4. Write the command to import the coolfunc() function from the neatomod module in the funpack package and rename it to coolf. Now write the command to import all functions from the neatomod module. What command can we use to obtain a list of all functions inside the module?

Answer:

A. -0.003

B. 0.02

Explanation:

Step 1: identify the given parameters

Giving the following parameters

Wing area (S)= 1.5 m²

Wing chord (C) = 0.45 m

Velocity (V) = 100 m/s

moment about center of gravity(Mcg) = -12.4 N-m

at another angle of attack, L = 3675 N and Mcg = 20.67 N-m

Step 2: calculate the value of the moment coefficient about the aerodynamic center (Cmcg)

[tex]q_{∞} =\frac{1}{2}\rho*v^{2}[/tex]

[tex]q_{∞} =\frac{1}{2}1.225*100^{2}[/tex]= 6125 N/m²

[tex]C_{mcg,w} =\frac{M_{cg,w} }{q_{∞}*S*C }[/tex]

[tex]C_{mcg,w} =\frac{-12.4}{6125*1.5*0.45 }[/tex] = -0.003  

[tex]C_{mcg,w}= C_{ac,w}= -0.003[/tex] at zero lift

Step 3: calculate coefficient of lift

Cl = L/q*s

Cl = 3675/6125*1.5 = 0.4

Step 4: calculate the location of the aerodynamic center

New moment coefficient about the aerodynamic center (Cmcg):

[tex]C_{mcg} =\frac{20.67}{6125*1.5*0.45}[/tex] = 0.005

[tex]C_{mcg,w} = C_{ac} ,w + C_{l}(h-h_{ac})[/tex]

[tex]h-h_{ac}= \frac{C_{mcg,w} -C_{ac,w}}{C_{l} }[/tex]

[tex]h-h_{ac}= \frac{0.005-(-0.003)}{0.4}[/tex]=0.02

the location of the aerodynamic center = 0.02

Answer:

7.65 mm

Explanation:

Stress, [tex]\sigma=\frac {F}{A}[/tex] where F is the force and A is the area

Also, [tex]\sigma=E\times \frac {\triangle L}{L}[/tex]

Where E is Young’s modulus, L is the length and [tex]\triangle L[/tex] is the elongation

Therefore,

[tex]\frac {F}{A}= E\times \frac {\triangle L}{L}[/tex]

Making A the subject of the formula then

[tex]A=\frac {FL}{E\triangle L}=\frac {6660\times 380}{110\times 10^{9}\times 0.5}=4.60145\times 10^{-5} m^{2}[/tex]

Since [tex]A=\frac {\pi d^{2}}{4}[/tex] then  

[tex]d=\sqrt {\frac {4A}{\pi}}=\sqrt {\frac {4\times 4.60145\times 10^{-5}}{\pi}}= 0.00765425m= 7.654249728 mm\approx 7.65 mm[/tex]

Answer:

Efficient when the marginal benefits of project = marginal costs of project.

Explanation:

Majority Decision Rule:

Majority decision rule is based on the notion of equality. An alternative is selected which has majority of votes. The simple majority decision rule may generate efficient results if the marginal benefits of a project are equal or greater than the marginal costs of the project.

Explanation:

It may be instructive to look at the opposite of the sentence here. Perhaps the smarter creature would be more unhappy when frustrated, recognizing how it gains from happiness relative to the creature with less experience and less knowledge of a situation that does not define it at the moment.  

Perhaps the argument is really about the fact that wisdom helps one to hypothetically live in multiple states and a lack of wisdom prevents or fails this possibility.

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.  

The Rankine passive earth pressure on a 12-ft high retaining wall with backfill of granular soil having an internal angle of friction of 30° and a unit weight of 125 pcf is calculated to be 500 psf.

