What is the maximum theoretical efficiency possible for an engine operating between 100°c and 400°c?

The amount of heat required to raise the temperature is 706.05J.

To find the answer, we need to know more about the specific heat capacity.

                                 Q=msΔT

                 [tex]m=2.50g=2.50*10^{-3}kg\\s=0.523 J/gK=\frac{0.523}{10^{-3}} J/kgK=523J/kgK\\T_1=65^0C=338K\\T_2=600^0C=873K\\[/tex]

            [tex]Q=2.5*10^{-3}*523*(873-333)=706.05 J[/tex]

Thus, we can conclude that, the amount of heat required to raise the temperature is 706.05J.

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The energy of the red light emitted by a neon atom is 2.827*10^-19J

Light produces photons which might be small packets of power. as the atoms of an object are heated, it results within the manufacturing of photons. The electrons locate pleasure from the warmth and produce extra energy. This power emits inside the shape of a photon, and extra photons are released because the cloth becomes hotter.

The SI unit of wavelength is metre typically denoted as m. at the same time as measuring wavelength the multiples or fractions of a metre is also used. extensively, exponential powers of 10 are used while wavelengths are of a huge property. while there are shorter wavelengths they're expressed as negative exponential.

c=3*10^8 h=6.626*10^-34

wavelength= 703.2*10^-9m

E=2.827*10^-19J

Hence energy of the red light is 2.827*10^-19J.

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The work done by the unstretched spring is 5 J.

Hooke's law states that the extension of a given material is directly proportional to the force applied as log as the elastic limit is not exceeded. Thus we have to know that; F = Ke

F = force applied

K = force constant

e = extension

Using the graph;

K = F/e

F = 200N

e = 5 cm or 5 * 10^-2 m

K = 200N/ 5 * 10^-2 m

K = 4000 N/m

Now;

Work = 1/2Ke^2

Work = 0.5 *  4000 N/m * (5 * 10^-2 m)^2

Work = 5 J

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Answer:

.513 Ω

Explanation:

The current through the 20 Ω load can be found to be

  (12-.3) v / 20 Ω  = .585 A

This current also flows through the internal resistance and produces the .3 v drop across this resistor

V/I  = R

.3 v / .585 A = .513 Ω  internal resistance

The distance should be increased by 0.4 mm.

Lens formula is used here.

For a spherical mirror, the focal length, image distance, and object distance are all related by the lens equation, often known as the lens formula.

It is given by,

[tex]\frac{1}{f} = \frac{1}{v} -\frac{1}{u}[/tex]

where,

f = focal length

u = distance of the object

v = distance of the image formed

Given,

f = 40 mm

u = 4 m = 4000 mm

v =?

Using the lens formula,

[tex]\frac{1}{f} = \frac{1}{v} -\frac{1}{u}[/tex]

[tex]\frac{1}{40} = \frac{1}{v} -\frac{1}{4000}\\[/tex]

[tex]\frac{1}{v} = \frac{1}{40} + \frac{1}{4000}[/tex]

v = 39.6 mm

Change in distance = 40 - 39.6

                                 = 0.4 mm

Hence, the distance should be increased by 0.4 mm.

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0.4 mm

The distance should be increased by 0.4 mm.

We use lens formula here,

[tex]\frac{1}{f}[/tex] = [tex]\frac{1}{v}[/tex] - [tex]\frac{1}{u}[/tex]

where,

f =  focal length

u = distance of the object

v = distance of the image formed

Given that,

f = 40 mm

u = 4 m = 4000 mm

v = ?

