A ball is thrown upward with an initial velocity of 20m/s
.a) How long is the ball in the air?
b) What is the greatest height reached by the ball?
c) When is the ball 15m above the ground?

Answer:

False

Explanation:

As seen in the diagram, electrons are rotating around the nucleus and are, therefore, not part of the nucleus.

Answer:

4 Wear personal protective equipment (PPE).

Depending on the job task performed, PPE when working near power lines includes safety glasses, face shields, hard hats, insulated boots, rubber gloves with leather protectors, insulating sleeves, and flame-resistant clothing to reduce the risk of electrocution.

The soil loss load would be reduced to 14,580 kgs if the runoff velocity decreased to 20km/hr.

Velocity is a vector quantity that describes the rate and direction of an object’s motion. It is a combination of speed and direction. Velocity is measured in units such as meters per second, feet per second, or miles per hour. Velocity can be constant, meaning it does not change, or it can be variable, meaning its magnitude or direction can change.

Soil loss load = Velocity x Weight

Soil loss load = 40m/hr x 729 kgs

Soil loss load = 29,160 kgs

If the runoff velocity decreased to 20km/hr, the soil loss load would be

Soil loss load = 20km/hr x 729 kgs

Soil loss load = 14,580 kgs

Therefore, the soil loss load would be reduced to half (14,580 kgs) if the runoff velocity decreased to 20km/hr.

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Answer:

A =41 .....

......

......

....C=21

Answer:

Given, diameter of pin head d = 1 mm = 1/25.4 = 0.0394 in Surface area of a pinhead, A = 4pr^2 =

Explanation:

eesh

Answer:

Primary cell with a nominal open circuit voltage of 1.5 Volts produced in very high volumes. Chemistry based on a zinc anode and a cathode/depolariser of manganese dioxide which absorbs the liberated hydrogen bubbles which would otherwise insulate the electrode from the electrolyte.

Answer:

Explanation:

(1) The electrolyte is a paste of ammonium chloride.

(2) The positive terminal (anode) is a carbon rod surrounded bymangAnese Dioxide as a depolarised .

(3) The negative terminal is zinc container.

Answer:

some part of your question is incomplete

attached below is the complete question

Answer :

F(0) = -5  < 0

F(1) = e - 1 > 0

since the functions : f(0) and f(1) have opposite signs then there is a 'c' whereby F(c) = 0 ( intermediate value theorem fulfilled )

Hence there is a root in the given equation : \(e^x = 6 - 5x\)

Explanation:

using Intermediate value Theorem

If F(x) is continuous and f(a) and f(b) have opposite signs then there will be a'c'E (a,b)  whereby F(c) = 0

given equation : \(e^x = 6 - 5x\)   on (0,1)

and F(x) = \(e^x - 6 + 5x = 0\)

This shows that the F(x) is continuous on (0,1)

F(0) = \(e^0 - 6 + 5(0)\)  = -5 which is < 0

F(1) = \(e^1 -6 + 5(1)\) = e -1 > 0  and e = 2.7182

since the functions : f(0) and f(1) have opposite signs then there is a 'c' whereby F(c) = 0 ( intermediate value theorem fulfilled )

Hence there is a root in the given equation : \(e^x = 6 - 5x\)

The first case where the power of synthesis was applied was in Chile when they had to put 100 families in houses around 40m².

It should be noted that the power of architecture is the fact that it can synthesize a complex entry to a problem.

Chilean architect Alejandro used this in building more than several houses for the poorest communities in Chile. His rigorous and innovative design approach used a social framework that laid a precedent within the profession.

During this period in Chile, a normal middle-class home was about 80m² but he built his in 40m² and it was a good home that managed the resources that were available.

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With a professional, do an automotive diagnostic

Verification and confirmation in

Step 1. This step in the vehicle diagnostic process enables the expert to look for potential issues.

Step 2: Identify the issue.

Step 3: Area isolation.