The question relates to calculating the Rankine passive earth pressure on a retaining wall that is 12-ft high with backfill of granular soil. The internal angle of friction (φ) provided is 30°, and the unit weight (γ) of soil is 125 pcf (pounds per cubic foot). First, to calculate the passive earth pressure (Ψp), we use the Rankine theory formula: Ψp = γh [1 - sin(φ)]/[1 + sin(φ)], where h is the height of the wall. Substituting the given values, Ψp = 125 * 12 * [1 - sin(30°)]/[1 + sin(30°)]. Since sin(30°) = 0.5, the calculation simplifies to: Ψp = 125 * 12 * [1 - 0.5]/[1 + 0.5], which further simplifies to Ψp = 125 * 12 * 0.5/1.5. Therefore, the Rankine passive earth pressure on the wall amounts to Ψp = 500 psf (pounds per square foot).

Answer:

Exfoliation.

Explanation:

Exfoliation is a form of chemical weathering that occurs when sheets of rocks peel from a massive rock's surface. It is also the unloading and sheeting of stress in a rock that expands the magma chamber.

Answer:

a) power = 929.78W

b) energy = 3,347,200J

c) 5 milkyway bars

Explanation:

a) Power is the rate of use of energy per unit time. This is given in the question as 800 kilocalories per hour. Part a) requires the power in Watts (W) which is equivalent to Joules per second. Thus, kilocalories per hour need to be converted to Joules per second:

[tex]1kcal = 4184J[/tex]

[tex]1hour = 3600s[/tex]

[tex]power=\frac{energy}{time}[/tex]

[tex]power=\frac{800kcal*\frac{4184J}{kcal}}{1hour*\frac{3600s}{hour}}[/tex]

[tex]power=\frac{3347200}{3600}[/tex]

[tex]power=929.78J/s=929.78W[/tex]

b) Total time required for a 15km run can be calculated by the speed of the runner (15km/h):

[tex]time=\frac{distance}{speed}[/tex]

[tex]time=\frac{15km}{15km/h}[/tex]

[tex]time=1hour=3600s[/tex]

The energy exerted over this time can be found by:

[tex]energy=power*time[/tex]

[tex]energy=929.78J/s*3600s[/tex]

[tex]energy=3,347,200J[/tex]

c) Total time required for a 13.1mile run can be calculated by the speed of the runner (15km/h):

[tex]1mile=1.6km[/tex]

[tex]time=\frac{distance}{speed}[/tex]

[tex]time=\frac{13.1mile*\frac{1.6km}{mile}}{15km/h}[/tex]

[tex]time=\frac{20.96}{15}[/tex]

[tex]time=1.40h=5030.4s[/tex]

The energy exerted over this time can be found by:

[tex]energy=power*time[/tex]

[tex]energy=929.78J/s*5030.4s[/tex]

[tex]energy=4,677,165J[/tex]

Assuming one Milky Way (Original Single 52.2g) has 240 kilocalories

[tex]number=\frac{energy}{energy_{milkyway}}[/tex]

[tex]number=\frac{4677165}{240kcal*\frac{4184J}{kcal}}[/tex]

[tex]number=\frac{4677165}{1004160}[/tex]

[tex]number=4.65[/tex]

5 milkyway bars are needed

Answer:

The distinction between real variables and nominal variables is known as inflation rate.

Explanation:

The inflation rate is what distinguishes real variables (such as increase or decrease in prices/price level of goods or services) from nominal variables (such as the quantity of available money: high or low money supply). Real variables, which are affected by nominal variables, are actually nominal variables that have been adjusted for inflation.

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:

Explanation:

As a security  professional, I will respond positively to the OSHA requirements overlap. OSHA guidelines are meant to provide general guidance to all members of various entities throughout the country, while local or state codes also ensure compliance with laws unique to their areas, taking into account workplace safety and security.

OSHA accepts the security codes of the state building To the degree that such codes comply with OSHA regulations, such as BOCA. All the codes and regulations for local, state-owned construction, electrical and life protection are under the same umbrella. Generally, all security protocols and specifications are in accordance with OSHA guidelines. Nonetheless, certain points will overlap, while localized codes will also be addressed to a particular community or state that may

Answer:

The temperature of the first exit (feed to water heater) is at 330.15ºC. The second exit (exit of the turbine) is at 141ºC. The turbine Power output (if efficiency is %100) is 3165.46 KW

Explanation:

If we are talking of a steam turbine, the work done by the steam is done in an adiabatic process. To determine the temperature of the 2 exits, we have to find at which temperature of the steam with 1000KPa and 200KPa we have the same entropy of the steam entrance.

