If the continuity equation is applicable, what will happen to the air density (rho) if the cross sectional area of a tube changes? (low speed, subsonic and incompressible flow:A) rho1 = rho2B) rho1 > rho2C) The density depends on the change of the tube area.D) rho1 < rho2

Answer:

0.35

Explanation:

We resolve the component of the weight of the car along and perpendicular to the grade. We have mgsinФ and mgcosФ where Ф = angle of grade.

Now, the frictional force f = μN = μmgcosФ where μ = coefficient of friction

So, the net force along the grade is F = mgsinФ - μmgcosФ.

The work done by this force moving a distance, d along the grade is

W = (mgsinФ - μmgcosФ)d

This work equals the change in kinetic energy of the car. So ΔK = 1/2m(v₂² - v₁²) = W = (mgsinФ - μmgcosФ)d

1/2m(v₂² - v₁²) = (mgsinФ - μmgcosФ)d

1/2(v₂² - v₁²) = (gsinФ - μgcosФ)d

(v₂² - v₁²)/2d = (gsinФ - μgcosФ)

dividing through by gcosФ, we have

(v₂² - v₁²)/2dgcosФ = (gsinФ/gcosФ) - μgcosФ/gcosФ

(v₂² - v₁²)/2dgcosФ = tanФ -  μ

μ = tanФ - (v₂² - v₁²)/2dgcosФ

given that tanФ = 3% = 3/100 and 1 + tan²Ф = 1/cos²Ф, cosФ = 1/(√1 + tan²Ф) = 1/(√1 + (3/100)²) = 1/(√1 + (9/10000)) = 1/(√10000 + 9/10000) = 1/√(10009/10000) = 100/√10009 = 100/100.05 = 0.9995.

Also, given that v₁ = 90 km/h = 90 × 1000/3600 m/s = 25 m/s and v₂ = 45 km/h = 45 × 1000/3600 m/s = 12.5 m/s, d = 75 m and g = 9.8 m/s².

So, substituting the values of the variables into the equation, we have

μ = tanФ - (v₂² - v₁²)/2dgcosФ

μ = 3/100 - ((12.5 m/s)² - (25 m/s)²)/(2 × 75 m × 9.8 m/s² × 0.9995)

μ = 3/100 - ((156.25 m/s)² - (625 m/s)²)/1,469.265 m²/s²

μ = 3/100 - (-468.75 m²/s²)/1,469.265 m²/s²

μ = 3/100 + 468.75 m²/s²/1,469.265 m²/s²

μ = 0.03 + 0.32

μ = 0.35

So, theoretical friction  coefficient is 0.35

Answer:

Engineering drawing abbreviations and symbols are used to communicate and detail the characteristics of an engineering drawing.

There are many abbreviations common to the vocabulary of people who work with engineering drawings in the manufacture and inspection of parts and assemblies.

Technical standards exist to provide glossaries of abbreviations, acronyms, and symbols that may be found on engineering drawings. Many corporations have such standards, which define some terms and symbols specific to them; on the national and international level, like BS8110 or Eurocode 2 as an example.

Hope that help :)

The force of attraction between a cation with a valence of +1 and an anion with a valence of -1 can be calculated using Coulomb's law. the force of attraction between a cation with a valence of +1 and an anion with a valence of -1, separated by a distance of 1.9 nm, is approximately 4.74 x 10^8 N.

What is Coulomb's law?

Coulomb's law is a fundamental law in electromagnetism that describes the interaction between electrically charged particles. It states that the force of interaction between two point charges is directly proportional to the magnitude of the charges and inversely proportional to the square of the distance between them. Mathematically, the law can be expressed as:

F = k * (q1 * q2) / r^2

The force of attraction between a cation with a valence of +1 and an anion with a valence of -1 can be calculated using Coulomb's law. Coulomb's law states that the force of attraction between two charged particles is proportional to the product of their charges and inversely proportional to the square of the distance between them.

The equation for the force of attraction between two charged particles (q1 and q2) separated by a distance of r is given by:

F = k * (q1 * q2) / r^2

where k is the Coulomb's constant (8.987551787 x 10^9 N * m^2 / C^2).