4. Make repairs.

The final step is a check.

the procedure of determining a diagnosis, disease, or injury based on its indications and symptoms. To aid in the diagnosis, testing like blood tests, imaging tests, and biopsies may be done in addition to a physical examination and health history.

Using a computer scan tool, the repairman will obtain the diagnostic issue code data from the computer. A fully qualified technician can more correctly identify and address the issue by using this information.

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Using a computer scan tool, the repair technician will retrieve the diagnostic trouble code information from the computer. A properly trained technician can use this information to more accurately locate and repair the problem.

As part of the standard operating procedure, many experts recommend that shops perform a pre-scan of all vehicle computer modules, as well as a scan after the vehicle is repaired (SOP). This is not only good business practice, but it also helps to communicate with the customer about potential vehicle flaws that may not have been part of the customer's original concern.

Pre-scan: This entails gaining access to all vehicle modules and downloading any or all stored Diagnostic Trouble Codes (DTCs), including pending codes. Any DTCs that have been stored will be recorded on the work order, and if they are related to a customer, it may be necessary to notify the customer and obtain approval before proceeding with the repair.

Post Scan: After the vehicle has been repaired and before it is released to the customer, a full module scan is performed to ensure that another DTC was not set during the repair in addition to the repair. procedure for repair The results of this follow-up scan should also be documented on the repair order so that they can be added to the driving history.

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The paragraph related to transistor technologies covers questions about the differences between FET and BJT, the physical structure of an n-channel JFET, drain characteristics, JFET parameters, and the comparison between depletion and enhancement MOSFETs.

This paragraph contains two separate questions related to transistor technologies, specifically Field Effect Transistors (FET) and Bipolar Junction Transistors (BJT).

In Question 2, the first part asks for three differences between FET and BJT, which could include variations in construction, operation principles, or characteristics. The second part requests a drawing of the physical structure and device symbol for an n-channel Junction Field Effect Transistor (JFET). Lastly, it inquires about the meaning of drain characteristics in relation to JFET.

Question 3 begins with the mention of four JFET parameters and requires an explanation of any two of them. The second part asks for a comparison between depletion and enhancement Metal-Oxide-Semiconductor Field Effect Transistors (MOSFET).

The third part requests the determination of current at different gate-source voltage values for an n-channel enhancement MOSFET with a given threshold voltage.The paragraph concludes by mentioning the examiners responsible for the questions.

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Answer:

To calculate the steady state flux of atomic hydrogen through a steel vessel, we need to use Fick's first law, which states that the flux (J) is equal to the diffusivity (D) multiplied by the concentration gradient (dC/dx).

First, we need to calculate the concentration gradient by dividing the difference in hydrogen concentration between the inside and outside surfaces by the wall thickness of the vessel. The inside surface is kept saturated with hydrogen at a concentration of 4.5 moles/m3, and the outside surface is exposed to the atmosphere, which has a hydrogen concentration of 0 moles/m3. Therefore, the concentration gradient is (4.5 - 0) moles/m3 / (4 mm) = 1.125 moles/m3 mm.

Next, we need to substitute this value into Fick's first law along with the diffusivity of hydrogen in steel, which is given as 0.1 mm2/s. This gives us the steady state flux as J = (-0.1 mm2/s) * (1.125 moles/m3 mm) = -0.01125 moles/s mm2.

Finally, we need to convert the units of the flux from moles/s mm2 to moles/s m2. To do this, we can multiply the flux by 1,000 to convert the units of millimeters to meters, giving us a final steady state flux of -0.01125 moles/s mm2 * 1,000 = -1.125 moles/s m2.

IF THE VESSEL CONTAINS 20 MOLES OF HYDROGEN, CALCULATE THE TIME TAKEN TO DISSIPATE ALL OF THE HYDROGEN OF THAT THE VESSEL HAS A SURFACE AREA OF 3 M2.

