The president of South Africa Mr.C.Ramaphosa and Minister of Energy Mr.Gwede Mantashe released a media statement that they have bought a compensator system to solve the loadsheding pandemic and the Eskom CEO tasked you to realize and implement the compensator G (s) to keep in order save your job and his job. G(s)= s+5.015s+0.5/s

Answer:

a)123.37 mW

b) 58.75 mJ

Explanation:

a) The current (i) flowing through the terminal is given as:

\(i=\frac{dq}{dt} \\i=\frac{d(5sin(4\pi t)}{dt} =20\pi cos(4\pi t)\ mA\\\)

The Power delivered to the element (P(t)) is given as:

\(P(t)=v(t)i(t)= 3 cos(4\pi t)*20\pi cos(4\pi t)*10^{-3}\\P(t)=60\pi cos^2(4\pi t)*10^{-3}\\at \ t \ =\ 0.3s,\ the \ power\ delivered\ is:\\P(0.3) = 60\pi cos^2(4\pi *0.3)*10^{-3}=123.37mW\)

b)  the energy delivered to the element (W) between 0 and 0.6 s is given as:

\(W = \int\limits^0_{0.6} {P(t)} \, dt \\W=\int\limits^0_{0.6} {60\pi cos^{2}(4\pi t)} \, dt =58.75mJ\)

Answer:

I think it is the 2,3,5 and 1 ones

The radius of the cylinder is the length that measures from the center to the parameter. The length of the cylinder is 1 cm.

The density of the given substance can be used as the measure of the mass and the volume of the substance.

Given,

Length of the aluminum cylinder = 5cm

Mass = 42 g

Density of aluminum = 2700 kg/ m³ = 2.7 gm/cm³

From the density the volume is calculated as:

Density = mass ÷ volume

2.7 = 42 ÷ V

V = 42 ÷ 2.7

= 15.56

The radius of the cylinder is calculated as:

(r) = √(V / π × h)

Here, V =  15.56, π = 3.14, h = 5 cm

Substituting values:

r = √(15.56 ÷ 3.14 × 5)

= √ (15.56 ÷ 15.7)

= 0.99

= 1

Therefore, 1 cm is the length.

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The program's output would be: c w, c x, d w, d x, e w

This program iterates over the letters 'c' through 'e' and 'w' through 'x' using nested loops. The outer loop loops over the letters 'c' through 'e,' incrementing the value of the letter1 each time. The inner loop loops over the letters 'w' through 'x,' incrementing the value of the letter2 each time.

The inner loop's print statement utilizes string concatenation to output the values of letter1 and letter2, separated by a space. Following the printing of the data, the inner loop increments the value of the letter2 and then repeats until the letter2 exceeds 'x'. When the inner loop is finished, the outer loop increases the value of the letter1 and repeats until the letter1 is larger than 'e'.

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The piano frequency.A key on a piano or other musical instrument has a frequency of f0, and the higher and lower white and black keys have frequencies of f0 * rn.

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Answer:

a)  Cr = Co - Fx / D

b)   ΔC / Δx = ( CR - Cr )  / ( xR - xRo )

Explanation:

A) Derive an expression for the profile c(r) inside the tissue

F = DΔC / X  = D ( Co - Cr ) / X   ------ 1

where : F = flux , D = drug diffusion coefficient

            X = radial distance between Ro and R

Hence : Cr = Co - Fx / D

B) Express the diffusive flux at outer surface of the balloon

Diffusive flux at outer surface =  ΔC / Δx = CR - Cr / xR - xRo

OSHA 300 Logs and OSHA 300A SummariesWorkers have the right to review OSHA 300 Logs and OSHA 300A Summaries.

These logs and summaries are part of the OSHA recordkeeping requirements and provide information about work-related injuries and illnesses in the workplace. The OSHA 300 Log is a detailed record of all reportable injuries and illnesses, while the OSHA 300A Summary is a summary of the annual totals from the OSHA 300 Log. Workers have the right to access and review these records to ensure transparency and awareness of workplace safety issues. However, it's important to note that specific rights may vary depending on the jurisdiction and applicable regulations.

