When the column was changed to a new Nova-Pak C18 Column, (new Column: 60Å, 3 µm, 3.9 mm X 150 mm) (old column: Nova-Pak C18, 60Å, 4 µm, 3.9 mm X 150 mm), the peak resolution increased. Which factor in the Van Deemter equation illustrates this phenomenon and explain how that works. Please elaborate in full :)

The force on charge q2 is approximately 179.751 N.

The force between two point charges can be found using Coulomb's law:
F = (k * q1 * q2) / r^2
Where F is the force between the charges, k is the Coulomb constant (8.98755 × 10^9 N·m^2/C^2), q1 and q2 are the magnitudes of the charges in Coulombs, and r is the distance between the charges in meters.
In this case, q1 = 2 µC and q2 = 10 µC. The distance between the charges is the distance between the origin and the point on the x-axis where q2 is located, which is 10 m.
So, we can calculate the force on q2 as follows:
F = (8.98755 × 10^9 N·m^2/C^2) * (2 µC) * (10 µC) / (10 m)^2
F = (8.98755 × 10^9 * 2 * 10) / 100
F = 1.79751 × 10^9 / 100
F = 1.79751 × 10^7 N
The force on charge q2, we can use Coulomb's Law. Coulomb's Law states that the force (F) between two point charges is directly proportional to the product of their charges (q1 and q2) and inversely proportional to the square of the distance (r) between them:
F = k * (q1 * q2) / r^2
In this case, q1 = 2 µC, q2 = 10 µC, r = 10 m, and the Coulomb constant (k) is 8.98755 × 10^9 N·m^2/C^2.
The charges to Coulombs: q1 = 2 × 10^-6 C and q2 = 10 × 10^-6 C.
F = (8.98755 × 10^9 N·m^2/C^2) * ((2 × 10^-6 C) * (10 × 10^-6 C)) / (10 m)^2
F = (8.98755 × 10^9 N·m^2/C^2) * (2 × 10^-5 C^2) / (100 m^2)
F = 179.751 N

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A proton of mass m and energy em collides with a stationary alpha particle that has a mass of approximately 4m, then available energy ea,  ea = em - (5m)[tex]c^{2}[/tex]

When the proton collides with the alpha particle, their masses combine and the resulting mass is (m + 4m) = 5m. The total energy before the collision is the sum of the proton's energy (em) and the alpha particle's rest energy (0), which is em + 0 = em. After the collision, the combined mass (5m) will be in motion with a new energy (ea). According to the conservation of energy, the total energy before and after the collision must be equal. Therefore, em = ea + rest energy of the combined mass (5m)[tex]c^{2}[/tex]. Solving for ea, we get ea = em - (5m)[tex]c^{2}[/tex]

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The fact that Iceland recently gave in to pressure from international groups opposed to the hunting of killer whales and will forbid the hunting of killer whales off its coast does not cast doubt on the conclusion of the argument.

It actually supports the conclusion that all wildlife parks will be forced to end their shows featuring killer whales once their current killer whales are unable to perform.

The argument states that wildlife parks rely on hunting killer whales exclusively in the waters of Iceland because they are forbidden to do so off the coast of North America. If Iceland also forbids hunting, the parks will no longer have a source of killer whales, which will ultimately lead to the end of their shows. The conclusion is directly supported by the premise, and there is noccontradictory information provided that would cast doubt on this logical chain of reasoning.

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Yes, there is a stagnation point in this flow field. The stagnation point is the point where the velocity of the fluid is zero. To find the stagnation point, we need to solve for when u = 0 and v = 0. From the velocity field given, we have:

u = 0.5x
v = -2y - 1.2x

Setting u = 0, we get:

0.5x = 0

Solving for x, we get:

x = 0

Setting v = 0, we get:

-2y - 1.2x = 0

Substituting x = 0, we get:

-2y = 0

Solving for y, we get:

y = 0

Therefore, the stagnation point is at (0,0).

