(Process scores in a text file)
Suppose that a text file contains an unspecified number of scores. Write a program that prompts the user to enter the filename and reads the scores from the file and displays their total and average. Scores are separated by blanks. Your program should prompt the user to enter a filename.
Sample Run
Enter a filename: scores1.txt
There are 24 scores
The total is 800
The average is 33.33.
In Python.

Answer:

An engineer using a decision matrix to determine which one of his/her solution ideas meets the established criteria and constraints.

Answer:

SECTION LEARNING OBJECTIVES

By the end of this section, you will be able to do the following:

Distinguish between static friction and kinetic friction

Solve problems involving inclined planes

Section Key Terms

kinetic friction static friction

Static Friction and Kinetic Friction

Recall from the previous chapter that friction is a force that opposes motion, and is around us all the time. Friction allows us to move, which you have discovered if you have ever tried to walk on ice.

There are different types of friction—kinetic and static. Kinetic friction acts on an object in motion, while static friction acts on an object or system at rest. The maximum static friction is usually greater than the kinetic friction between the objects.

Imagine, for example, trying to slide a heavy crate across a concrete floor. You may push harder and harder on the crate and not move it at all. This means that the static friction responds to what you do—it increases to be equal to and in the opposite direction of your push. But if you finally push hard enough, the crate seems to slip suddenly and starts to move. Once in motion, it is easier to keep it in motion than it was to get it started because the kinetic friction force is less than the static friction force. If you were to add mass to the crate, (for example, by placing a box on top of it) you would need to push even harder to get it started and also to keep it moving. If, on the other hand, you oiled the concrete you would find it easier to get the crate started and keep it going.

Figure 5.33 shows how friction occurs at the interface between two objects. Magnifying these surfaces shows that they are rough on the microscopic level. So when you push to get an object moving (in this case, a crate), you must raise the object until it can skip along with just the tips of the surface hitting, break off the points, or do both. The harder the surfaces are pushed together (such as if another box is placed on the crate), the more force is needed to move them.

Answer:

  both are

Explanation:

It depends on what the symptoms of the "basic circuit problem" are. Both overvoltage and shorts are the kinds of things that can cause circuit damage, and either can be the cause of the other.

Answer:

13.9357 horse power

Explanation:

Annealed copper

Given :

Width, b = 9 inches

Thickness, \($h_0=2.2$\) inches

K= 90,000 Psi

μ = 0.2, R = 14 inches, N = 150 rpm

For the maximum possible draft in one pass,

\($\Delta h = H_0-h_f=\mu^2R$\)

     \($=0.2^2 \times 14 = 0.56$\) inches

\($h_f = 2.2 - 0.56$\)

     = 1.64 inches

Roll strip contact length (L) = \($\sqrt{R(h_0-h_f)}$\)

                                             \($=\sqrt{14 \times 0.56}$\)

                                             = 2.8 inches

Absolute value of true strain, \($\epsilon_T$\)

\($\epsilon_T=\ln \left(\frac{2.2}{1.64}\right) = 0.2937$\)

Average true stress, \($\overline{\gamma}=\frac{K\sum_f}{1+n}= 31305.56$\) Psi

Roll force, \($L \times b \times \overline{\gamma} = 2.8 \times 9 \times 31305.56$\)

                                 = 788,900 lb

For SI units,

Power = \($\frac{2 \pi FLN}{60}$\)  

           \($=\frac{2 \pi 788900\times 2.8\times 150}{60\times 44.25\times 12}$\)

           = 10399.81168 W

Horse power = 13.9357

Answer:

Weight Percentage of Ash = 3.4

Explanation:

Given - Coal containing 21% ash is completely combusted, and the ash is 100% removed in a water contact scrubber. If 10,000 kg of coal is burned per hour with a scrubber flow rate of 1.0 m3/min.

To find - the weight percentage of the ash in the water/ash stream leaving the scrubber is most nearly ?