In this case for steam at 3000 kPa, 500°C, s= 7.2345Kj/kg K. i=3456.18 KJ/Kg

For steam at 1000 kPa and s= 7.2345Kj/kg K → T= 330.15ºC i=3116.48KJ/Kg

For steam at 200 kPa and s= 7.2345Kj/kg K → T= 141ºC i=2749.74KJ/Kg

For the power output, we have to multiply the steam flow with the enthalpic jump.

The addition of the 2 jumps is the total power output.

Answer:

New slip = 0.145

Explanation:

Revolutions per minute rpm = 1996

Synchronous speed  Ns= 120 * f / p

= 120 * 75 / 4 = 2250

Hence initial slip S1 = 2250 - 1996 / 2250 = 0.113

From output power = [tex] 3*S*E^2*R2 [/tex]

Hence at constant E and R, output power depends on slip

So for new power and new slip to new initial power and initial slip,

Hence 2334/1816 = S2/ S1

New slip S2 = 2334 * 0.113 / 1816 = 0.145

Hence, new speed = Ns ( 1 - S2 )

= 2250*(1 - 0.145)

= 1924 rpm

Answer:

The temperature and relative humidity when it leaves the heating section = T2 = 19° C and ∅2 = 38%

Heat transfer to the air in the heating section = Qin = 420 KJ/min

Amount of water added =  0.15 KG/min

Explanation:

The Property of air can be calculated at different states from the psychometric chart.

At T1 = 10° C  and ∅ = 70%

h1 = 87 KJ/KG of dry air

w1 = 0.0053  kg of moist air/ kg of dry air

v1 = 0.81 m^3/kg

AT T3 = 20° C ,  3  ∅ = 60%

h3 = 98 KJ/KG of dry air

w3 = 0.0087 kg of moist air/ kg of dry air

The moisture in the heating system remains the same when flowing through the heating section hence, (w1 = w2)

The mass flow rate of dry air,

m1 = V'1/V1 = 35/0.81

m1 = 43.21 kg/min

By balancing the energy in heating section we get:

mwhw + ma2h2 = mah3

(w3 -w2)hw + h2 = h3

h2 = h3 - (w3 -w2)hw @ 100 C

Hence, hw = hg @ 100 C and w2 = w1

h2 = h3 - (w3 -w2) hg @ 100 C

h2 = 98 - ( 0.0087 - 0.0053) * 2676

h2 = 33.2 KJ/KG

The exit temperature and humidity will be,

T2 = 19.5° C and  2 ∅ = 37.8%

(b) Calculating the transfer of heat in the heating section

Qin = ma(h2 -h1) = 43.21(33.2 - 23.5)

Qin = 420 KJ/min

(c) Rate at which water is added to the air in the humidifying section,

mw = ma(w3 - w2) = (43.2)(0.0087 - 0.0053)

mw = 0.15 KG/min

Answer:

d) A shear stress that varies linearly over the section.

Explanation:

Given that shaft is under pure torsion

As we know that relationship between shear stress in the shaft and radius given as

[tex]\dfrac{T}{J}=\dfrac{\tau}{r}[/tex]

For solid shaft

[tex]J=\dfrac{\pi}{64}d^4[/tex]

Therefore shear stress given as

[tex]\tau=\dfrac{T}{J}\times r[/tex]

T=Applied torque on the shaft

τ=Shear stress at any radius r

From the above equation we can say that shear stress is vaying linearly with the radius of the shaft

Therefore the answer will be d.

Answer: I have answered the questions in rephrased sentences as below;

When implementing a safety and health program, management leadership need employee participation. Effective management of worker safety and health programs has improved employee productivity and morale in the workplace.

Nearly a third of all serious occupational injuries and illnesses stem from overexertion of repetitive motion.

Training is a way for employers to provide tools to enable employees to protect themselves and others from injuries.

Under OSHA, employees are protected from discrimination when reporting a work-related injury, illness, or fatality.