Using the lens formula, we get:

[tex]\frac{1}{f}[/tex] = [tex]\frac{1}{v}[/tex] - [tex]\frac{1}{u}[/tex]

[tex]\frac{1}{40}[/tex] = [tex]\frac{1}{v}[/tex] - [tex]\frac{1}{4000}[/tex]

[tex]\frac{1}{v}[/tex] = [tex]\frac{1}{40}[/tex] + [tex]\frac{1}{4000}[/tex]

v = 39.6 mm

Change in the distance = 40 - 39.6

                                        = 0.4 mm

Therefore, the distance has to be increased by 0.4 mm

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[tex]\huge\underline{\underline{\boxed{\mathbb {SOLUTION:}}}}[/tex]

Apply the Conservation of Linear Momentum:

[tex]\longrightarrow \sf{m_1 u_1 +1 + m_2 u_2 = m_1v_1 + m_2v_2}[/tex]

[tex]\longrightarrow \sf{(7 \cdot 4) + (1 \cdot 0) = (7 \cdot 3) + 1v_2}[/tex]

Solving further:

[tex]\longrightarrow \sf{28+0 = 21+ v_2}[/tex]

[tex]\longrightarrow \sf{28 = 21+ v_2}[/tex]

[tex]\longrightarrow \sf{v_2= 28- 21}[/tex]

▪ [tex]\longrightarrow \sf{v_2= 7 \dfrac{m}{s} }[/tex]

[tex]\huge\underline{\underline{\boxed{\mathbb {ANSWER:}}}}[/tex]

[tex] \boxed{\bm{v_2= 7 \dfrac{m}{s}}}[/tex]

The probability that she makes 6 of them is 0.242

Simply put, the probability is the likelihood that something will occur. When we don't know how an event will turn out, we can discuss the likelihood or likelihood of several outcomes. Statistics is the study of events that follow a probability distribution.

A basketball player has a probability = 0. 603

The player shoots 10 free throws.

Let x be a random variable that represents the number of free throws made by a basketball player.

x ≈ Binomial (n,p)

p = 0.603

n= 10

P(x=k) = [tex]_nC_k p^k (1-p)^{n-k}[/tex]

P(6)    [tex]= _1_0C_6 (0.603)^6 (1-0.603)^{10-6}[/tex]

           [tex]=210 (0.048) (0.397)^4[/tex]

           [tex]=210 (0.048)(0.024)[/tex]

           [tex]= 0.242[/tex]

Therefore, the probability that she makes 6 of them is 0.242

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The volume of a unit cell is 4.3 × 10⁻¹⁰ pm³.

We need to start with the features of a face-centered cubic system in order to be able to determine the edge length of the unit cell.

As we know, a face-centered cubic system is characterized by a unit cell that has a total of 14 lattice points.

Now, these lattice points will contain

Assume that r is the edge length of the unit cell and that r is the radius of an iron atom. You'll see that we can represent x in terms of the cell's diagonal, d, using Pythagoras' Theorem.

        [tex]d^2 = x^2+ x^2[/tex]

        [tex]d^2=2x^2[/tex]

The cell's diagonal will be equal to one radius from the atom in the top corner, two from the atom in the face, and one from the atom in the lower corner.

[tex]d= r+2r+r = 4r[/tex]

This means that we have,

[tex](4r)^2=2x^2[/tex]

[tex]16.r^2 = 2x^2 \\x=\sqrt{8r^2}[/tex]

This is equivalent to

[tex]x=2\sqrt{2} r = 2\sqrt{2} .124= 350.7 pm[/tex]

So, the edge length of the unit cell is equal to 350.7 pm.

The volume of a unit cell. As we know, the volume of a cube is given by the formula

V = [tex]l[/tex] × [tex]l[/tex] × [tex]l[/tex]

[tex]l[/tex] = the edge length of the cube

Volume = (350.7)³ pm³

            = 4.3 × 10⁻¹⁰ pm³

            = 4.3e-30 cm³

Therefore, the volume of the unit cell is 4.3 × 10⁻¹⁰ pm³.

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Answer:

ΔV=Vi + Vf

           2

ΔV= 11m/s + 19m/s

                 2

ΔV= 15m/s

Explanation:

The sun would look like from earth if the entire photosphere were the same temperature as a sunspot then wouldn't see anything from Earth but darkness

The lowest layer of the sun is the photosphere. It is the outer layer of the sun. It is the deepest region of the luminous object. It is composed of convection cells which are called granules . The Sun's photosphere has a temperature between 4,500 and 6,000 K

Sunspots are areas where the magnetic field is about 2,500 times stronger than Earth's, much higher than anywhere else on the Sun.  Because of the strong magnetic field, the magnetic pressure increases while the surrounding atmospheric pressure decreases.