To determine if there is a stagnation point in the given two-dimensional velocity field, v = (u, v) = (0.5 + 1.2x)i + (-2.0 - 1.2y)j, follow these steps:

1. A stagnation point occurs when both the u and v components of the velocity field are zero. So, we need to solve the following system of equations:

0.5 + 1.2x = 0
-2.0 - 1.2y = 0

2. To find the x-coordinate of the stagnation point, rearrange the first equation and solve for x:

3. To find the y-coordinate of the stagnation point, rearrange the second equation and solve for y

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The force exerted by the water on the bottom of the pool is 4,269,312 N and on each end of the pool is 227,711 N.

(a) The force exerted by the water on the bottom of the pool is equal to the weight of the water above it.

The weight of the water is equal to its mass multiplied by the acceleration due to gravity (g = 9.81 m/s^2), and its mass is equal to its volume multiplied by its density (ρ = 1000 kg/m^3 for fresh water at standard conditions):

[tex]Volume\ of\ water = length * width * depth = 32.0 m * 8.0 m * 1.70 m = 435.2 m^3\\Mass of water = Volume x Density = 435.2 m^3 * 1000 kg/m^3 = 435200 kg\\Weight of water = Mass x g = 435200 kg x 9.81 m/s^2 = 4,269,312 N[/tex]

Therefore, the force exerted by the water on the bottom of the pool is 4,269,312 N.

(b) The force exerted by the water on each end of the pool can be calculated by considering the pressure of the water on the end. The pressure of the water at a depth of 1.70 m is equal to the height of the water column (1.70 m) multiplied by the density of water and the acceleration due to gravity:

[tex]Pressure = depth * density * g = 1.70 m * 1000 kg/m^3 * 9.81 m/s^2 = 16,743 Pa[/tex]

The force exerted by the water on each end is equal to the pressure multiplied by the area of the end:

[tex]Force = Pressure * Area = 16,743 Pa * 8.0 m * 1.70 m = 227,711 N[/tex]

Therefore, the force exerted by the water on each end of the pool is 227,711 N.

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(a) The dc source voltage is 61.2 V.

(b) The THD of the load current is approximately 33.2%.

(a) To calculate the dc source voltage required to produce a load current of 2.0 A rms, we first need to calculate the impedance of the load at the fundamental frequency. The impedance can be calculated as Z = R + jωL, where R is the resistance of the load, L is the inductance of the load, and ω is the angular frequency.

ω = 2πf

ω = 2π x 120 Hz

ω = 753.98 rad/s

Z = 25 + j(753.98 x 0.025)

Z = 25 + j18.85 Ω

The rms value of the load current is given by I = V/Z, where V is the rms value of the voltage supplied by the inverter.

I = 2.0 A rms, Z = 25 + j18.85 Ω

Therefore, V = IZ

V = (2.0 A rms) x (25 + j18.85 Ω)

V = 61.2 + j45.35 V rms

The dc source voltage is the average value of the voltage waveform, which is equal to the rms value multiplied by π/2.

Vdc = (π/2) x 61.2 V rms ≈ 96.2 Vdc

(b) The total harmonic distortion (THD) of the load current is a measure of the distortion of the current waveform from a perfect sinusoid. It is defined as the square root of the sum of the squares of the harmonic components of the current waveform, divided by the rms value of the fundamental component.

THD = √[(I2² + I3² + ... + In²)/I1²] x 100%

where I1 is the rms value of the fundamental component, and I2, I3, ..., In are the rms values of the second, third, ..., nth harmonic components.

For a square-wave inverter, the load current waveform contains only odd harmonic components. The rms value of the nth harmonic component can be calculated as

In = (4Vdc/(nπZ)) x sin(nπ/2)

where n is the harmonic number.

Using this equation, we can calculate the rms values of the first three harmonic components of the load current.