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The immediate value 01h represents the bitmask with a 1 in the least significant bit.

To toggle the least significant bit of the second and third elements in the array, you can use the XOR (exclusive OR) instruction with an immediate value of 01h. Here is the instruction:

xor word ptr arr+2, 01h

xor performs the bitwise XOR operation between the second and third elements of the array and the immediate value 01h.

word ptr is used to specify that we are operating on a word-sized (2-byte) element.

arr+2 is the memory address of the second element in the array.

The immediate value 01h represents the bitmask with a 1 in the least significant bit.

After executing this instruction, the least significant bit of the second and third elements in the array will be toggled. For example, if the initial array is 0F8h, 16h, 0DBh, 77h, it will be modified to 0F8h, 17h, 0DAh, 77h, as required.

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The value of pv is 0.101

Data given and notation  

X = 788 represent the average life for the sample

θ  = 40 represent the population standard deviation

n =  30 sample size

µ = 800 represent the value that we want to test

α = represent the significance level for the hypothesis test.  

z would represent the statistic (variable of interest)  

pv  represent the p value for the test (variable of interest)  

State the null and alternative hypotheses.  

We need to apply a two tailed  test.  

What are H0 and Ha for this study?  

Null hypothesis: µ = 800

Alternative hypothesis :µ  ≠ 800

Compute the test statistic

The statistic for this case is given by:  

z = X -µo/θ  √n

z-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".  

Calculate the statistic  

z = 788 - 800/40√30 = - 1.64

Give the appropriate conclusion for the test

Since is a one side left tailed test the p value would be:  

pv = 2 × P ( z< - 1.64 ) = 0.101

Therefore the value of pv is 0.101

The complete question is : An electrical firm manufactures light bulbs that have a lifetime that is approximately normally distributed with a mean of 800 hours and a standard deviation of 40 hours. Test the hypothesis that μ = 800 hours against the alternative, μ = 800 hours, if a random sample of 30 bulbs has an average life of 788 hours. Use a P-value in your answer.

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The creation of a strategy is guided by the readiness evaluations. Managing change: Some project plans incorporate change management strategies.

The process of managing change in an organised, deliberate manner in order to fulfil organisational goals, objectives, and missions is known as strategic change management.

Organizations must adapt if they are to survive and grow in the face of fierce industry rivalry.

Operating from 1998 to 2018, BionX was a Canadian company that produced bicycle conversion kits and electric motors for bicycles under the moniker EPS.

The BionX board of directors hired executive and made strategic changes to the senior leadership team in 2013 and 2014 in response to a 55% warranty return rate.

Thus, these are some changes that can be made.

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When repairing puncture-type damage of a metal faced laminated honeycomb panel, the edges of the doubler should be tapered to A-two times the thickness of the metal.

This ensures a proper and secure bond between the doubler and the panel, providing strength and stability to the repair. Tapering the edges also helps to distribute stress evenly across the panel, reducing the likelihood of future damage. It is not recommended to taper the edges based on appearance alone, as this can compromise the integrity of the repair.

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Maximum quantization error is calculated using the formula Maximum quantization error = (VH - VL)/2(2n), where n is the ADC's resolution in bits.

The largest inaccuracy that can occur corresponds to a voltage that is (1/2) of an LSB. This quantization would be the sole cause of error in the recorded samples for a perfect digital recorder. This mistake establishes the absolute upper bound on the readings' accuracy for a real digital recorder. The difference between the analog signal and the nearest available digital value at each sampling instant from the A/D converter is known as the quantization error. Quantization error also introduces noise to the sample signal known as quantization noise.

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Answer:

hello the diagram related to your question is missing attached below is the required diagram.