To solve this problem, we need to first calculate the flux of atomic hydrogen through the vessel using Fick's first law:

J = -D(dC/dx)

where J is the flux, D is the diffusivity of hydrogen in steel, and dC/dx is the concentration gradient.

Given that the diffusivity of hydrogen in steel is 0.1 mm2/s, the inside concentration is 4.5 moles/m3, and the outside concentration is 0, the concentration gradient is 4.5 moles/m3.

Plugging these values into the equation above, we get:

J = -0.1 mm2/s * 4.5 moles/m3 = -0.45 moles/s-m2

Next, we need to calculate the time it takes to dissipate all 20 moles of hydrogen from the vessel. We can do this by dividing the total number of moles of hydrogen by the flux:

t = 20 moles / (-0.45 moles/s-m2) = 44.44 s

So it would take approximately 44.44 seconds to dissipate all of the hydrogen from the vessel.

Explanation:

SELF EXPLANATORY

The time taken is 44.44 seconds to dissipate all of the hydrogens from the vessel.

To solve this problem, we need to first calculate the flux of atomic hydrogen through the vessel using Fick's first law:

J = -D(dC/dx)

where J is the flux, D is the diffusivity of hydrogen in steel, and dC/dx is the concentration gradient.

Given that the diffusivity of hydrogen in steel is 0.1 mm²/s, the inside concentration is 4.5 moles/m³ and the outside concentration is 0, the concentration gradient is 4.5 moles/m³.

Plugging these values into the equation above, we get:

J = -0.1 mm²/s * 4.5 moles/m³ = -0.45 moles/s-m²

Next, we need to calculate the time it takes to dissipate all 20 moles of hydrogen from the vessel. We can do this by dividing the total number of moles of hydrogen by the flux:

t = 20 moles / (-0.45 moles/s-m2) = 44.44 s

So it would take approximately 44.44 seconds to dissipate all of the hydrogen from the vessel.

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Magnitude of average acceleration  \(a_{avg}\)=2.75 units

Average acceleration \(a_{avg}=\) \(2.265 i- 1.58j\)

We're given that \(s=\frac{t^2}{4}\)

And partice is at A at t=2 s and at B at t= 2.2s

Instantaneous velocity \(\frac{ds}{dt} =\frac{t}{2}\)

Velocity at A  (\(V_{a}\)) at 2s = \(\frac{t}{2}\) or  \(\frac{t}{2}Cos \frac{\pi}{3} i+ \frac{t}{2}Sin \frac{\pi}{3}j=\frac{i+\sqrt{3} j}{2}\)

Velocity at B  (\(V_{b}\)) at 2.2s =\(\frac{t}{2}\) or \(\frac{t}{2}Cos \frac{\pi}{6} i+ \frac{t}{2}Sin \frac{\pi}{6}j=\frac{\sqrt{3}i+j }{2}\)

Average acceleration = \(\frac{V_{b}-V{a}}{t}\) =\(2.265 i- 1.58j\)

magnitude = \(\sqrt{(2.265)^2- (1.58)^2}\) = \(2.75\)

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I will agree with option 1 that says Realistic people favour concrete methods for solving problems.

In real sense, there is no evidence to suggest that realistic people are more likely to become engineers than architects but base on certain characteristics we can just make a logical assumption.

Some of these characteristics are:

Interesting part is that these characteristics are of equal beneficial in both fields of Engineering and Architect.

The personality trait of "realistic" is not directly linked to a preference for working with people over things. It is possible for someone who is realistic to enjoy working with both people and things, or to prefer working with things over people.

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The background-attachment property that is similar to scroll, but is used for elements, such as scroll boxes, to allow the element background to scroll along with the content within a box is 'local.'

The background-attachment property sets whether a background image is fixed or scrolls along with the containing element's text. The background-attachment property sets how the background image will be positioned in the viewport.The background-attachment property has four values, which are:scroll: The image will move as the user scrolls the element, and it will scroll along with the textfixed: The image will not move when the user scrolls the element; instead, it will remain fixed in place, just like a background image in a picture frame. The fixed value fixes the image in a specific location of the page, regardless of the scrolling of the rest of the page.