Solution -

Given that,

Coal Burned Rate = 10,000 kg/hr

                              = \(\frac{10,000}{60 min} * 1 hr *\frac{kg}{hr}\)

                              = 166.6666 kg/min

⇒Coal Burned Rate = 166.6666 kg/min

Now,

Given that,

Ash content in coal = 21 %

⇒Ash in (coal that burned) = 166.6666 × \(\frac{21}{100}\) kg/min

                                             = 34.9999 ≈ 35 kg/min

⇒Ash in (coal that burned) = 35 kg/min

Now,

We know,

Density of water = 1000 kg/m³

Now,

Water flow Rate = \(1\frac{m^{3} }{min} * density\)

                           = 1000 kg/min

⇒Water flow Rate = 1000 kg/min

Now,

Total Mass flow Rate of (Water + Ash stream) = ( 1000 + 35) kg/min

                                                                           = 1035 kg/min

⇒Total Mass flow Rate of (Water + Ash stream) = 1035 kg/min

So,

Weight Percentage of Ash = (Weight of Ash ÷ Total weight of Stream) × 100

                                            = (35 ÷ 1035) × 100

                                            = 3.38 ≈ 3.4

∴ we get

Weight Percentage of Ash = 3.4

Answer:

1. TURN ON YOUR HAZARD/EMERGENCY LIGHTS

Turn on your hazard lights to warn other drivers as soon as you sense something's wrong. Keep them on until help arrives, recommends the National Motorists Association (NMA).

2. SLOW DOWN AND PULL OFF THE ROAD

Aim for the right shoulder of the road. Consumer reports recommends that you pull over to a safe, flat location that is as far away from moving traffic as possible.

3. TURN YOUR WHEELS AWAY FROM THE ROAD AND PUT ON THE EMERGENCY BRAKE

The California Department of Motor Vehicles (DMV) recommends pulling your emergency brake, sometimes called the parking brake. If you have to park on a hill or slope, turn the car's wheels away from the road to help prevent the care from rolling into traffic, says the California DMV.

4. STAY IN YOUR VEHICLE

If you're on a highway or crowded road, the Insurance Information Institute (III) recommends that you avoid getting out of your vehicle to look at the damage or fix a mechanical problem. If you need to get out of the car, get your vehicle to a safe place and make sure the road around you is completely clear. If you're stopped on the right-hand side of the road, get out through the passenger-side door.

5. BE VISIBLE

Once you're safely out of the vehicle, prop up your hood to let other drivers know they should proceed with caution. This will alert other drivers that you're broken down, according to the NMA.

6. SET UP FLARES OR TRIANGLES

Place flares or triangles with reflectors behind your car to alert other drivers to the location where you've stopped, says the III.

7. CALL FOR HELP

Call or use an app to get a tow truck, mechanic or roadside assistance to come help. your insurance company or other provider who may be able to help. If you're in an emergency situation or are not sure who to contact, call 911 or the local police for help.

Hope this helps :)

Slide Pins as the fixed caliper has components that are consider all of the following, excluding slide pins.

Caliper mountings is important to follow proper safety procedures when it is dealing with hazardous materials such as the asbestos dust. This may be include using the protective equipment, such as the gloves and the respirators, as well as following specific cleaning procedures to be minimize the risk of the exposure.

The hazardous dust has been properly dealt with it as it may be necessary to use a brake or the cleaning solvent to further clean the caliper mountings and slides or the pins. It has been important to follow the manufacturer's instructions and just to use caution when it is using any cleaning solvent.

Technician would says that one should clean the caliper mountings and slides or the pins using the equipment which is used in the procedures for dealing with asbestos or the hazardous dust. Technician b would says that once the dust has been taken care of one may need to use a brake cleaning solvent to clean the components further.

Therefore, Slide Pins as the fixed caliper has components that are consider all of the following, excluding slide pins.