Sunspots tend to occur in pairs that have magnetic fields pointing in opposite directions.  A typical spot consists of a dark region called the umbra, surrounded by a lighter region known as the penumbra.  

The sunspots appear relatively dark because the surrounding surface of the Sun (the photosphere) is about 10,000 degrees F, while the umbra is about 6,300 degrees F.  Sunspots are quite large as an average size is about the same size as the Earth.

If the photosphere were of the same temperature as the sunspot, then we wouldn't see anything from Earth but darkness. We were only able to see and recognize the sunspot because it is surrounded by the visible photosphere.

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The total work done of 0.018 joules is needed to move the charges apart and double the distance between them.

We have two electric charges q(A) = 1μc and q(B) = -2μc kept at a distance 0.5 meter apart.

We have to calculate  much work is needed to move the charges apart and double the distance between them.

The potential energy of the system of two charges separated by a distance is given by -

[tex]U = \frac{1}{4\pi\epsilon_{o} } \frac{qQ}{r}[/tex]

In order to solve this question, it is important to remember the work - energy theorem which states -

"The change in the energy of the body is equal to work done on it"

Hence, using this work -energy theorem in the question given to us we get -

[tex]U_{f} -U_{i} =W_{net}[/tex]

In our case -

[tex]U_{f} = \frac{1}{4\pi\epsilon_{o} } \frac{qQ}{2r}\\\\U_{i} = \frac{1}{4\pi\epsilon_{o} } \frac{qQ}{r}\\\\W=\frac{1}{4\pi\epsilon_{o} } \frac{qQ}{2r} - \frac{1}{4\pi\epsilon_{o} } \frac{qQ}{r}\\\\W = \frac{qQ}{4\pi\epsilon_{o}r} (\frac{1}{2} -1)\\\\W = 9\times 10^{9}\times \frac{1 \times 10^{-6} \times 2\times10^{-6} }{0.5} \times \frac{-1}{2}[/tex]

W = 0.018 joules

Hence, the total work done should be 0.018 joules.

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The time period of the spring - mass system undergoing simple harmonic motion is 5.024 seconds.

We have a mass weighing 16 pounds which is attached to a spring whose spring constant is 25 lb/ft . This complete spring - mass system is undergoing simple harmonic motion.

We have to calculate the time period of this simple harmonic motion.

The formula to calculate the time period of a spring - mass system undergoing simple harmonic motion is -

[tex]T=2\pi \sqrt{\frac{m}{k} }[/tex]

Where -

T is the time period of spring - mass simple harmonic motion.

m is the mass of body

k is the spring constant

In the question given -

mass (m) = 16 pounds

Spring constant (k) = 25 lb/ft

Substituting the values in the formula above -

T =  [tex]2\pi \sqrt{\frac{16}{25} }[/tex]  = [tex]\frac{2\pi \times4}{5}[/tex]  = 5.024 seconds

Hence, the time period of the spring - mass system undergoing simple harmonic motion is 5.024 seconds.

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The hydroxide ion concentration in a solution is 3.31*10^-8M.

To find the answer, we have to know more concentration of solution.

                     [tex]pH=-log[H_3O^+][/tex]

                   [tex][H_3O^+]=-10^{pH}=10^{-6.25}=5.6*10^{-7}M[/tex]

                        [tex]K_w=[H_3O^+][OH^-]\\Where,\\K_w=10^{-14}[/tex]

                         [tex][OH^-]=\frac{K_w}{[H_3O^+]} =\frac{10^{-14}}{10^{-6.52}}\\\\[/tex]

                           [tex][OH^-]=3.31*10^{-8}M[/tex]

Thus, we can conclude that, the hydroxide ion concentration in a solution is 3.31*10^-8M.