I1 = 2.0 A rms (given)

I3 = (4 x 96.2 Vdc / (3π x 25 Ω)) x sin(3π/2)

I3 ≈ 0.632 A rms

I5 = (4 x 96.2 Vdc / (5π x 25 Ω)) x sin(5π/2)

I5 ≈ 0.254 A rms

The THD can now be calculated as

THD = √[(0.632² + 0.254²)/2.0²] x 100%

THD ≈ 33.2%

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The de Broglie wavelength of an electron traveling at a speed of 5.0 x 10^6 m/s is approximately 0.145 picometers (pm).

The de Broglie wavelength of an electron is given by the equation:

λ = h/mv

Where λ is the de Broglie wavelength, h is Planck's constant, m is the mass of the electron, and v is the velocity of the electron.

Substituting the given values, we get:

λ = h/(mv) = (6.626 x 10^-34 J s)/(9.11 x 10^-31 kg x 5.0 x 10^6 m/s)

λ = 0.145 pm (rounded to three significant figures)

Therefore, the de Broglie wavelength of an electron traveling at a speed of 5.0 x 10^6 m/s is approximately 0.145 picometers (pm).

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The incident angle θ for which the beam of light in the figure will hit the indicated point on the screen is 60 degrees.

In this question, we need to find the incident angle for which the beam of light in the figure will hit the indicated point on the screen. We have a semicircular disk of glass with an index of fraction of n = 156 (Figure 1). We are given that the refractive index of the glass is n = 1.56. Using Snell's law,n1sinθ1=n2sinθ2where, n1= refractive index of the incident medium, n2= refractive index of the refracted medium, θ1= angle of incidence, θ2= angle of refraction. As air is the incident medium, the refractive index of air is 1.n1 = 1 and n2 = 1.56 sin(θ1) = 1.56sin(θ2)

As the angle of incidence (i) and the angle of reflection (r) are equal,i = rso, the angle between the incident ray and the normal, θ1 = 60°

Thus, sin(60) = 1.56sin(θ2)sin(θ2) = 0.63θ2 = 40.94°

As the light is refracted away from the normal, the angle of incidence is greater than the angle of refraction.

Hence, the incident angle of the beam of light is 60°.

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We found a solution to x^2 + 1 ≡ 0 (mod 113) with |x| < 113/2, and we used a sum of two squares representation to express 229 as a sum of two squares.

To find a solution to x^2 + 1 ≡ 0 (mod p) for primes p = 113 and p = 229, we can use quadratic reciprocity and observe that:

- For p = 113, we have (-1/113) = (-1)^56 = 1, which means that x^2 + 1 ≡ 0 (mod 113) has a solution. One such solution is x = 21, which satisfies 21^2 + 1 ≡ 0 (mod 113) and |21| < 113/2.
- For p = 229, we have (-1/229) = (-1)^114 = -1, which means that x^2 + 1 ≡ 0 (mod 229) does not have a solution.

However, we can use the fact that 229 ≡ 1 (mod 4) to write:
229 = a^2 + b^2,
where a = 15 and b = 14. This is a sum of two squares representation of 229.

To see why this works, note that any prime of the form p = 4n + 1 can be expressed as a sum of two squares. Specifically, we can use the following identity: (2m+1)^2 + (2n)^2 = 4(m^2 + m + n^2)
to write any odd number as a sum of two squares. For example, we have:
229 = 15^2 + 14^2,
because 229 = 4(56) + 1 and we can set m = 7 and n = 0 in the above identity.

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In the circuit shown in Figure P7.129, there is a small sine-wave signal called vsig with an average value of zero. The transistor in the circuit has a beta value of 100. The value of RE is 1.26 k, Rc is 50 k, and the output signal is -0.165(100t)V with an overall voltage gain of 0.11.  