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To prove the relationship between the mean coefficient of friction (μ) and the shear angle (x), given that the frictional force (F) on the tool rake face is equal to KtA, where K is a constant, A is the area of the cross-section of the chip, and a is the rake angle, we can follow these steps:

Step 1: Express the frictional force in terms of the mean coefficient of friction:

The frictional force, F, is equal to the product of the mean coefficient of friction (μ) and the normal force (N), which is equal to KtA:

F = μN = μKtA.

Step 2: Calculate the normal force, N:

The normal force can be determined using trigonometry. Since a is the rake angle, the component of the normal force in the direction of the shear force is N cos(x - a).

Step 3: Equate the frictional force with the calculated normal force:

Setting F = N cos(x - a), we get:

μKtA = N cos(x - a).

Step 4: Substitute the expression for the normal force:

μKtA = (N cos(x - a)).

Step 5: Rearrange the equation to solve for μ:

Divide both sides by N cos(x - a):

μ = KtA / (N cos(x - a)).

Step 6: Express N in terms of K and A:

Using trigonometry, we find that N = K sin(x - a).

Step 7: Substitute N into the equation:

μ = KtA / ((K sin(x - a)) cos(x - a)).

Step 8: Simplify the expression:

μ = K cos^2(x - a) / (K sin(x - a) cos(x - a) + 1).

Therefore, we have derived the relationship between the mean coefficient of friction (μ) and the shear angle (x) as μ = K cos^2(x - a) / (K sin(x - a) cos(x - a) + 1), where K is a constant, A is the area of the cross-section of the chip, and a is the rake angle.

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Answer:

They are configured this way to avoid overlap issues.

Explanation:

The concentration ratio for the PTC is approximately 1.48, and the length of the parabolic surface is approximately 5.2 meters.

To determine the concentration ratio and length of the parabolic surface for a Parabolic Trough Collector (PTC) with the given parameters, we can use the following formulas:

Concentration Ratio (CR) = Rim Angle / Aperture Angle

Length of Parabolic Surface (L) = Aperture^{2} / (16 * Focal Length)

First, let's calculate the concentration ratio:

Given:

Rim Angle (θ) = 80º

Aperture Angle (α) = 5.2 m

Concentration Ratio (CR) = 80º / 5.2 m

Converting the rim angle from degrees to radians:

θ_rad = 80º * (π / 180º)

CR = θ_rad / α

Next, let's calculate the length of the parabolic surface:

Given:

Aperture (A) = 5.2 m

Receiver Diameter (D) = 50 mm = 0.05 m

Focal Length (F) = A^{2} / (16 * D)

L = A^{2} / (16 * F)

Now we can substitute the given values into the formulas:

CR =\((80º * (π / 180º)) / 5.2 m\)

L = \((5.2 m)^2 / (16 * (5.2 m)^2 / (16 * 0.05 m))\)

Simplifying the equations:

CR ≈ 1.48

L ≈ 5.2 m

Therefore, the concentration ratio for the PTC is approximately 1.48, and the length of the parabolic surface is approximately 5.2 meters.

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On freezing, wet days, roadways on bridges and overpasses are most likely to hide spots of ice. Option 2 is the answer.

Roadways on bridges and overpasses are most likely to hide spots of ice because these elevated structures are exposed to cold air from both above and below, causing them to cool more rapidly and retain colder temperatures.

As a result, any moisture on the surface of the bridge or overpass is more likely to freeze, creating hazardous icy patches that may not be immediately visible to drivers. It's important to exercise caution when driving on bridges and overpasses during such conditions and adjust your speed accordingly.