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The frequency of a photon is  1.06 * [tex]10^{18}[/tex] Hz.

The mass of a neutron is:

m = 1.67 * [tex]10^{-27}[/tex] kg

Since we know its speed, we can calculate the neutron's momentum:

p = mv = (1.67 * [tex]10^{-27}[/tex] kg)(1400 m/s) = 2.34 * [tex]10^{-24}[/tex] kg m /s

The problem says the photon has the same momentum of the neutron, p.  The photon momentum is given by

                                               p = h / λ              

where h is the Planck constant, and  λ  is the photon wavelength. If we re-arrange the equation and we use the momentum we found before, we can calculate the photon's wavelength:

p = h / λ =  [6.6 * [tex]10^{-34}[/tex] J/s]  /  [2.34 * [tex]10^{-24}[/tex] kg m /s ]

             = 2.82 * [tex]10^{-10}[/tex] m

And since we know the photon travels at speed of light c, we can now calculate the photon frequency:

f = c /  λ  =  [3 * [tex]10^{8}[/tex] m/s] / [ 2.82 * [tex]10^{-10}[/tex] m ]

              = 1.06 * [tex]10^{18}[/tex] Hz.

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The angle of refraction is a. 15°

Snell's law is described as “The ratio of the sine of the angle of incidence to the sine of the angle of refraction is a constant, for the light of a given colour and for the given team of media”.

By Snell's law:

η = sin i / sin r. i = 25, η = 1.6

1.6 = sin25° / sin r

sin r = sin25° / 1.6

= 0.4226/1.6= 0.2641

calculating

r = sin⁻¹( 0.2641 )

sin−1(0.264) is ≈ 15.3075

the angle of refraction is a. 15°

So the angle associated with the incidence and refraction can have a relation to the refractive index of the two mediums. Snell's First Law: It states that the incident ray, the refracted ray and the standard lie on the same plane.

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The cannonball will travel 311.8 m  horizontally. Among the given options, 310 m is the correct answer.

What is a Projectile ?

A projectile is any object like stone, ball or javelin that take a trajectory path when thrown into the air.

The following parameters are given from the question

Resolve the initial velocity into horizontal and vertical components. That is,

To calculate the range that is, the horizontal distance, will can first calculate the total time take by the ball to reach the ground.

Total time T = 2usinФ / g

Substitute all the parameters

T = (2 x 60sin30) / 10

T = (2 x 30) / 10

T = 60 / 10

T = 6 s

The horizontal distance R = ucosФ x T

R = 60 x cos 30 x 6

R = 311.77 m

Therefore, the cannonball will travel 312 m approximately horizontally.

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A drift velocity in physics is the average speed that charged particles—like electrons—achieve within a material as a result of an electric field. An electron will typically move randomly across a conductor at the Fermi velocity, producing an average velocity of zero. Applying an electric field causes a little net flow in one direction to be added to this random motion; this is the drift.

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Drift velocity

it is a average speed of the particle reached in a random motion

There are three basic steps in a nuclear fission chain reaction are Initiation, Propagation, Termination.

1. Initiation

The nuclear fission  of [tex]U_{235} ^{92}[/tex] is started by the absorption of a neutron; a single atom must react in order for the chain reaction to begin.

2. Propagation:-

With each stage producing additional product, this component of the process is repeated repeatedly. When [tex]U_{236} ^{92}[/tex] splits apart, neutrons are released, which start the nuclear fission of more uranium atoms.

3. Termination:

The chain will come to an end eventually. Termination might take place if the reactant [tex]U_{235} ^{92}[/tex] is exhausted or if the subsequent neutrons in the chain leave the sample without being absorbed by [tex]U_{235} ^{92}[/tex].

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Electric utility power blackouts will result in short electric power outages for the organization.

When the electrical power network, often known as the electrical grid, is down, there are power outages. Power outages can have a variety of causes, but they always tend to affect entire regions or cities rather than simply a single residence. For example, the Northeast blackout of 2003 left 50 million people without power from New York to Canada’s capital Ottawa.  