The circuit involves a small sine-wave signal with an average of zero and the transistor has a beta value of 100, then answer to the following questions are as follows:

(a) The value of RE can be found using Ohm's law and Kirchhoff's law as follows:

VE = VBE + IE*RE

[tex]0.5 \, \text{mA} = \left(\frac{0.7 \, \text{V}}{100}\right) + \left(\frac{V_E}{R_E}\right)[/tex]

RE ≈ 1.26 kΩ

(b) Rc can be found using Kirchhoff's law as follows:

VCC - IC*Rc - VCE ≈ 0

[tex]R_c \approx \frac{{V_{CC} - V_{CE}}}{{I_C}} = \frac{{3 \, \text{V} - 0.5 \, \text{V}}}{{0.5 \, \text{mA} \times 100}} \approx 50 \, \text{k}\Omega[/tex]

(c) The small-signal equivalent circuit is shown in the image below. The overall voltage gain can be calculated as:

[tex]A_v = -\frac{\beta(R_c \parallel R_L)}{(r_{\pi} + R_E) + \beta(R_c \parallel R_L)}[/tex]

where [tex]r_{\pi} = \frac{25 \, \text{mV}}{0.5 \, \text{mA}} = 50 \, \Omega[/tex] (assuming VT ≈ 25 mV).

For RL = 10 kΩ, Av ≈ -0.11.

(d) The output signal can be found using the equation:

[tex]v_o(t) = -0.165 \sin(1000t) \, \text{V}[/tex]

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a) The amplitude modulation ratio is 0.5625.

b) The total harmonic distortion of the load current is 40.53%.

c) The real power absorbed by the load is 405.36 W.

a) The amplitude modulation ratio (m) can be calculated using the equation m = Erms/Emax, where Erms is the desired rms voltage and Emax is the maximum voltage. Here, Erms is given as 54 V and Emax can be calculated as Emax = √(2) × 96 V = 135.76 V. Substituting these values, we get m = 0.5625.

b) The frequency modulation ratio (f) is given as 17. The total harmonic distortion (THD) can be calculated using the equation THD = √((Vthd² - V1²)/V1²) * 100%, where Vthd is the total harmonic voltage and V1 is the fundamental voltage. Here, V1 is given as 54 V. The total harmonic voltage can be calculated as Vthd = sqrt(sum of squares of all harmonic voltages) = sqrt((V3² + V5² + V7² + ...)/2), where V3, V5, V7, etc., are the harmonic voltages. Substituting the given values, we get Vthd = 43.72 V. Substituting these values, we get THD = 40.53%.

c) The real power absorbed by the load can be calculated using the equation P = V1²/R, where V1 is the fundamental voltage and R is the resistance of the load. Substituting the given values, we get P = 405.36 W.

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The required current for the superconducting solenoid is approximately 9.0E+2 A.

To calculate the required current for the superconducting solenoid, we can use the formula for the magnetic field strength (B) produced by a solenoid:
B = μ₀ * n * I
where B is the magnetic field strength (3.50 T), μ₀ is the permeability of free space (417 x 10^-7 T-m/A), n is the number of turns per meter (984 turns/m), and I is the current in amperes (A).
Rearranging the formula to solve for I:
I = B / (μ₀ * n)
Plugging in the given values:
I = 3.50 T / ((417 x 10^-7 T-m/A) * (984 turns/m))
I ≈ 9.0E+2 A
So, the required current for the superconducting solenoid is approximately 9.0E+2 A.

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To determine the required current for the superconducting solenoid, we need to use the formula for the magnetic field generated by a solenoid: B = u * n * I, where B is the magnetic field, u is the permeability of free space (given as Mo in this case), n is the number of turns per unit length (984 turns/m), and I is the current.

Rearranging the formula, we get : I = B / (u * n)

Plugging in the given values, we get : I = 3.50 T / (417x10^-7 T-m/A * 984 turns/m) = 2.8E+3 A

Therefore, the required current for the superconducting solenoid to generate a magnetic field of 3.50 T with 984 turns/m is 2.8E+3 A.

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The period of the rocking chair is approximately 1.06 seconds per cycle, and the frequency is approximately 0.94 cycles per second (Hz).

The period is the time it takes to complete one cycle. To find the period, divide the total time by the number of cycles:
Period = Total time / Number of cycles
Period = 18 seconds / 17 cycles
Period ≈ 1.06 seconds per cycle

The frequency is the number of cycles completed in one second. To find the frequency, divide the number of cycles by the total time:
Frequency = Number of cycles / Total time
Frequency = 17 cycles / 18 seconds
Frequency ≈ 0.94 cycles per second (or Hertz, Hz)

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The period of rocking is 1.06 s and frequency of the rocking is 0.94 Hz (cycles per second).

The period (T) of a periodic motion is the time it takes for one complete cycle of the motion. i.e.,

T = (Total time taken) / (Number of cycles)

The frequency (f) of the motion is the number of cycles per unit time or in one second. i.e.,

f = (Number of cycles) / (Total time taken)

Frequency is also considered to be reciprocal of the period.

∴ [tex]f=\frac{1}{T}[/tex]

Now, given

No. of cycles completed = 17

Time taken for 17 cycles = 18 second.

Therefore,

Period, T = (total time) / (number of cycles)

               = 18 s / 17

               ≈ 1.06 s

And,

Frequency, f = (number of cycles) / (total time)

                    = 17 / 18 s

                    ≈ 0.94 Hz

Or, frequency can also be obtained as,

[tex]f=\frac{1}{T}[/tex]

  = 1/1.06

  ≈ 0.94 Hz.

Therefore, the period of rocking is 1.06 s and frequency of the rocking is 0.94 Hz (cycles per second).

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The angular magnification of the eyepiece is 4. The other possible total magnifications are 400 and 1600.

To find the angular magnification of the eyepiece, we need to use the formula:
Overall Magnification = Objective Magnification x Eyepiece Magnification
We know that the overall magnification of the microscope is 800, and the objective magnification is 200. Therefore, we can rearrange the formula to solve for the eyepiece magnification:
Eyepiece Magnification = Overall Magnification / Objective Magnification
Plugging in the values we know, we get:
Eyepiece Magnification = 800 / 200 = 4
Therefore, the angular magnification of the eyepiece is 4.

To find the other total magnifications possible with the two other objectives, we can use the same formula as before:                                  Overall Magnification = Objective Magnification x Eyepiece Magnification
For the first objective with a magnification of 100, we can plug in the values we know: Overall Magnification = 100 x 4 = 400
For the second objective with a magnification of 400, we can plug in the values we know: Overall Magnification = 400 x 4 = 1600

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Experiment 2 is the better test of how force affects an object's motion.

Experiment 2 is the better test of how force affects an object's motion because it involves testing the effect of force on the motion of a heavier object, whereas in experiment 1, the force applied to the ball was only changed by half.

In experiment 2, the same force was applied to two different masses, allowing the student to compare the effect of force on different objects.

This is important because the mass of an object affects its motion.

Therefore, the results from experiment 2 will provide a better understanding of the relationship between force and motion, which is the goal of the experiment.

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The density of the liquid is 0.5 g/cm³. to calculate the density, we use the formula: density = mass/volume.

The mass of the liquid can be found by subtracting the mass of the empty density bottle from the mass of the bottle filled with liquid.

For water: mass of liquid = 70g - 20g = 50g.

For the second liquid: mass of liquid = 55g - 20g = 35g.

Since the volume of the density bottle remains the same for both liquids, the density can be calculated as mass of liquid/volume of the bottle.

For water: density = 50g/100cm³ = 0.5 g/cm³.

For the second liquid: density = 35g/100cm³ = 0.35 g/cm³.

Therefore, the density of the liquid is 0.5 g/cm³.

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The spring constant for a mass of 51.7 grams on a spring that undergoes 21 cycles with a small amplitude in 3.42 seconds is 76.8 N/m.

The value of k for a mass on a spring can be determined using the formula T=2π√(m/k), where T is the period of oscillation, m is the mass, and k is the spring constant. In this problem, we know that the mass is 51.7 grams and that 21 cycles took 3.42 seconds, which means that the period of oscillation is T=3.42/21=0.163 seconds. Since the amplitude is small, we can assume that the motion is simple harmonic, which means that T=2π√(m/k) can be used. Rearranging this formula gives k=m(2π/T)^2, which gives k=51.7(2π/0.163)^2=76.8 N/m.

This value was calculated using the formula k=m(2π/T)^2, where m is the mass and T is the period of oscillation.

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A diffraction grating has several advantages over a double slit when it comes to dispersing light into a spectrum. Its higher resolution, ability to disperse light over a larger angle, and accuracy in measuring wavelengths make it a valuable tool in scientific research.

A diffraction grating and a double slit are both devices used to disperse light into a spectrum. However, there are some advantages that a diffraction grating has over a double slit.
One advantage of a diffraction grating is that it has a much higher resolution than a double slit. This is because a diffraction grating has many more slits than a double slit, allowing for more diffraction and a sharper, more detailed spectrum.
Another advantage of a diffraction grating is that it can disperse light over a larger angle than a double slit. This means that it can separate colors more effectively and provide a clearer spectrum.
Additionally, a diffraction grating can be used to measure the wavelengths of light with great accuracy. By measuring the angles at which different colors are dispersed, scientists can determine the exact wavelengths of the different colors.

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If a person goes to the bottom of a very deep mine shaft on a planet of uniform density, then the person's weight is exactly the same as at the surface. Option(A) is true.

The force of gravity is directly proportional to the mass of the planet and inversely proportional to the square of the distance between the person and the center of the planet.

Gravity is a fundamental force that governs the motion of objects in the universe. It is an attractive force between any two objects with mass or energy, and its strength depends on the mass and distance between the objects.

Since the planet has uniform density, the mass beneath the person cancels out, resulting in no change in weight.

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David Bowie changed his original name (David Robert Jones) to avoid confusion with Davy Jones, a member of the popular band The Monkees.

Bowie didn't want to be associated with Davy Jones and sought a distinct identity for his own career in music. Davy Jones was a British singer and actor who gained fame as a member of The Monkees in the 1960s. As David Robert Jones began his own musical journey, he decided to adopt the stage name "David Bowie" to prevent any potential confusion between the two artists. Bowie's new name not only provided him with a unique identity but also allowed him to craft a distinct image and persona that would define his groundbreaking and influential career in music and art.

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Assumptions:

The cylinder is thin-walled, which means that the thickness of the cylinder wall is much smaller than the radius of the cylinder (t << R).

The material of the cylinder is linear elastic, which means that Hooke's law applies to it.

The cylinder is in a state of static equilibrium, which means that the internal pressure is balanced by the forces in the wall of the cylinder.

Analysis:

Consider a small segment of the cylinder wall with a length of "dl" and an angle of "dθ" as shown in the figure below:

Thin-walled cylinder diagram

The forces acting on this segment are:

The force due to the internal pressure, which acts perpendicular to the segment and has a magnitude of Pdl.

The force due to the stress in the circumferential direction, which acts tangentially to the segment and has a magnitude of σθdl.

The force due to the stress in the axial direction, which acts parallel to the segment and has a magnitude of σzdl.

Using the equilibrium conditions, we can write:

∑Fx = 0 ==> σθ dl - σθ (dθ + dl) + σz (R + t/2) dθ - σz (R - t/2) dθ = 0

∑Fy = 0 ==> Pdl - σzdl + σzdl = 0

Simplifying these equations and dividing by dl, we get:

σθ - σθ' + σz(R/t + 1/2) - σz(R/t - 1/2) = 0

P - σz = 0

where σθ' is the circumferential stress on the opposite side of the cylinder wall.

We can solve these equations for the stresses in terms of the pressure P, the radius R, and the wall thickness t:

σz = P(R/t)/2

σθ = P(R/t)

σT0 = 0 (there is no radial stress)

Therefore, the principal stresses in a thin-walled closed-end, linear elastic cylinder subjected to internal pressure P in equilibrium are given by:

σz = P(R/t)/2

σθ = P(R/t)

σT0 = 0

These equations are valid under the assumptions stated above.

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