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The power supply can be designed using a bridge rectifier, capacitors, and Zener diode.

Here is the design of a regulated power supply:

Type 1: 6V

Given Source Voltage= 110±10 Vrms

= 110 + 10

= 120V (Maximum)

Given Output Voltage = 6V ± 5%

= 6V ± 0.3V

Minimum Output Voltage = 5.7V

Maximum Output Voltage = 6.3V

So, taking the maximum voltage into account, the output voltage is 6.3V.

Let's find out the value of the Capacitors and Zener diode

Resistor value for the Zener diode:

For 6.3V, we can take a 5.1V zener diode.

A 1W zener diode can take maximum power = 1W.

If R is the load resistance, V is the input voltage and Vz is the zener voltage, then the resistor R required is given by:

\(R = \frac{{({V_s} - {V_z})^2}}{P}\)

Where

P = 1W, Vs = 120V, Vz = 5.1V

\(R = \frac{{(120 - 5.1)^2}}{1}\)

= 14,141\Omega

So, a 14k resistor will be used.

Capacitor value for rectification:

Now, we need to find the value of the capacitor for rectification.

Average DC Voltage = (Vmax-Vmin)/piFor full-wave rectifier, the average voltage is given by:

\(V_{avg} = \frac{{{V_s}}}{\pi }\)

For given source voltage,

Vavg = 38V

I = \(\frac{{P}}{{{V_s}}}\)

= \(\frac{1}{{120}}\)

= 0.008A

Where, P = 1W, Vs = 120V

Current through the diode = 0.008A.

Capacitance required is given by:

\(C = \frac{I \times T}{V_{ripple}}\)

Where T = 1/f = 1/60 = 0.0167s (time period) and Vripple = 1.4V.

1.4V ripple is taken for full-wave rectification.

\(C = \frac{0.008 \times 0.0167}{1.4}\)

= 0.000095

F = 95µF

Therefore, a 14k resistor, 95µF capacitor, and 5.1V zener diode are used in the design of the power supply.

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B. Electro-energized is not an example of an electronically-controlled suspension system.

Electro-energized is not a common term used to describe suspension systems, and it is not clear what it refers to in the context of suspension systems.

The other options, levelling, active, and magneto-rheological, are all examples of electronically-controlled suspension systems:

Levelling suspension systems use sensors to detect the ride height of the vehicle and adjust the suspension to maintain a consistent height.

Active suspension systems use sensors and electronic control systems to adjust the suspension in real-time to provide a smoother ride and better handling.

Magneto-rheological suspension systems use magnetically controlled fluid to adjust the damping rate of the suspension, providing a more responsive and customizable ride.

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An ideal vapor-compression refrigeration cycle that uses Ammonia as the working fluid is modified to include a counterflow heat exchanger, also known as a suction line heat exchanger.

Here, the refrigerant exits the evaporator as a saturated vapor at 1.5 bars and is heated at a constant pressure to 10°C before entering the compressor, which is isentropic and compresses the refrigerant to 20 bars. The refrigerant enters the condenser, where it is cooled at a constant pressure to 49°C, 20 bars. Finally, the refrigerant liquid flows through the heat exchanger, and the mass flow rate of refrigerant is 10 kg/min.The degree of subcooling is the distinction between the refrigerant's actual temperature and its saturation temperature at the pressure of the valve.

In the present case, the pressure of the valve is 20 bars, which is also the pressure of the refrigerant as it enters the valve. Since the refrigerant is a liquid as it passes through the heat exchanger and there is no indication of any pressure loss or superheating, it must still be a liquid as it enters the expansion valve. The saturation temperature of ammonia at 20 bars can be found using tables, which is 33.34°C. As a result, the refrigerant has a subcooling degree of 33.34 - 20 = 13.34°C. Therefore, the subcooling degree of the refrigerant at the expansion valve is 13.34°C.