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Answer:

Explanation:

For first isobaric process (expansion with constant pressure):

\(W_{1}=P_{1}(V_{2}-V_{1})=(0.43\times 10^{6})(2.9-0.28)=1.1266 \times 10^{6} J=1.1266 MJ(+) (Expansion)\)

Now, for the first isochoric: \(W_{2}=0\), and \(Q_{2}=\Delta U=U_{3}-U_{2}\)

For the second isobaric process (compression with constant pressure):

\(W_{3}=P_{2}(V_{1}-V_{2})=(4.3\times 10^{6})(0.28-2.9) =-11.266\times 10^{6}J=-11.266 MJ (Compression)\)

For the last isochoric: \(W_{4}=0, Q_{4}=\Delta U=U_{4}-U_{3}\)

So, the total work per cycle: \(W=W_{1}+W_{2}+W_{3}+W_{4}=1.1266+0-11.266+0=-10.1394 MJ\)

Answer:

a) the mass flow rate of the steam is  \(\mathbf{m_1 =6.92 \ kg/s}\)

b) the exit velocity of the steam  is \(\mathbf{V_2 = 562.7 \ m/s}\)

c) the exit area of the nozzle is  \(A_2\) = 0.0015435 m²

Explanation:

Given that:

A steam with 5 MPa and 400° C enters a nozzle steadily

So;

Inlet:

\(P_1 =\) 5 MPa

\(T_1\) = 400° C

Velocity V = 80 m/s

Exit:

\(P_2 =\) 2 MPa

\(T_2\) = 300° C

From the properties of steam tables  at \(P_1 =\) 5 MPa and \(T_1\) = 400° C we obtain the following properties for enthalpy h and the speed v

\(h_1 = 3196.7 \ kJ/kg \\ \\ v_1 = 0.057838 \ m^3/kg\)

From the properties of steam tables  at \(P_2 =\) 2 MPa and \(T_1\) = 300° C we obtain the following properties for enthalpy h and the speed v

\(h_2 = 3024.2 \ kJ/kg \\ \\ v_2= 0.12551 \ m^3/kg\)

Inlet Area of the nozzle = 50 cm²

Heat lost Q = 120 kJ/s

We are to determine the following:

a) the mass flow rate of the steam.

From the system in a steady flow state;

\(m_1=m_2=m_3\)

Thus

\(m_1 =\dfrac{V_1 \times A_1}{v_1}\)

\(m_1 =\dfrac{80 \ m/s \times 50 \times 10 ^{-4} \ m^2}{0.057838 \ m^3/kg}\)

\(m_1 =\dfrac{0.4 }{0.057838 }\)

\(\mathbf{m_1 =6.92 \ kg/s}\)

b) the exit velocity of the steam.

Using Energy Balance equation:

\(\Delta E _{system} = E_{in}-E_{out}\)

In a steady flow process;

\(\Delta E _{system} = 0\)

\(E_{in} = E_{out}\)

\(m(h_1 + \dfrac{V_1^2}{2})\) \(= Q_{out} + m (h_2 + \dfrac{V_2^2}{2})\)

\(- Q_{out} = m (h_2 - h_1 + \dfrac{V_2^2-V^2_1}{2})\)

\(- 120 kJ/s = 6.92 \ kg/s (3024.2 -3196.7 + \dfrac{V_2^2- 80 m/s^2}{2}) \times (\dfrac{1 \ kJ/kg}{1000 \ m^2/s^2})\)

\(- 120 kJ/s = 6.92 \ kg/s (-172.5 + \dfrac{V_2^2- 80 m/s^2}{2}) \times (\dfrac{1 \ kJ/kg}{1000 \ m^2/s^2})\)

\(- 120 kJ/s = (-1193.7 \ kg/s + 6.92\ kg/s ( \dfrac{V_2^2- 80 m/s^2}{2}) \times (\dfrac{1 \ kJ/kg}{1000 \ m^2/s^2})\)

\(V_2^2 = 316631.29 \ m/s\)

\(V_2 = \sqrt{316631.29 \ m/s\)

\(\mathbf{V_2 = 562.7 \ m/s}\)

c) the exit area of the nozzle.