For a cation with a valence of +1 and an anion with a valence of -1, the force of attraction can be calculated as follows:

q1 = +1, q2 = -1, r = 1.9 x 10^-9 m

F = k * (q1 * q2) / r^2

= 8.987551787 x 10^9 N * m^2 / C^2 * (1 * -1) / (1.9 x 10^-9 m)^2

= 4.74 x 10^8 N

So, the force of attraction between a cation with a valence of +1 and an anion with a valence of -1, separated by a distance of 1.9 nm, is approximately 4.74 x 10^8 N.

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The precursor to modern operating systems that allowed programs to be processed without human interaction is B) batch processor.

A batch processor is a system that processes a collection of jobs or programs in a sequential manner, without requiring constant human intervention. It allows multiple programs or jobs to be submitted for execution as a batch, and the system automatically executes them one after another without the need for manual intervention between jobs.

In a batch processing environment, programs are typically stored on punched cards, magnetic tapes, or other storage media. The batch processor reads the programs and data from the storage media, executes the programs, and produces the desired output. This process continues until all the jobs in the batch have been processed.

Batch processing was a significant advancement in the early days of computing when computers were primarily used for scientific and business applications. It allowed programs to be executed without the need for constant human attention, thereby improving efficiency and productivity.

Resident monitor (A) refers to a portion of the operating system that remains in memory at all times and provides basic system services. Middleware (C) refers to software that acts as an intermediary between different applications or systems, facilitating communication and data exchange. Spooling (D) stands for "Simultaneous Peripheral Operation On-Line" and refers to a technique used to improve input/output performance by storing data in a buffer before it is processed.

Among the given options, batch processor (B) is the best choice as the precursor to modern operating systems that enabled automated, non-interactive processing of programs.

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If the oil pressure gauge fluctuates over a wide range from zero to normal operating pressure, the most likely cause is air lock in the scavenge pump intake.

Oil pressure gauge fluctuation is a common problem for the engine. It can occur due to various reasons such as faulty oil pressure gauge, oil pump failure, improper maintenance of the engine, oil leakage, and much more. But if the oil pressure gauge fluctuates over a wide range from zero to normal operating pressure, the most likely cause is air lock in the scavenge pump intake.The scavenge pump in the engine is used to remove the oil from the engine's crankcase and delivers it back to the oil tank. If there is an air lock in the scavenge pump intake, then it will not pump the oil properly from the crankcase and deliver it back to the oil tank. It will cause the oil pressure to fluctuate over a wide range from zero to normal operating pressure.To fix this issue, you should first check the oil level in the engine and make sure that it is at the proper level. After that, you can check the scavenge pump intake for any air lock or blockage. If there is an air lock, then you need to remove it. If there is a blockage, then you need to remove the blockage to get the oil pump working again properly.

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Answer:

the rate of flow = 29.28 ×10⁻³ m³/s or 0.029 m³/s

Explanation:

Given:

Diameter of the pipe = 100mm = 0.1m

Contraction ratio = 0.5

thus, diameter at the throat of venturimeter = 0.5×0.1m = 0.05m

The formula for discharge through a venturimeter is given as:

Where,

is the coefficient of discharge = 0.97 (given)

A₁ = Area of the pipe

A₁ =  

A₂ = Area at the throat

A₂ =  

g = acceleration due to gravity = 9.8m/s²

Now,

The gauge pressure at throat = Absolute pressure - The atmospheric pressure

⇒The gauge pressure at throat = 2 - 10.3 = -8.3 m (Atmosphric pressure = 10.3 m of water)

Thus, the pressure difference at the throat and the pipe = 3- (-8.3) = 11.3m

Substituting the values in the discharge formula we get

or

or

Q = 29.28 ×10⁻³ m³/s

Hence, the rate of flow = 29.28 ×10⁻³ m³/s or 0.029 m³/s

Hope This Helps :D

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.

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Bend strength and rupture modulus are other names for flexural strength.

The range of the flexural strength of flax and kenaf is observed to be between 167 and 169 MPa, which is almost the same. They are bast fibers, which are multicellular layers from 40% to 45% of plants that have been knocked together. Flexural strength reveals the amount of force needed to fracture a test sample with a certain measurement diameter. When this limit is reached, the test specimen cracks. The material can withstand more impacting forces the higher the value. Flexural testing gauges a material's stiffness or resistance to bending by measuring the amount of force needed to bend a plastic beam.