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Answer:

The plotted diagram of line distance versus the transmission speed is uploaded below.

Explanation:

Given the data in the question;

Bit length B = 1000 bits

propagation velocity V = 2 × 10⁸ m/s

now, we know that the bit length of a link is expressed as;

B = R × d/V

where V is propagation velocity

d is the distance

R is the transmission speed

B is bit length

so we substitute

1000 = R × d/(2 × 10⁸)

1000 = Rd/(2 × 10⁸)

2 × 10¹¹ = Rd

R = 2 × 10¹¹ / d

R = 2E+11

Hence, we plot the transmission speed versus line distance; as shown in the image BELOW.

From the plot, if the transmission speed increases, the distances between stations decreases and vise versa.

Hence, both are inversely proportional.

The graph plot of line distance versus the transmission speed for a bit length of 1000 bits is; plotted below with Lν = 2 × 10¹¹ m/s

We are given;

Bit length; B = 1000 bits

Propagation velocity; V = 2 × 10⁸ m/s

Formula for bit length of a link is expressed as;

B = ν × L/V

where;

V is propagation velocity

L is the distance

ν is the transmission speed

B is bit length

Thus, plugging in the relevant values gives;

1000 = ν × L/(2 × 10⁸)

Thus;

Lν = 1000 × 2 × 10⁸

Lν = 2 × 10¹¹ m/s

Thus, find the attached image of a graph showing the line distance versus the transmission speed for a bit length of 1000 bits.

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Answer:

A) condenser pressure = 9.73 kPa,

B) 10242 kw

C) 36.9%

Explanation:

given data

entrance pressure of steam = 12.5 MPa

temperature of steam = 550⁰c

flow rate of steam = 7.7 kg/s

outer pressure = 2 MPa

reheated steam temperature = 450⁰c

isentropic efficiency of turbine( nt ) = 85% = 0.85

isentropic efficiency of pump = 90% = 0.90

From steam tables

at 12.5 MPa and 550⁰c ; h3 = 3476.5 kJ/kg,  S3 = 6.6317 kJ/kgK

also for an Isentropic expansion

S4s = S3 .

therefore when S4s = 6.6317 kJ/kg and P4 = 2 MPa

h4s = 2948.1 kJ/kg

nt = 0.85

nt (0.85) = \(\frac{h3-h4}{h3-h4s}\) = \(\frac{3476.5 - h4}{3476.5 - 2948.1}\)

making h4 subject of the equation

h4 = 3476.5 - 0.85 (3476.5 - 2948.1)

h4 = 3027.3 kj/kg

at P5 = 2 MPa and T5 = 450⁰c

h5 = 3358.2 kj/kg,  s5 = 7.2815 kj/kgk

at P6 , x6 = 0.95  and s5 = s6

using nt = 0.85 we can calculate for h6 and h6s

from the chart attached below we can see that

p6 = 9.73 kPa, h6 = 2463.3 kj/kg

B) the net power output

solution is attached below

c) thermal efficiency

thermal efficiency = 1 - \(\frac{qout}{qin}\) = 1 - ( 2273.7/ 3603.8) = 36.9% ≈ 37%

Assuming  the wedge has an angle of 5°.The minimum value of the force P that is required to begin moving the block up the incline is: 322.84 N.

First step

Using this formula to find the weight of the block

W=mg

W=27×9.81

W=264.87 N

Second step

Angles of friction ∅A and ∅B

∅A=tan^-1(μA)

∅A=tan^-1(0.70)

∅A=34.99°

∅B=tan^-1(μB)

∅B=tan^-1(0.40)

∅B=21.80°

Third step

Equate the sum of forces in m-direction to 0 in order to find the reaction force at B.