The value of AC = 6m

answer 2.25 N

Explanation:

From the diagram attached below

average intensity of loading = w/2

Total load due to linearly varying load = \(\frac{wL}{2}\)

w = maximum intensity of load , L = span of varying load

since AC = 6m( L )  then

Total load due to linearly varying load  = 3w  ------ ( 1 )

attached below is the detailed solution of the given problem

Answer:

what's the question?...........

The sample of 81 students was selected with a mean score of 90, this illustrates an example of a right tailed one sample z test.

From the information given, the teacher claims that the mean score of students in his class is greater than 82 with a standard deviation of 20. In this case, the right tailed one sample z test is used.

The z value will be:

= (90 - 82)/20/✓81

= 3.6

Since 3.6 > 1.645, the null hypothesis will be rejected as there's enough evidence to support the teacher's claim.

When an online medicine shop claims that the mean delivery time for medicines is less than 120 minutes with a standard deviation of 30 minutes, the left tailed one sample test is used.

The z value will be:

= (100 - 120)/(30/✓49)

= -4.66

The null hypothesis is rejected as there is enough evidence to support the claim of the medicine shop.

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Known :

V1 = 10 L

T1 = 27°C

P1 = 12 atm

P2 = 3 atm

Solution :

Work done in the process

W = P1 • V1 ln (P1 / P2)

W = (12 atm) • (10 L) ln (12 / 3)

W = 166.35 L • atm

W = 168.55 kJ

Answer: d consequences

Explanation:

Explanation:

Fire doors are closed and paths are clear

Answer:

A) About \(9.273 \times 10^{-9}\) newtons

B) 76.518 newtons

C) 111.834 newtons

Explanation:

A) \(F_g=\dfrac{GM_1M_2}{r^2}\) , where G is the universal gravitational constant, M 1 and 2 are the masses of both objects in kilograms, and r is the radius in meters. Plugging in the given numbers, you get:

\(F_g=\dfrac{(6.67408 \times 10^{-11})(7.8)(11.4)}{(0.8)^2}\approx 9.273 \times 10^{-9}\)

B) You can find the weight of each object on Earth because you know the approximate acceleration due to gravity is 9.81m/s^2. Multiplying this by the mass of each object, you get a weight for the first particle of 76.518 newtons.

C) You can do a similar thing to the previous particle and find that its weight is 11.4*9.81=111.834 newtons.

Hope this helps!

The command used on a UNIX/Linux system to display the path taken across a network to reach a remote node, including the hops and timing encountered on the path, is the traceroute command.

The traceroute command is used to display the route (path) and measure transit delays of packets across an Internet Protocol (IP) network. It is typically used to diagnose network connectivity issues by showing the path that packets take across a network from one system to another. The command displays the IP addresses and hostnames of each hop, as well as the time taken to reach each hop.

Here is an example of how to use the traceroute command:

This command will display the path taken across the network to reach the www.example.com system, including the hops and timing encountered on the path.

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Dynamic viscosity is greater than kinematic viscosity for air, water, and mercury at 1 atm and 20 degrees Celsius, due to their varying densities and fluid properties.

Dynamic viscosity (μ) is the measure of a fluid's internal resistance to flow, while kinematic viscosity (ν) is the ratio of dynamic viscosity to density.

At 1 atm and 20 degrees Celsius, the dynamic viscosity of air is the smallest at around 1.8 x 10^-5 Pa·s, followed by water at around 8.9 x 10^-4 Pa·s, and then mercury at around 1.55 x 10^-3 Pa·s.

However, the kinematic viscosity of air is much larger than water and mercury due to its low density, at around 1.5 x 10^-5 m^2/s compared to water at around 1.0 x 10^-6 m^2/s and mercury at around 1.1 x 10^-6 m^2/s.

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Answer:

Explanation:

In Shunt DC Motors, the speed can be controlled using the following techniques:

Armature voltage control - By varying the armature voltage, the magnetic field produced by the armature current can be controlled, which affects the speed of the motor. Thus, the speed of the motor can be regulated by controlling the armature voltage.