Explanation: All personnel including management & employees must be directly involved when workplace HSE policies are being made & reviewed. This is because everyone in the work environment is impacted one way or the other when incidents occur.

Training & Reporting are key responsibilities of managers, employers & supervisors, so it is mandatory to be done without discrimination so as to foster employees happiness which ultimately lead to zero incidents & increased productivity & profit.

Answer:

The answer will be Rule 61G15-23 F.A.C, relating to Seals.

Explanation:

According to the description given by: Florida administrative code&Florida administrative register the Minimum requirements for engineering documents are in the section 'Final 61G15-23'  from 11/3/2015. This document provides specifications of materials required for the safe operation of the system that is the result of engineering calculations, knowledge and experience.

Answer:

false jdbebheuwowjwjsisidhhdd

Answer:

a.6531.53 mole/hr

b. 32.76 degC

c.  3.78 deg.c

Explanation:A gas mixture containing 85.0 mole% N2 and the balance n-hexane flows through a pipe at a rate of 100.0 m3/h. The pressure is 2.00 atm absolute and the temperature is 140.0°C.

a. What is the molar flow rate of the gas? kmol/h

b. To what temperature would the gas have to be cooled at constant pressure in order to begin condensing hexane? °C

c. To what temperature would the gas have to be cooled at constant pressure in order to condense 75.0% of the hexane? °C

Given V= Volume of gas mixture= 100m3/hr=10^5 Lt/hr P= 2.0 atm and T= 100 deg.c =100+273.15= 373.15K

n= moles of mixture= PV/RT , where  R=0.08206 L.atm/mole.K

n= 2.0*10^5/0.08206*373.15)=6531.53 mole/hr

b. To what temperature would the gas have to be cooled at constant pressure in order to begin condensing hexane? A gas mixture containing 85 mole% N2 and the°C

Condensation begins at a point at which the partial pressure of vapor =vapor pressure of liquid at the given temperature

Partial pressure of  hexane in the mixture= 0.15*2.0= 0.3 atm

so for saturation to begin, the vapor pressure shoudl correspond to 0.3 atm=30.39 Kpa

Antoine constant for Hexane

lnP  (Kpa)= 13.82- 2696/(T-48.833)  ( T is in K

ln(30.39 )= 13.8193- 2696/ (T-48.833)

, 3.414= 13,82-2696/(T-48.333)

2696/(T-48.833)= 13.82-3.414=10.405

T-46.833= 2696/10.414=259.08

259.08+46.833 K=305.91k

305.91k-273.15K

32.76C

c. To what temperature would the gas have to be cooled at constant pressure in order to condense 75% of the hexane? °C?

Vapor present in the gas mixture= 25%, Its partial pressure=0.15* 0.25*2.0 =0.075 atm= 7.59Kpa

from ln(7.59)= 13.82- 2696/ (T-48.333)

2.02 = 13.82- 2696/(T-48.333)

2696/(T-48.333)= 11.179

T-48.333= 2696/11.79=228.60

T= 228.6+48.333= 276.93 K= 3.78 deg.c

Answer:

c. The sum of the two and the direction of  either.

Explanation:

If both sources are linear, we can apply the superposition theorem, which in this case, states simply that the resulting current is just the algebraic sum of both currents.

This is due to any of them, is independent from the other, so we can calculate her influence without any other source present.

Assuming that both sources are ideal (no shunt resistances) , we can apply superposition, removing all sources but one each time, just replacing the sources by an open circuit.

Let's suppose that we have a 5A source flowing to the rigth, and a 3A source in the opposite direction.

Applying superposition, we have:

I₁ = 5A (the another source is removed)

I₂ = -3A (The minus sign indicates that is flowing in the opposite direction, 5A source is removed)

I = I₁ + I₂ = 5A + (-3A) = 2 A

As it can be seen , the resulting current is the sum (algebraic) of both current sources, and has the direction of either of the sources, depending on the absolute value of the current sources.

(If the directions of I₁ and I₂ were inverted, with the same absolute values, the direction would be the opposite).

We could have arrived to the same result applying KCL.

Answer:

The answer is C): Stereospecific and concerted.