The exit of the nozzle can be determined by using the expression:

\(m = \dfrac{V_2A_2}{v_2}\)

making \(A_2\) the subject of the formula ; we have:

\(A_2 = \dfrac{ m \times v_2}{V_2}\)

\(A_2 = \dfrac{ 6.92 \times 0.12551}{562.7}\)

\(A_2\) = 0.0015435 m²

No, a lift truck with a 6000 lb capacity at a 24" load center cannot lift a 6000 lb load positioned at 60".

The load center refers to the horizontal distance from the vertical face of the forks to the center of gravity of the load being lifted. In this case, the load center is 24".

To determine if the lift truck can handle a particular load, you need to consider the load weight and its distance from the load center. The load moment (weight x distance) should not exceed the truck's maximum load moment capacity.

In this scenario, the load moment would be 6000 lb x 60" = 360,000 lb-inches. However, since the lift truck has a 6000 lb capacity at a 24" load center, its load moment capacity would be 6000 lb x 24" = 144,000 lb-inches.

As the load moment (360,000 lb-inches) exceeds the load moment capacity (144,000 lb-inches) of the lift truck, it would not be able to safely lift the 6000 lb load positioned at 60". It would exceed the lifting capabilities of the lift truck, posing a risk of instability and potential accidents.

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The seg7 .v, fa .v, and add4bit .v modules will be designed to convert a seven-bit binary number into its seven-segment display representation, implement a 4-bit full-adder, and implement a 4-bit adder, respectively. Details not specified will be chosen and explained in the Lab report.

To complete a preliminary design of your seg7. v, fa.v, and add4bit .v (a draft of your Verilog code), you need to consider the purpose of each module. The seg7 .v module will convert a seven-bit binary number into its seven-segment display representation, the fa.v module will implement a 4-bit full-adder, and the add4bit .v module will implement a 4-bit adder. Any details not specified in this Lab should be chosen by you and plainly explained in your Lab report.

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Answer:

1. All engines  exhibit wear over time ⇒ Synthetic Improves engine protection by resisting oil breakdown.

Synthetic oil does not breakdown so easily which means that it protects the engine more and protects it from wearing.

2. Engines are cold at start-up and not while running ⇒ Synthetic provides maximum protection in extreme hot and cold temperature conditions.

By providing protection for the engine during cold and hot conditions, the engine will not be too cold when the car is started up.

3.  If oil is thicker, engines lose power and efficiency ⇒ Synthetic has greater resistance to oil thickening to maintain engine efficiency.

Part of the characteristics of synthetic oil is that it does not get as thick as regular oil which means that the adverse effects of thick oil are spared on the engine.

Using the knowledge in computational language in python it is possible to write a code that mouse is pressed, a circle is created with a fill matching the color at the current index in app.colorlist.

import numpy as np

import argparse

import cv2

# construct the argument parse and parse the arguments

ap = argparse.ArgumentParser()

ap.add_argument("-i", "--image", help = "path to the image")

args = vars(ap.parse_args())

# load the image

image = cv2.imread(args["image"])

boundaries = [

([17, 15, 100], [50, 56, 200]),

([86, 31, 4], [220, 88, 50]),

([25, 146, 190], [62, 174, 250]),

([103, 86, 65], [145, 133, 128])

]

for (lower, upper) in boundaries:

# create NumPy arrays from the boundaries

lower = np.array(lower, dtype = "uint8")

upper = np.array(upper, dtype = "uint8")

# find the colors within the specified boundaries and apply

# the mask

mask = cv2.inRange(image, lower, upper)

output = cv2.bitwise_and(image, image, mask = mask)

# show the images

cv2.imshow("images", np.hstack([image, output]))

cv2.waitKey(0)

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Answer: The process of data analysis uses analytical and logical reasoning to gain information from the data. The main purpose of data analysis is to find meaning in data so that the derived knowledge can be used to make informed decisions.