∑fm=0

W sin (∅A+20°)  + RB cos (∅B+∅A)=0

264.87 sin(34.99°+20°) + RB cos (21.80°+34.99°)=0

216.94+0.5477Rb=0

RB=216.94/0.5477

RB=396.09 N

Fourth step

Equate the sum of forces in x-direction to 0 in order to find force Rc.

∑fx=0

RB cos (∅B) - RC cos (∅B+ 5°)=0

396.09 cos(21.80°) - RC cos (21.80°+5°)=0

RC=396.09 cos(21.80°)/cos(26.80°)

RC=412.02 N

Last step

Equate the sum of forces in y-direction to 0 in order to find force P required to move the block up the incline.

∑fy=0

RB sin (∅B) + RC sin (∅B)-P=0

P=Rb sin (∅B) + RC sin (5°+∅B)

P=396.09 sin(21.80°) +412.02sin (5°+21.80°)

P=322.84 N

Inconclusion the minimum value of the force P that is required to begin moving the block up the incline is: 322.84 N.

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False. In a balanced transportation model, supply may not always equal demand, and constraints cannot be treated as equalities.

The model aims to minimize transportation costs while meeting supply and demand requirements. It involves assigning quantities of goods from supply sources to demand destinations, considering constraints such as capacity limits and availability. Inequality constraints arise due to variations in supply and demand quantities. The model may use linear programming techniques to optimize the transportation plan, ensuring that constraints are satisfied while minimizing costs. Therefore, supply does not necessarily equal demand, and constraints are not treated as equalities in a balanced transportation model.

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Answer:

599.7 m

approximately 600 m

Explanation:

initial speed of the rocket = 0

net acceleration upwards = 7 m/s^2

the engine cuts out 10 sec after take off

maximum height reached = ?

we neglect air resistance

To get the velocity of the rocket at the point where the engine cuts off, we use the equation

v = u + at

where

v is the velocity at this point where the engine stops = ?

u is the initial velocity of the rocket from rest = 0 m/s

a is the net acceleration upwards = 7 m/s^2

t is the time the engine runs = 10 s

substituting, we have

v = 0 + (7 x 10)

v = 70 m/s

to get the distance from the ground to this point, we use the equation

\(v^{2}\) = \(u^{2}\) + 2as

where

v is the final velocity at the the height where the engine is cut out = 70 m/s

u is the initial speed at the ground = 0 m/s

a is the net acceleration on the rocket = 7 m/s^2

s is the distance from the ground to this point

substituting, we have

\(70^{2}\) = \(0^{2}\) + 2(7 x s)

4900 = 14s

s = 4900/14 = 350 m

After this point when the engine cuts out, the rocket experiences an acceleration proportional to the acceleration due to gravity 9.81 m/s^2 downwards, and slows down gradually before coming to a stop at the maximum height.

To get this height, we use the equation

\(v^{2}\) = \(u^{2}\) - 2gs   (the negative sign is due to the downward direction of the acceleration g)

where

v is the final velocity at the maximum height = 0 m/s (it comes to a stop)

u is the speed at the instance that the engine is cut out = 70 m/s

g is the acceleration due to gravity = 9.81 m/s^2

s is the distance from this point to the maximum height

substituting values, we have

\(0^{2}\) = \(70^{2}\) - 2(9.81 x s)

0 = 4900 - 19.62s

4900 = 19.62s

s = 4900/19.62 = 249.7 m

The maximum height that will be reached = 350 m + 249.7 m = 599.7 m

approximately 600 m

To find how many ways, you must simply multiply 10 tasks and 5 exercises

    Ways: = 10 * 5 = 50 ways

Hope that helps!

Answer:

10

Explanation:

The speed of the chain is the same for both sprockets.

vᵢ = vₒ

ωᵢ rᵢ = ωₒ rₒ

The sprockets have the same pitch, so the radius is proportional to the number of teeth.  So we can say:

ωᵢ nᵢ = ωₒ nₒ

Plugging in values:

(60 rpm) (40) = (240 rpm) nₒ

nₒ = 10

The output sprocket has 10 teeth.  This makes sense, if the output sprocket is 4 times smaller, it will turn 4 times faster than the input sprocket.