Explanation:

An E2 mechanism is a single step elimination reaction that is both stereospecific and concerted: it is stereospecific because it can convert the components of the reaction into stereoisomeric products; it is concerted because all bonds are broken and formed during the single step.

Answer:

a) [tex] \dot Q_{out} = (20 + 273} \times 0.008 = 2.344 kW[/tex]

b) T_2 = 36.4 degree C

c) [tex]\Delta S_{gen} = 0.006512 KW/K[/tex]

Explanation:

Given data:

[tex]P_1 = 160 kPa[/tex]

volumetric flow [tex]V_1 = 0.03 m^3/s[/tex]

[tex]P_2 = 800 kPa[/tex]

power input [tex]W_{in} = 10 kW[/tex]

[tex]T_{surr} = 20 degree C[/tex]

entropy = 0.008 kW/K

from refrigerant table for P_1 = 160 kPa and x_1 = 1.0

[tex] v_1 = 0.12355 m^3/kg[/tex]

[tex]h_1 = 241.14 kJ/kg[/tex]

[tex]s_1 = 0.94202 kJ/kg K[/tex]

a) mass flow rate [tex] \dot m = \frac{V_1}{v_1}[/tex]

[tex]\dot m = \frac{0.03}{0.12355} = 0.2428 kg/s[/tex]

heat loss[tex] = T_{surr} \times entropy[/tex]

heat loss[tex] \dot Q_{out} = (20 + 273} \times 0.008 = 2.344 kW[/tex]

b) from energy balance equation

[tex]W_{in} 0 \dot Q_{out} = \dot m (h_2 -h_1}[/tex]

[tex]10 - 2.344 = 0.2428 (h_2 - 241.14}[/tex]

[tex]h_2 = 272.67 kJ/kg[/tex]

from refrigerant table, for P_2 = 800 kPa and h_2 = 272.67 kJ/kg

T_2 = 36.4 degree C

c) from refrigerant table P_2 = 800 kPa and h_2 = 272.67 kJ /kg

[tex]s_2 = 0.93589 kJ/kg K[/tex]

rate of entropy

[tex]\Delta S_R =  \dot m =(s_2 -s_1)[/tex]

[tex]\Delta S_R = 0.2428 \times (0.93589 -0.94202) = - 0.0014884 kW/K[/tex]

rate of entropy for entire process

[tex]\Delta S_{gen} = \Delta _S_R + \Delta_{surr}[/tex]

[tex]\Delta S_{gen} = 0.0014884 + 0.008 = 0.006512 KW/K[/tex]

To solve this problem it is necessary to apply the concepts related to temperature stagnation and adiabatic pressure in a system.

The stagnation temperature can be defined as

[tex]T_0 = T+\frac{V^2}{2c_p}[/tex]

Where

T = Static temperature

V = Velocity of Fluid

[tex]c_p =[/tex] Specific Heat

Re-arrange to find the static temperature we have that

[tex]T = T_0 - \frac{V^2}{2c_p}[/tex]

[tex]T = 673.15-(\frac{528}{2*1.005})(\frac{1}{1000})[/tex]

[tex]T = 672.88K[/tex]

Now the pressure of helium by using the Adiabatic pressure temperature is

[tex]P = P_0 (\frac{T}{T_0})^{k/(k-1)}[/tex]

Where,

[tex]P_0[/tex]= Stagnation pressure of the fluid

k = Specific heat ratio

Replacing we have that

[tex]P = 0.4 (\frac{672.88}{673.15})^{1.4/(1.4-1)}[/tex]

[tex]P = 0.399Mpa[/tex]

Therefore the static temperature of air at given conditions is 72.88K and the static pressure is 0.399Mpa

Note: I took the exactly temperature of 400 ° C the equivalent of 673.15K. The approach given in the 600K statement could be inaccurate.

Answer:

The coattail effect

Explanation:

Clearly, the Daggies franchise were a huge part of Marco's success and attracted many customers for the business. This is why Marco deteriorated as soon as many Daggies franchisees closed.