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The Rule That Packets Not Originating From Inside Your LAN Should Not Be Forwarded Relates to routers

In order to transfer data between two or more packet-switched computer networks, a router—either real or virtual—is used.. The Internet Protocol address (IP address) of the destination is examined by a router, which then determines the optimal path for the data packet to take to get there.

According to the various application categories, there are five different types of routers available. They include VPN routers, core routers, edge routers, wireless routers, and wired routers. The aforementioned fundamental details can be used to aid in making the best router selection possible.

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The minimum pressure at which the water-supplying pump must operate is approximately 2.45 x 10⁶ Pa or 2450 bar.

To calculate the minimum pressure at which the water-supplying pump must operate, we need to use the formula for the volumetric flow rate of a fluid:

Q = A * v

where,

Q =the flow rate

A = the cross-sectional area of the hole

v = the velocity of the fluid.

We can use this formula to determine the velocity of the water, and then use the Bernoulli equation to determine the minimum pressure.

From the question:

ṁ=0.05 kg/s

ρ= 996 kg/m

d= 0.02 cm

Calculating the area of the hole:

A = (π/4) * (d²)

where,

d is the diameter of the hole, and pi is approximately 3.14.

d = 0.02 cm = 0.0002 m

A = (3.14/4) * (0.0002 m)² = 1.57 x 10⁻⁷ m²

Using the mass flow rate to calculate the velocity of water:

Q = ṁ / ρ

where,

Q = the volumetric flow rate

ṁ = the mass flow rate

ρ = the density of water.

Q = (0.05 kg/s) / (996 kg/m³)

Q = 5 x 10⁻⁵ m³/s

Using the area of the hole to find the velocity of water

v = Q / A

v = (5 x 10⁻⁵ m³/s) / (1.57 x 10⁻⁷m²)

v = 318.8 m/s

Using the Bernoulli equation to find the pressure:

P = Patm + (1/2) * ρ * v²

where,

P = the pressure

Patm = the atmospheric pressure

ρ =the density of water

v = the velocity of the water.

P = Patm + (1/2) * (996 kg/m³) * (318.8 m/s)²

Hence, the minimum pressure at which the water-supplying pump must operate is approximately 2.45 x 10⁶ Pa or 2450 bar.

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Answer:true

Explanation:

Answer:

A power supply unit (PSU) converts mains AC to low-voltage regulated DC power for the internal components of a computer. Modern personal computers universally use switched-mode power supplies

Answer:

Volt is the SI (Standard International) unit of electrical potential of the..

Explanation:

Answer:

This problem has been solved!

See the answer

7. The voltage that must be less than the breakdown voltage of the diode in order to prevent damage to the diode is the ________.

peak inverse voltage

maximum diode voltage

reverse surge voltage

maximum peak voltage

8. Assume the input signal to a rectifier circuit has a peak value of Vm = 12 V and is at a frequency of 60 Hz. Assume the output load resistance is R = 2kΩ and the ripple voltage is to be limited to Vr= 0.4 V. Determine the capacitance required to yield this specification for a (a) full-wave rectifier and (b) half-wave rectifier. Show all work.

9.A full-wave rectifier is to be designed to produce a peak output voltage of 12 V, deliver 120 mA to the load, and produce an output with a ripple of not more than 5 percent. An input line voltage of 120 V (rms), 60 Hz is available. Consider a bridge type rectifier. Specify the transformer ratio and the size of the required filter capacitor. Show all work.

Explanation:

Experiential estimating is being used to determine the project duration and total cost.

What is Experiential?
Experiential learning is an educational philosophy that focuses on learning through experience. It involves actively engaging in a hands-on, sensory-based activity that encourages and facilitates learning. This type of learning is often associated with the physical environment, where the learner interacts directly with the environment and the activity to gain knowledge. Experiential learning can also involve activities such as role-playing, simulations, and problem-solving. This approach to learning is based on the idea that learning is most effective when it is meaningful and relevant to the learner. It promotes active learning, reflection, and the development of critical thinking skills.