The true power is obtained as 70.7 VA.

We define the power a the rate of doing work, we know that the power in a circuit is the product of the current and the voltage. In this case, we want to find the true power thus we have to involve the use of the phase shift in degrees.

Thus;

True power = PcosΦ

P =  100 VA

Φ = 45 degrees

True power =  100 VA * cos 45 degrees

True power = 70.7 VA

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Answer:

Explanation:

Automotive Engineering is one of the most challenging and comprehensive fields of engineering today. The automotive sector has come a long way since the age of steam powered flat engines to the age of fast racing Formula 1 car, light and heavy commercial vehicles with lots of technological developments being done every day. “A good scientist is a person with original ideas. A good engineer is a person who makes a design that works with as few original ideas as possible”.

Engineering has been my passion from the start. I took the first step to become an engineer by securing 93.2% in SSC and 82.8% in HSC. I also secured 8259th rank among 10 lakhs students in AIEEE and 6979th rank among 4.7 lakhs in JEE in 2010 so that I could get a seat in one…show more content…

Additionally, Germany avails the advantages of securing the best of knowledge, quality of education, and above all the international exposure. The quality of German Mechanical Engineering has always been much vaunted. The research carried out on structural design of vehicles, alternative and electrified vehicle propulsion at the University is appealing. The breadth and depth of the courses coupled with a stimulating research environment seem to me the right mix for seminal work and pioneering research. I believe that with variety of courses offered and with highly knowledgeable faculty and excellent facilities, the University will provide a perfect environment to focus all my resources towards my goal.

In summary, I am fully committed to a career in automotive sector. My long term goal is to pursue my research interests in the field of automobiles and strive to emerge as an active contributor to the field in the World in general and India in particular.  Hence I am eagerly looking forward to spending the next few years in a structured Master’s program in Automotive Engineering. The Mechanical Department of RWTH Aachen has World renowned Faculty besides academically brilliant and motivated graduates in whose association my understanding of the area will be much.

Answer:

What force is requiere to move the cylinder along

Explanation:

The kinematic viscosity of the ol is 0.006 fr2

During normal operation, refrigerant pumped into a condenser is in the form of a gas.

The process of maintaining the temperature of the desired space lower than that of the surroundings is called refrigeration. In the process of refrigeration, the chemical used is called refrigerant.

In the cycle of refrigeration, the liquid form of the refrigerant took heat from the desired space in the evaporator and is taken by the compressor where it is compressed at a high temperature and pressure greater than that of an atmosphere.

The refrigerant now is pumped to the condenser in the form of high-pressure and high-temperature gas. Here it loses its heat to the atmosphere and condenses in the form of liquid.

Therefore, during normal operation, refrigerant pumped into a condenser is in the form of a gas.

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If the diameter of the bar is 0.490 at this load, determine I. the engineering stress and strain, and [2] II. the true stress and strain is 1561. 84 MPa.

Strain is a unitless degree of ways a great deal an item receives larger or smaller from an implemented load. Normal stress happens while the elongation of an item is in reaction to an everyday pressure (i.e. perpendicular to a surface), and is denoted via way of means of the Greek letter epsilon.

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When a section of land is 19.83ch x 19.09ch x 19.31ch x 20.14ch. The area of the lot is most nearly D. 37.38ac

In order to calculate the area of the lot, we need to multiply the length and width of the lot. In this case, the length is 19.83ch and the width is 19.09ch.

Therefore, the area of the lot is 19.83ch x 19.09ch

= 373.8347 square chains.

To convert square chains to acres, we divide by 10. Therefore, the area of the lot in acres is:

= 373.8347 square chains / 10

= 37.38347 acres.

The answer is D.

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Answer:

Zero

Explanation:

For this, it is necessary to use the  the Hagen-Poiseuille equation, which is a law of physics that describes an incompressible fluid of low viscosity through a tube of constant circular cross-section.