The coattail effect is the phenomenon where an influencing member in a party (franchisee in this case)  contributes largely to the success of another, which is the case with Marco and Daggies

Answer:

Length of stilling basin = 32.9 feet

Height of end sill = 6.58 feet

Explanation:

Discharge = Q = 400 ft^3 /sec

Slope = 2.5 ft

Width = 20 feet

n = 0.013

we will assume the depth of flow as "d"

Q = 1/n (R)^2/3 (slope)^1/2 A   ( here R is the hydraulic Radius)

by substituting the given data in above formula we get:

400 = 1/0.013 * (R)^2/3 * sqrt (2.5/100) * 20d

R = A/P

here, A is the flow area and P is the wetted perimeter

400*0.013 = (20d/(20+20d))^2/3 * sqrt(2.5/100) * 20d

d = 1.42 feet

Depth of stilling channel before the jump will be = d1 =  8 feet

Depth of stilling channel after the jump will be = d2 = 1.42 feet

Length of stilling basin = 5(d2 - d1)

                                      = 5( 8 - 1.42)

Length of stilling basin = 32.9 feet

Now calculating the height of end sill:

Jump height = (8 - 1.42)

Height of end sill = 6.58 feet

Matrices, particularly jagged and rectangular ones, serve different functions in programming. Jagged matrices handle non-uniform data and graphs, while rectangular matrices are necessary for graphics and linear algebra. Both should be included in programming languages to provide flexibility and cater to various applications.

Matrices are a fundamental aspect of both mathematics and computer science. They are primarily used for representing and manipulating linear transformations, data, and relations. Matrices come in various forms, with two common types being jagged matrices and rectangular matrices.

When considering whether to include jagged or rectangular matrices, or both, in a programming language, it's essential to recognize the different use cases each type serves. Inclusion of both types allows for greater flexibility and functionality. A programming language that supports both can cater to a wider range of applications and user needs. Nullable matrices can be used when necessary to mimic jaggedness within rectangular matrix structures, offering a middle ground.

Including both jagged and rectangular matrices in a programming language offers flexibility for diverse applications, optimizing memory usage and performance across various data structures and operations.

Applications of Matrices

Applications that Require Jagged Matrices

1. Sparse Data Representation: Efficient storage of data where the majority of elements are zero, such as social network adjacency matrices.

2.Graph Representation: Storing graphs where the number of edges varies widely between nodes.

3. Image Processing: Non-uniform image sampling or varying resolution images.

4. Variable-Length Data: Managing records of varying lengths in databases or documents.

5. Numerical Solutions: Adaptive mesh refinement in computational fluid dynamics.

Applications that Require Rectangular Matrices

1. Linear Algebra: Solving systems of linear equations, eigenvalue problems.

2. Data Analysis: Handling datasets in machine learning, where data points have the same number of features.

3. Computer Graphics: Transformations and projections in 2D and 3D graphics.

4. Control Systems: State-space representations of dynamic systems.

5. Signal Processing: Filtering, convolution operations in audio and image processing.

Arguments for Including Matrix Types in a Programming Language

Just Jagged Matrices

- Pros:

 - Flexibility in handling datasets with variable row lengths.

 - Efficient memory usage for sparse or highly irregular data.

- Cons:

 - Increased complexity in implementation and usage, as many linear algebra operations assume rectangular structure.

 - Can be less performant for operations that benefit from contiguous memory storage.

Just Rectangular Matrices

- Pros:

 - Simplicity in usage and implementation for most standard applications.

 - Better support for linear algebra operations and optimizations in libraries.

- Cons:

 - Inefficient memory usage for sparse or variable-length data, leading to wasted space.

 - Lack of flexibility for applications requiring variable row lengths.

Both Jagged and Rectangular Matrices

- Pros:

 - Flexibility to choose the most appropriate type based on the specific application requirements.

 - Optimal memory usage and performance by using jagged matrices for sparse data and rectangular matrices for linear algebra operations.

- Cons:

 - Increased complexity in the programming language and potential confusion for users in choosing the correct type.

 - Possible performance overhead in managing two types of matrices.