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Answer:

A, Hazardous Material

Explanation: Because that defines a hazardous material. A Safety Data Sheet isn't a material, so it can cause harm.

The use GCF method is correctly illustrated recursively below

The GCF is the greatest common factor that can be used to correctly divide two or more numbers without a reminder.

For example,

The parameters 20 and 30, to know the greatest common factor, write out each factors of 20 and 30

The factors of 20 = 1, 2, 4, 5, 10, and 20

The factors of 30 = 1, 2, 3, 5, 6, 10, 15, and 30

Therefore the GCF = 10

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The process comprises breaking the truss into individual pieces and analyzing each piece as a separate rigid body.

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Job enlargement, also known as horizontal job loading, involves increasing the number of tasks that an employee is responsible for within their existing job role. This is in contrast to job enrichment, which involves increasing the level of responsibility or autonomy of an employee within their job role.

Some other terms that are related to job enlargement and job enrichment include:

Optimal job loading: This refers to the optimal balance of tasks and responsibilities within a job role, which can vary depending on the individual employee and the requirements of the job.

Lateral job loading: This refers to the addition of tasks that are similar in nature to the employee's existing responsibilities, but do not involve an increase in responsibility or autonomy.

Parallel job loading: This refers to the addition of tasks that are similar in nature to the employee's existing responsibilities, and may involve an increase in responsibility or autonomy.

Vertical job loading: This refers to the addition of tasks that involve an increase in responsibility or autonomy within the employee's job role. This can also be referred to as job enrichment.

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To calculate the time required for the center of the sphere to reach 335 o C, we need to use the following formula:

()=∞+(0−∞)−ℎ/

where T(t) is the temperature at time t, T0 is the initial temperature, T∞ is the temperature of the surrounding fluid, h is the convective heat transfer coefficient, A is the surface area of the sphere, m is the mass of the sphere, c is the specific heat of the sphere material, and t is time.

Using the given values, we can calculate the time required for the center of the sphere to reach 335 o C:

A = πd^2/4 = 7.85x10^-5 m^2

m = ρV = (4/3)π(d/2)^3ρ = 1.178 kg

t = (1/α) ln[(T0 - T∞)/(T(t) - T∞)] = 79.26 seconds

To estimate the time required for the center of the sphere to reach 50 o C in the water bath, we use the same formula, but with a different value of h:

h = 6000 W/m^2-°C

T∞ = 20 °C

T0 = 335 °C

m = 1.178 kg

c = 1000 J/kg-°C (since C = 1 kJ/kg-°C)

T(t) = 50 °C

t = (1/α) ln[(T0 - T∞)/(T(t) - T∞)] = 242.18 seconds

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The utility control center calculates ACE based on tie-line flows, and the AGC module sends control signals to generators.

One of its main jobs is to figure out the Area Control Error (ACE) using the amounts of electricity flowing between different areas. The ACE shows the contrast between the planned and real power trade between various parts or control areas in the power grid.

The AGC module tells the generators to increase or decrease their power. The control signals are usually sent through a communication network. This helps the AGC adjust the power levels of generators based on the changing load conditions.

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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

a. Based on the 43 years of record and the fact that the 1972 flood occurred once during this interval, it can be estimated that a flood of this magnitude would be expected to occur once every 43 years, on average.

b. Using the graph paper, plot a recurrence curve for the Rapid Creek flood data. Ignoring the 1972 flood, an eyeball, best fit, and a straight line should be plotted through the points. The 100-year flood discharge can be determined by looking at the point on the graph which is at the 100-year recurrence interval.

c. The recurrence interval of the 1972 flood, using the curve drawn, can be determined by looking at the point on the graph which is nearest to the 1972 flood discharge.

d. Given the range of values from a and c above, it can be stated that floods as large as the one which occurred in 1972 are rare events and occur infrequently.

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Fault location techniques are used in power systems for accurate pinpointing of the fault position.

This paper presents a comparative study between two fault location methods in distribution network with Distributed Generation (DG). Both methods are based on computing the impedance using fundamental voltage and current signals. The first method uses one-end information and the second uses both ends

This function takes in three ints, a, b, and c, and uses the Math.abs() method to calculate the absolute value of the difference between each pair of ints (a and b, a and c, b and c).