A place experiences a blackout when all power is lost. Power outages of this kind generally affect a sizable population over perhaps extraordinarily broad areas. Blackouts are particularly difficult to quickly fix because they typically result from significant damage to electrical generation facilities (such as structural damage from violent wind storms or lightning strikes). As a result, these types of outages can, in the worst-case scenarios, last for several weeks.

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A low partial pressure of oxygen promotes hemoglobin binding to carbon dioxide. This is an example of the  Haldane effect

The less saturated hemoglobin is and the lower the partial pressure of oxygen in the blood is, the more readily hemoglobin binds to carbon dioxide. This is an example of the Haldane effect.

The Haldane effect is a property of hemoglobin first described by John Scott Haldane, within which oxygenation of blood in the lungs displaces carbon dioxide from hemoglobin, increasing the removal of carbon dioxide. Consequently, oxygenated blood has a reduced affinity for carbon dioxide.

The Haldane effect is a property of hemoglobin first described by John Scott Haldane, within which oxygenation of blood in the lungs displaces carbon dioxide from hemoglobin, increasing the removal of carbon dioxide. Consequently, oxygenated blood has a reduced affinity for carbon dioxide.

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When a proton slows from 0. 8c up to 0. 6c, it’s kinetic energy decreases by what factor of 1/4.

Kinetic energy of a proton having a charge Q and Voltage V is given by:

Kinetic energy = Q×V

Where, Q is the charge

            V is the Voltage

Kinetic energy is directly proportional to the charge on a proton.

Let the kinetic energy of proton having charge 0.8C be KE₁ and kinetic energy of proton having charge 0.6C be KE₂

Hence KE₁ = 0.8 × V

KE₂ = 0.6 × V

Now, KE₁ / KE₂ = 0.8 V / 0.6 V

⇒          KE₁ / KE₂ = 4 / 3

⇒         KE₂ / KE₁ = 3/4

⇒         KE₂ = 3/4 times KE₁

Hence, the final kinetic energy becomes 3/4 times the initial kinetic energy.

The kinetic energy decreases by a factor of:

Change = KE₁ - KE₂

Change = KE₁ - 3/4KE₁

Change = 1/4 KE₁

Hence, the kinetic energy decreases by a factor of 1/4

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The power expended is calculated as 91,100 W.

We know that the acceleration can be obtained by the use of the equation;

v = u + at

v = final velocity

u = initial velocity

t = time

a = acceleration

a= v - u/t

a =  27 - 0/ 5.2 = 5.2 m/s^2

Thus;

v^2 = u^2 + 2as

but u= 0 m/s

v^2 = 2as

s =  v^2/2a

s = (27)^2/2 * 5.2

s = 70 m

Now;

Work done =  1300 kg * 5.2 m/s^2 * 70 m = 473200 J

Power = 473200 J/5.2 seconds

Power =  91,100 W

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The answer is (d) class

One or more objects may be created from a class.

Class objects normally have methods that perform useful operations on their data.

What is class?

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One or more objects may be created from a class.

Class objects contains methods which perform useful operations on their data.

A class in object-oriented programming refers to a template specification of the methods and variables in a a type of object.

An object is a specific instance of a class that holds real values rather than variables.

The class is a key concept in object-oriented programming.

We may also construct a class object and use it in another class. It is frequently used to improve the structure of class.

Hence, one or more objects may be created from a class.

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Ritemade machinery, inc., designs, makes, and sells a drill press. steel equipment company copies the design without ritemade's permission. steel's conduct is actionable provided that RiteMade's design is patented.

A design patent is a form of legal protection of the unique visual qualities of a manufactured item. A design patent may be granted if the product has a distinct configuration, distinct surface ornamentation or both or A patent is an exclusive right granted for an invention.

In other words, a patent is an exclusive right to a product or a process that generally provides a new way of doing something, or offers a new technical solution to a problem.