Including both jagged and rectangular matrices in a programming language provides the greatest flexibility and efficiency across a wide range of applications. While it adds some complexity, the benefits of being able to choose the optimal structure for specific tasks outweigh the drawbacks. This approach ensures that the language can handle diverse data structures and operations effectively, catering to both irregular and uniform datasets.

The mass of the car plus its occupants is calculated using Newton's second law of motion. After determining the net force by subtracting the force of friction from the force exerted by the wheels, it is divided by the given acceleration to find the mass, which is approximately 1027.78 kg.

The question pertains to the application of Newton's laws of motion and forces to determine the mass of a vehicle, given the acceleration, force exerted, and the opposing forces of friction. The calculation is based on Newton's second law, which states that the net force acting on an object is equal to the mass of the object times its acceleration (F = ma). To find the mass of the car plus its occupants, we start with calculating the net force (Fnet) which is the difference between the force the wheels exert backward on the road and the force of friction. By applying Newton's second law (Fnet = m × a), we rearrange it to solve for mass (m = Fnet / a).

Net Force Equation: Fnet = Force exerted - Friction
= 2100 N - 250 N
= 1850 N

Now, we can determine the mass using the net force and the given acceleration.

Mass (m) = Fnet / a
= 1850 N / 1.80 m/s2
= 1027.78 kg

Therefore, the mass of the car plus its occupants is approximately 1027.78 kg.

Employers aren't required to keep records specifically for an employee with the flu, unless under certain conditions related to workplace illnesses. Reasonable exceptions to the FOIA include the protection of sensitive personal information such as government employees' medical records.

Employers are not generally required to keep records of an employee who has simply contracted the flu. However, certain regulations may apply if the illness could be work-related or if it pertains to a larger public health concern that requires tracking. In the context of the Freedom of Information Act (FOIA), the question of keeping medical records would fall under exemptions related to personal privacy. For instance, medical records for government employees would be a reasonable exception to FOIA, as they contain sensitive personal information that is protected from public disclosure.

Answer:

Explanation:

Given

[tex]T_h=250^{\circ}C\approx 523\ K[/tex]

[tex]T_L=30^{\circ}C\approx 303\ K[/tex]

[tex]Q_1=6 kW[/tex]

From Clausius inequality

[tex]\oint \frac{dQ}{T}=0[/tex]  =Reversible cycle

[tex]\oint \frac{dQ}{T}<0[/tex]  =Irreversible cycle

[tex]\oint \frac{dQ}{T}>0[/tex]  =Impossible

(a)For [tex]P_{out}=3 kW[/tex]

Rejected heat [tex]Q_2=6-3=3\ kW[/tex]

[tex]\oint \frac{dQ}{T}= \frac{Q_1}{T_1}-\frac{Q_2}{T_2}[/tex]

[tex]=\frac{6}{523}-\frac{3}{303}=1.57\times 10^{-3} kW/K[/tex]

thus it is Impossible cycle

(b)[tex]P_{out}=2 kW[/tex]

[tex]Q_2=6-2=4 kW[/tex]

[tex]\oint \frac{dQ}{T}= \frac{Q_1}{T_1}-\frac{Q_2}{T_2}[/tex]

[tex]=\frac{6}{523}-\frac{4}{303}=-1.73\times 10^{-3} kW/K[/tex]

Possible

(c)Carnot cycle

[tex]\frac{Q_2}{Q_1}=\frac{T_1}{T_2}[/tex]

[tex]Q_2=3.47\ kW[/tex]

[tex]\oint \frac{dQ}{T}= \frac{Q_1}{T_1}-\frac{Q_2}{T_2}[/tex]

[tex]=\frac{6}{523}-\frac{3.47}{303}=0[/tex]

and maximum Work is obtained for reversible cycle when operate between same temperature limits

[tex]P_{out}=Q_1-Q_2=6-3.47=2.53\ kW[/tex]

Thus it is possible

Answer: Metal has a oxide layer when in air. Before an experiment cleaning it with sand paper will remove the leyer of oxide and ensure truer results.  Cleaning with tissue paper will not result in removing the layer of oxide.

Shows how active H2 can be.

Explanation:

Following are the responses to the given question:

For question 1:

For question 2:

For question 3:

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