Under what conditions will a heat
exchanger be a
non-dissipative element ?

A) The resolution limit set by diffraction for f/2.8 is 1.5 lines mm RP.

B) The resolution limit set by diffraction for f/2.8 is 1.1 lines mm RP.

Part A: The resolution limit set by diffraction for f/2.8 is as follows:

The formula for diffraction is given as;

                                   sinθ = 1.22 λ / d

Where, λ is the wavelength of light

            d is the diameter of the aperture.

From the above formula, we can say that resolution is inversely proportional to the diameter of the aperture.

Therefore, the smaller the aperture diameter, the greater the resolution.

As per the problem statement, A = 560 nm and diameter at f/2.8 is 18 mm.

The radius of the aperture is half of the diameter, therefore;

                          Radius, r = 9 mm

                                d = 2 × r

                                   = 18 mm

Putting the values in the formula of diffraction;

                                    sinθ = 1.22 × 560 × 10⁻⁹ / 18

                                            = 0.0375

                                        θ = sin⁻¹(0.0375)

                                            = 2.15°

The angle θ is formed between the center of the lens and the edge of the lens.

In the above image, ABC represents the lens and the angle θ is formed between the lines AO and BO.

The angle AOB is equal to 2θ.

The distance between A and B is the diameter of the aperture.

Therefore, AB = 18 mm.

Using simple trigonometry, we can find the length of the chord AB as follows:

             Length of chord AB = 2 × r × sinθ

                                               = 2 × 9 × sin(2.15)

                                               = 0.335 mm

Hence, the number of lines per millimeter resolved on the film = 1/ (2 × length of chord AB)

                                                                                                         = 1/ (2 × 0.335)

                                                                                                          = 1.49

                                                                                                           ≈ 1.5 lines mm

                              RP(f/2.8) = 1.5 lines mm

Part B: The resolution limit set by diffraction for f/22 is as follows:

From the problem statement, A = 560 nm and diameter at f/22 is 2.3 mm.

The radius of the aperture is half of the diameter, therefore;

                                       Radius, r = 1.15 mm

                                                   d = 2 × r

                                                      = 2.3 mm

Putting the values in the formula of diffraction;

                                              sinθ = 1.22 × 560 × 10⁻⁹ / 2.3

                                                      = 0.303

                                                  θ = sin⁻¹(0.303)

                                                    = 17.4°

Using simple trigonometry, we can find the length of the chord AB as follows:

                                                 Length of chord AB = 2 × r × sinθ

                                                                                   = 2 × 1.15 × sin(17.4)

                                                                                    = 0.437 mm

Hence, the number of lines per millimeter resolved on the film = 1/ (2 × length of chord AB)

                                                                                                         = 1/ (2 × 0.437)

                                                                                                         = 1.14

                                                                                                         ≈ 1.1 lines mm

                                                RP(f/22) = 1.1 lines mm

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The legal position in the above instance is that Stanly can refuse to pay for the services rendered by NIC Construction (pvt) Ltd due to the construction company's failure to construct the common dining area as specified in the contract.

According to the scenario, NIC Construction (pvt) Ltd was hired by Stanly to build a building on his land.

He clearly stated all the specifications for the building and gave all the necessary instructions to the construction company.

One of the major conditions of the contract was that the building should have a common dining area and six rooms for guests. NIC Construction (pvt) Ltd handed over the building on the agreed date.

However, the construction company failed to build the common dining area.

As a result, Stanly refuses to pay for the services rendered by NIC Construction (pvt) Ltd.

Legal position in the above instance: In the above scenario, NIC Construction (pvt) Ltd had a binding contract with Stanly, and it was agreed that the construction company would build a building with a common dining area and six rooms for guests.

However, NIC Construction (pvt) Ltd failed to construct the common dining area as specified in the contract.

In this regard, the failure of NIC Construction (pvt) Ltd to complete the building to the agreed specifications amounts to a material breach of the contract.

In such an instance, Stanly has the legal right to refuse to pay for the services rendered by NIC Construction (pvt) Ltd.

Therefore, the legal position in the above instance is that Stanly can refuse to pay for the services rendered by NIC Construction (pvt) Ltd due to the construction company's failure to construct the common dining area as specified in the contract.

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The Trapezoid Rule and Corrected Trapezoid Rule can be used to approximate the integral of ₁²e[tex]^(-2x²)[/tex] dx with a given interval width of h = 1. The Trapezoid Rule approximates the integral by summing the areas of trapezoids, while the Corrected Trapezoid Rule improves accuracy by considering additional midpoint values.

To estimate the minimum number of subintervals needed for desired accuracy, one typically iterates by gradually increasing the number of intervals until the desired level of precision is achieved.

a) Using the Trapezoid Rule:

The Trapezoid Rule estimates the integral by approximating the area under the curve with trapezoids. The formula for the Trapezoid Rule with interval width h is:

∫(a to b) f(x) dx ≈ h/2 * [f(a) + 2f(a+h) + 2f(a+2h) + ... + 2f(b-h) + f(b)]

In this case, we have a = 1, b = 2, and h = 1. The function f(x) = [tex]e^(-2x^2)[/tex].

b) Using the Corrected Trapezoid Rule:

The Corrected Trapezoid Rule improves upon the accuracy of the Trapezoid Rule by using an additional midpoint value in each subinterval. The formula for the Corrected Trapezoid Rule with interval width h is:

∫(a to b) f(x) dx ≈ h/2 * [f(a) + 2f(a+h) + 2f(a+2h) + ... + 2f(b-h) + f(b)] - (b-a) * [tex](h^2 / 12)[/tex] * f''(c)

Here, f''(c) is the second derivative of f(x) evaluated at some point c in the interval (a, b).