If you create original sketches of your designs, those sketches are protected by copyright law. That means that no one can copy, distribute, publicly display, etc. your sketch without your permission.

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The cost of operating the iron is $2186.49.

To find the answer, we have to know about the power.

              [tex]cost=power(kW)*t(hours)*(price/ kWh)[/tex]

                        [tex]P=IV=5.2*240=1248W[/tex]

                     [tex]t=2*365*24hours=17520hours.[/tex]

                  [tex]cost=1248*10^{-3}*17520*0.10} =\$2186.49[/tex]

Thus, we can conclude that the cost of operating the iron is $2186.49.

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From the first law of Thermodynamic, the change in internal energy of the system is 83 KJ.

The Internal energy of a system is the sum of the random distribution of kinetic and potential energies of its atoms or molecules. Internal energy can be increased by

Looking for the change in internal energy of the system If 71 KJ of heat energy is absorbed by the system and 12 KJ of work is done on the system for a certain process,

From the first law of Thermodynamic,

ΔU = Q + W

Where

Substitute all the parameters into the formula

ΔU = 71 + 12

ΔU = 83 KJ

Therefore, the change in internal energy of the system is 83 KJ.

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The four pieces of information that you need to gather while you’re stopped and waiting to turn at a traffic light include the following Check rear, Check for a gap, Check path and Check signal.

Traffic light are those set of light signals found on the highway road side that helps in controlling and monitoring the moving vehicles.

The traffic light is made up of three lights which include the following:

When stopped by the traffic light you need to gather the following information to avoid tragic situations;

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~400 nm to ~700 nm are the wavelengths of electromagnetic wave in free space.

What are the wavelengths of electromagnetic wave in free space?

Electromagnetic waves are categorized on the basis of their frequency f, which is equivalent to wavelength λ = c/f. Visible light has a wavelength that ranges from ~400 nm to ~700 nm. Violet light has a wavelength of ~400 nm, and having a frequency of ~7.5*1014 Hz.

So we can conclude that: ~400 nm to ~700 nm are the wavelengths of electromagnetic wave in free space.

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The wavelengths of the light are 4.3 * 10^-12 m and 0.2 m respectively.

The term wavelength has to do with the horizontal distance that is covered by a wave. We know that a long wavelength implies that the wave is able to travel a long distance from one point to another.

Given that;

c = λf

c = speed of light

λ = wavelength of ight

f = frequency of light

Thus;

λ = 3 * 10^8/ 7.00 x 10^19

λ = 4.3 * 10^-12 m

λ = 3 * 10^8/1.50 x 10^9

λ = 2 * 10^-1 or 0.2 m

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Missing parts:

What are the wavelengths of electromagnetic wave in free space that have the following frequencies? (a) 7.00 x 10^19 Hz______ pm (b) 1.50 x 10^9 Hz__________ cm

The area of a second piston that exerts a force of 24,000 N is 0.04 [tex]m^{2}[/tex].

Gravitational force is measured in newtons. Utilizing the physics formula pressure = force/area, you can compute pressure if you know the value of a force. Both pounds per square inch (psi) and newtons per square meter (N/m2) are acceptable units of measurement for this pressure.

Given force 1 =  12,000 N

The pressure which has produced a force of 12,000 N at the piston 1 = Force/ area

= 12,000N / 0.020 [tex]m^{2}[/tex]

= 6,00,000 N/[tex]m^{2}[/tex]

Now force at piston 1 must be equal to force at piston 2. thus, force at piston 2 is given by:

Force at piston 2 = pressure × area

24,000 N = 6,00,000 × area

area = 0.04 [tex]m^{2}[/tex]

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565.2 lbs pressure can apply to a plunger.

It is given that,

The air supplied to a diaphragm is 20 psig.

Diameter is 6

Then r=3

We know that,

Area=

πr²

=(3.14×3²)

= 28.26

The formula is (3.14 ×radius of diaphragm squared) ×psig.

=( 28.26 ×20)

= 565.2 lbs of pressure.

Hence, the pressure is  565.2 lbs.

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