Field resistance control - The speed of the motor can also be controlled by varying the field resistance. By increasing the field resistance, the flux produced by the field winding is decreased, and the speed of the motor increases. Conversely, reducing the field resistance increases the flux and decreases the speed.

Armature resistance control - In this method, the resistance of the armature circuit is varied. By increasing the armature resistance, the current in the motor reduces, leading to a decrease in the speed of the motor.

In Series DC Motors, the speed can be controlled using the following techniques:

Armature voltage control - Similar to shunt motors, the speed of the series motor can be regulated by varying the armature voltage.

Field flux control - As the name suggests, the flux produced by the field winding can be controlled by varying the field winding resistance. By increasing the field resistance, the flux reduces, leading to an increase in speed. Conversely, decreasing the field resistance increases the flux and decreases the speed.

Armature resistance control - By varying the armature resistance, the current in the motor can be controlled. Increasing the resistance decreases the current, leading to an increase in speed.

Overall, these techniques enable the speed control of shunt and series DC motors, allowing them to be used in a wide range of applications where variable speed control is required.

PLS MARK ME BRAINLIEST

The statement that The different taste qualities (sweet, sour, bitter, salt, umami) each have their own area of the tongue that is most sensitive to them is False.

Gate Theory states that competing signals such as rubbing your elbow can make you more sensitive to pain. This theory suggests that physical signals such as massage, pressure, heat, or cold can block the transmission of pain messages through the spinal cord and prevent them from reaching the brain. Therefore, when people rub their elbows or apply pressure to the area, this stimulates the non-painful touch fibers and decreases the transmission of the painful stimuli.

As per the statement, patients who are allowed to self-administer morphine use less than when they receive injections from healthcare providers is True. Olfaction is a very important sense for humans, and they often become depressed when they lose their sense of smell (olfaction). The statement that The different taste qualities (sweet, sour, bitter, salt, umami) each have their own area of the tongue that is most sensitive to them is False.

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Answer:

D. Collects radiation reflected from Earth.

Explanation:

Answer:

shear strength = 2682.31 Ib/ft^2

Explanation:

major principal stress = 100 Ib / in2

minor principal stress = 20 Ib/in2

Normal stress = 3000 Ib/ft2

Determine the shear strength when direct shear test is performed

To resolve this we will apply the coulomb failure criteria relationship between major and minor principal stress a

for direct shear test

use Mohr Coulomb criteria relation between normal stress and shear stress

Shear strength when normal strength is 3000 Ib/ft  = 2682.31 Ib/ft^2

attached below is the detailed solution

Answer:

Explanation:

In this context, we are dealing with the cooling of hot air as it flows through tubes exposed to atmospheric air. Based on this, we can make the following statements:

a. Heat transfer coefficient inside the tube will be higher - This statement is false. As the hot air flows through the tube and comes into contact with the cooler atmospheric air outside the tube, heat transfer occurs between the two fluids. In this process, the temperature difference between the two fluids drives the heat transfer. Therefore, the heat transfer coefficient inside the tube will be lower because the hot air will have a higher temperature than the atmospheric air, and the temperature difference will be smaller.

b. Heat transfer coefficient inside the tube will be lower - This statement is true, as explained in the previous statement.

c. Fins should be attached to the outside of the tube - This statement is true. Fins are used to increase the surface area of the tube, which in turn increases the rate of heat transfer between the hot air inside the tube and the atmospheric air outside. Attaching fins to the outside of the tube will increase the surface area exposed to the atmospheric air, facilitating faster cooling of the hot air.

d. Fins should be attached to the inside of the tube - This statement is false. Attaching fins to the inside of the tube will actually reduce the effective cross-sectional area of the tube, restricting the flow of air and hindering the heat transfer process. Therefore, fins should be attached to the outside of the tube to maximize the surface area and facilitate efficient heat transfer.