The function that takes in three ints, a, b, and c, and uses the Math.abs() method to calculate the absolute value of the difference between each pair of ints (a and b, a and c, b and c).

The function then compares the absolute value of each difference to 10 and returns true if one of them is greater or equal than 10. If none of the differences are greater than or equal to 10, the function returns false.

Therefore, This function takes in three ints, a, b, and c, and uses the Math.abs() method to calculate the absolute value of the difference between each pair of ints (a and b, a and c, b and c).

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Answer:

The meole fraction of acetone in the outlet liquid is \(x_1 = 0.0072\)

Explanation:

1.

The schematic diagram to represent this process is shown in the diagram attached below:

2.

the mole fraction of acetone in the outlet liquid is determined as follows:

solute from Basis Gas flow rate \(G_s = 10(1-0.012) =9.88 kmol/hr\)

Let the entering mole be :\(y_1 = 1.2\) % = 0.012

\(y_1 =(\dfrac{y_1}{1-y_1})\)

\(y_1 =(\dfrac{0.012}{1-0.012})\)

\(y_1 =0.012\)

Let the outlet gas concentration be \(y_2\) = 0.1% = 0.001

\(y_2 = 0.001\)

Thus; the mole fraction of acetone in the outlet liquid is:

\(G_s y_1 + L_s x_2 = y_2 L_y + L_s x_1\)

\(9.88(0.012-0.001)=15*x_1\)

\(9.88(0.011) = 15x_1\)

\(x_1 = \dfrac{0.10868}{15}\)

\(x_1 = 0.0072\)

The mole fraction of acetone in the outlet liquid is \(x_1 = 0.0072\)

The frequency  is 6.67 kHz and the duty cycle is:16.67%.

​Given:

Pulse width=25

Period=150

Frequency:

Frequency=1/(150×10^-6)

Frequency=1/0.00015

Frequency=6.666 kHz

Frequency=6.67 kHz (Approximately)

Duty cycle:

Duty cycle=(25×10^-6)/ (150×10^-6)×100%

Duty cycle=16.67%

Therefore the frequency  is 6.67 kHz and the duty cycle is:16.67%.

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When required to drill holes on a roof that has no power supply the best drill for the job would be a cordless drill.

It should be noted that a power supply is an electrical device which supplies electric power to an electrical load.

In this case, the main purpose of a power supply is simply to be able to convert electric current from the source to the correct current, and frequency.

In this case, the power supply unit converts the main AC to a low-voltage regulated DC power. Therefore, When required to drill holes on a roof that has no power supply the best drill for the job would be a cordless drill.

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Answer:

Explanation:

A scenario is an actual sequence of interactions (i.e., an instance) describing one specific situation; a use case is a general sequence of interactions (i.e., a class) describing all possible scenarios associated with a situation. Scenarios are used as examples and for clarifying details with the client. Use cases are used as complete descriptions to specify a user task or a set of related system features.

A function named reverse that accepts a dictionary from integers to strings as a parameter and returns a new dictionary of strings to integers that is the original's "reverse" is given below:

# Function to create an empty stack. It

# initializes size of stack as 0

def createStack():

   stack = []

   return stack

# Function to determine the size of the stack

def size(stack):

   return len(stack)

# Stack is empty if the size is 0

def isEmpty(stack):

   if size(stack) == 0:

       return true

# Function to add an item to stack . It

# increases size by 1

def push(stack, item):

  stack.append(item)

# Function to remove an item from stack.

# It decreases size by 1

def pop(stack):

   if isEmpty(stack):

       return

   return stack.pop()

# A stack based function to reverse a string

def reverse(string):

   n = len(string)

   # Create a empty stack

   stack = createStack()

   # Push all characters of string to stack

   for i in range(0, n, 1):

       push(stack, string[i])

   # Making the string empty since all

   # characters are saved in stack

   string = ""

   # Pop all characters of string and put

   # them back to string

   for i in range(0, n, 1):

       string += pop(stack)

   return string

# Driver code

s = "Geeksforgeeks"

print("The original string is : ", end="")

print(s)

print("The reversed string(using stack) is : ", end="")

print(reverse(s))

The original string is : Geeksforgeeks

The reversed string(using stack) is : skeegrofskeeG

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