Answer: it is 10 teeth and the guy under me already explained soo

Explanation:

Answer:

2)

a)  to the right of the dipole    E_total = kq [1 / (r + a)² - 1 / r²]

b)To the left of the dipole      E_total = - k q [1 / r² - 1 / (r + a)²]

c) at a point between the dipole, that is -a <x <a  

      E_total = kq [1 / x² + 1 / (2a-x)²]

d) on the vertical line at the midpoint of the dipole (x = 0)

E_toal = 2 kq 1 / (a ​​+ y)² cos θ

Explanation:

2) they ask us for the electric field in different positions between the dipole and a point of interest. Using the principle of superposition.

This principle states that we can analyze the field created by each charge separately and add its value and this will be the field at that point

Let's analyze each point separately.

The test charge is a positive charge and in the reference frame it is at the midpoint between the two charges.

a) to the right of the dipole

The electric charge creates an outgoing field, to the right, but as it is further away the field is of less intensity

           E₊ = k q / (r + a)²

where 2a is the distance between the charges of the dipole and the field is to the right

the negative charge creates an incoming field of magnitude

           E₋ = -k q / r²

The field is to the left

therefore the total field is the sum of these two fields

           E_total = E₊ + E₋

           E_total = kq [1 / (r + a)² - 1 / r²]

we can see that the field to the right of the dipole is incoming and of magnitude more similar to the field of the negative charge as the distance increases.

b) To the left of the dipole

The result is similar to the previous one by the opposite sign, since the closest charge is the positive one

E₊ is to the left and E₋ is to the right

          E_total = - k q [1 / r² - 1 / (r + a)²]

We see that this field is also directed to the left

c) at a point between the dipole, that is -a <x <a

In this case the E₊ field points to the right and the E₋ field points to the right

                      E₊ = k q 1 / x²

                      E₋ = k q 1 / (2a-x)²

                      E_total = kq [1 / x² + 1 / (2a-x)²]

in this case the field points to the right

d) on the vertical line at the midpoint of the dipole (x = 0)

    In this case the E₊ field points in the direction of the positive charge and the test charge

    in E₋ field the ni is between the test charge and the negative charge,

the resultant of a horizontal field in zirconium on the x axis (where the negative charge is)

                      E₊ = kq 1 / (a ​​+ y) 2

                      E₋ = kp 1 / (a ​​+ y) 2

                      E_total = E₊ₓ + E_{-x}

                      E_toal = 2 kq 1 / (a ​​+ y)² cos θ

e) same as the previous part, but on the negative side

                        E_toal = 2 kq 1 / (a ​​+ y)² cos θ

When analyzing the previous answer there is no point where the field is zero

The different configurations are outlined in the attached

3) We are asked to repeat part 2 changing the negative charge for a positive one, so in this case the two charges are positive

a) to the right

in this case the two field goes to the right

           E_total = kq [1 / (r + a)² + 1 / r²]

b) to the left

            E_total = - kq [1 / (r + a)² + 1 / r²]

c) between the two charges

E₊ goes to the right

E₋ goes to the left

            E_total = kq [1 / x² - 1 / (2a-x)²]

d) between vertical line at x = 0

E₊ salient between test charge and positive charge

           E_total = 2 kq 1 / (a ​​+ y)² sin θ

In this configuration at the point between the two charges the field is zero

Answer: quarter turn

Explanation:  There are two basic types of valves ball valves and quarter turn valves or unblocks the hole, either allowing or preventing fluid flow.

The maximum peak output voltage and current if the supply voltages are changed to +15 V and --1V will be 15V and 0.1A.

From the information given, the supply voltages are +15V and -15V. The maximum peak output voltage will be 15V.

The maximum peak current will be:

= 15/150

= 0.1A

In conclusion, the maximum peak output voltage and current will be 15V and 0.1A.

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