By derivating the Pressure over x it becomes zero

\(\frac{dP}{dx} = 0\)

the torque is required to rotate the inner cylinder at 12 r min, the outer cylinder remaining stationary is 11.0807 N-m.

we have left out some other important details, such as the time required to reach that rpm, whether the beginning state is at rest, and the axis around which the cylinder is revolving. However, we'll suppose that it will be 60 seconds (time is necessary to convert rpm to angular acceleration)

rotation around the center axis

Angular acceleration = torque + MOI

Currently, angular acceleration equals 2 rpm/(t 60).

In order to avoid getting an extremely high figure, we estimated that the diameter of the cylinder was 600 mm rather than 600 meters: angular acceleration= 1.047 rad/s² MOI for cylinder across center dia= 1/4MR²+ 1/12ML²

MOI= 10.5833 kg-m

Now, the product of these two is torque.

11.0807 N-m of torque

Various MOI equations can be used to compute for different axes of rotation.

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The complete question is:

Crude oil at 20 c fills the space between two concentric cylinders 250 mm high and with diameters of 150 mm and 156 mm. Find the torque is required to rotate the inner cylinder at 12 r min, the outer cylinder remaining stationary.

Answer:

The answer is "\(\bold{e^{-t} \sin t}\)"

Explanation:

Given equation:

\(\to x+2x+2x=0.............(a)\)

Let x= e^{mt} be a solution of the equation will be:

\(\to m^2+2m+2=0\)

compare the value with the  \(am^2+bm+c=0\)

\(\to a= 1\\\to b= -2\\\to c= 2\\\)

Formula:

\(\bold{=\frac{-b \pm \sqrt{b^2-4ac}}{2a}}\\\\=\frac{-2 \pm \sqrt{(-2)^2-4\times 1 \times 2}}{2 \times 1}\\\\=\frac{-2 \pm \sqrt{4-8}}{2}\\\\=\frac{-2 \pm \sqrt{-4}}{2}\\\\=\frac{-2 \pm 2i}{2}\\\\\)

Calculate the g.s at equation (a):

\(\to x = e^{-t} (c_1 \cos t+ c_2 \sin t)............(b)\\\\\to x = e^{-t} (c_2 \cos t -c_1 \sin t) - e^{-t} (c_1 \cos t + c_2\sin t) ...............(c)\\\\\)

\(_{when} \\\\\to t= 0\\\\\to x=x_0\\\\\to x= v_0 =1 \\\\ \ then \ form \ equation \ (b) \ and \ equation \ (c)\)

\(\to a= 0 \\ \to c_2 -c_1 =1 \\ \to c_2=1\)

The value of \(x = e^{-t} ( 0 \times \cos t + 1 \times \sin t)\)

\(\boxed{\bold{x = e^{-t} (\sin t)}}\)

The load resistance R0 is approximately 645.95 ohms, and the current I0 delivered by the ideal battery is approximately 0.536 A.

In the first step, we are given the power range of the refrigerator load (100-250 W) and the chosen power P0 of 185 W. Since the load is connected in a one-loop circuit, the current flowing through each part of the circuit remains the same.

In the second step, an ideal battery with no internal resistance is connected to the load. We need to calculate the load resistance R0 and the current I0 delivered by the battery. The power received by the load is given as P0 = 185 W.

To calculate R0, we can use the formula P0 = I0^2 * R0, where I0 is the current and R0 is the resistance. Rearranging the formula, we have R0 = P0 / I0^2. Plugging in the values, we get R0 = 185 / (0.536^2) ≈ 645.95 ohms.

For the current I0, we can use Ohm's Law, which states that I0 = ℇ0 / R0, where ℇ0 is the emf (electromotive force) of the battery. Given ℇ0 = 345 V and R0 ≈ 645.95 ohms, we can calculate I0 as I0 = 345 / 645.95 ≈ 0.536 A.

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Answer:

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Explanation: