currency(monthly_pay))

print()

print('Continue? (y/n): ')

choice = input().strip()

if choice.lower() == 'n':

break

if __name__=='__main__':

main()

Unix Terminal>

Code output screenshot:

Answer:

89.967°C

Explanation:

According to Stefan-Boltzmann's law,

E = μΩA(Ts⁴ - Ti⁴)

Where,

Emissitivity, μ = 0.6,

Stefan-boltzmann's constant, Ω = 5.67 x 10^-8 W/m²K⁴

Surface area, A = 1.5 x 2.5 x 10-⁴ = 3.75 x 10-⁴ m²

Satellite temperature, Ti = -100°C = 173K

Power dissipated, E = 0.42 W

Since we're evaluating the temperature for both the top and bottom surfaces,

E = 2 x μΩA(Ts⁴ - Ti⁴)

0.42 = 2 x 0.6 x 3.75 x 10-⁴ x 5.67 x 10^-8 x (Ts⁴ - 173⁴)

0.0165 x 10^-12 = Ts⁴ - 8.957 x 10^8

Ts⁴ = 173.57 x 10^8

Taking 4th root of both sides,

Ts = 362.97 K

Ts = (362.97 - 273)°C

Ts = 89.967°C

Answer

The answer and procedures of the exercise are attached in the following archives.

Step-by-step explanation:

You will find the procedures, formulas or necessary explanations in the archive attached below. If you have any question ask and I will aclare your doubts kindly.  

Answer

The answer and procedures of the exercise are attached in the following archives.

Step-by-step explanation:

You will find the procedures, formulas or necessary explanations in the archive attached below. If you have any question ask and I will aclare your doubts kindly.  

Answer:b

Explanation:

A beam is subjected to an upward load in the middle and supported in both ends. The upper fibers i.e. a topmost layer of the beam is in tension while bottom fiber i.e. fiber of bottom layer is in compression because they are under compressive force.

Fibers above the neutral axis are in tension and fibers below the neutral axis are in compression.    

So option b is correct              

Answer:FALSE

Explanation:

Answer:

The layering and mixing effect of ice cubes in salt water and fresh water respectively.

Explanation:

This test helps to explain the melting of ice cubes which is more obvious when colored ice cubes are used, as it reveals the layering and mixing effect of ice cubes in salt water and fresh water respectively.

The mixing effect shows that ice cubes melt faster in fresh water (low density) than in salt water(high density).