To estimate the minimum number of subintervals needed for a desired level of accuracy, you would typically start with a small number of intervals and gradually increase it until the desired level of precision is achieved.

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The minimum flow rate required to meet the net crop ET needs of a 77-acre center pivot system with crop peak ET of 0.33 inches/day and DU of 0.87, with pre-infiltration losses of 9%, is 16.93 gpm.

A center pivot system is an irrigation technique that rotates around a pivot point and spreads water over a crop area. The system has several advantages, including reducing labor costs and increasing efficiency. A 77-acre center pivot system can irrigate an area of 77 acres using 0.33 inches per day and a DU (deep percolation) of 0.87. Pre-infiltration losses of 9% are also present. The irrigation system must meet the crop's net ET needs, which are determined by subtracting pre-infiltration losses from peak crop ET.

Net crop ET = crop peak ET - pre-infiltration losses.

The minimum flow rate the system can have to meet net crop ET needs is:

Minimum flow rate = (Area x Net crop ET x 0.62) / DU

In this equation: Area = 77 peak ET = 0.33 inches/day DU = 0.87 Pre-infiltration losses = 9%

Net crop ET = crop peak ET - pre-infiltration losses

Net crop ET = 0.33 - (0.09 x 0.33)

Net crop ET = 0.3 inches/day

Minimum flow rate = (77 x 0.3 x 0.62) / 0.87

Minimum flow rate = 16.93 gpm

Net crop ET is the amount of water needed for a crop to reach maturity and produce the best yield. In this question, the net crop ET is determined by subtracting the pre-infiltration losses from the peak crop ET. In addition, the DU, which is the percentage of water that is not lost to deep percolation, is factored into the calculation. The minimum flow rate required to meet net crop ET needs can be calculated using the minimum flow rate equation, which takes into account the area, net crop ET, and DU. The minimum flow rate equation is (Area x Net crop ET x 0.62) / DU.

In conclusion, the minimum flow rate required to meet the net crop ET needs of a 77-acre center pivot system with crop peak ET of 0.33 inches/day and DU of 0.87, with pre-infiltration losses of 9%, is 16.93 gpm.

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The average power dissipated in the 2 ohms resistor is 651.6 V.

The average power dissipated in the 2 ohms resistor is calculated by applying the following formula.

P = IV

P = (V/R)V

P = V²/R

The given parameters include;

The root - mean - square voltage is calculated as follows;

Vrms = 0.7071V₀

Vrms = 0.7071 x 51 V

Vrms = 36.1 V

The average power dissipated in the 2 ohms resistor is calculated as;

P = (36.1 V)² / 2Ω

P = 651.6 V

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The complete question is below:

This electric circuit is described by the following equation: [tex]\[ E=\varepsilon_{n} \rho^{i \omega t} \][/tex] What is the average power (in W/) dissipated by the [tex]2 \Omega \)[/tex] resistor in the circuit if the peak voltage E₀ = 51 V?

The total power needed for the airplane to climb at a 10° angle to the horizontal with a speed of 110 knots and a drag of 36.0 kN is approximately X horsepower.

To calculate the total power needed, we need to consider the forces acting on the airplane during the climb. The force of gravity acting on the airplane is given by the weight, which is the mass (12000 kg) multiplied by the acceleration due to gravity (9.8 m/s²).

The component of this weight force parallel to the direction of motion is counteracted by the thrust force of the airplane's engines. The component perpendicular to the direction of motion contributes to the climb.

This climb force can be calculated by multiplying the weight force by the sine of the climb angle (10°).Next, we need to calculate the power required to overcome the drag.

Power is the rate at which work is done, and in this case, it is given by the product of force and velocity. The drag force is 36.0 kN, and the velocity of the airplane is 110 knots.

However, we need to convert the velocity from knots to meters per second (1 knot = 0.5144 m/s) to maintain consistent units.Finally, the total power needed is the sum of the power required to overcome the climb force and the power required to overcome drag.

The power required for climb can be calculated by multiplying the climb force by the velocity, and the power required for drag is obtained by multiplying the drag force by the velocity. Adding these two powers together will give us the total power needed.

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The dissolution of a solid into a liquid is the process that increases the entropy of a system. Hence, option a) is the correct answer.

Dissolution of a solid into a liquid is the process that increases the entropy   because when a solid dissolves in a liquid, the particles of the solid break apart and become more spread out in the liquid. This increases the number of possible arrangements of particles, leading to an increase in entropy.

The other processes, deposition, crystallization, and freezing, all involve a decrease in entropy as the particles become more ordered and arranged in a regular structure.

hence, the correct answer is a) dissolution.

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(i) The energy eigenstates of the atomic electron are usually described by wave functions ne m(r). Relate each of n, l, and m to the eigenvalue of a specific operator by giving the eigenvalue equation for this operator acting on yne m(r).The values of n, l, and m are known as quantum numbers. n is the principle quantum number which is the energy level of the hydrogen atom.

It is also the number of nodes the wave function has. It can take any positive integer value (n = 1, 2, 3...).l is the azimuthal quantum number that describes the angular momentum of the electron. It is also known as the orbital quantum number. It can take values of 0 to n-1 for a given n value.m is the magnetic quantum number that is related to the magnetic moment of the electron. It ranges from -l to l.The Hamiltonian operator of a hydrogen atom is H = - (h^2/2m) * Δ^2 - e^2/(4πε0r). The operator corresponding to the principle quantum number is H = E * n^2, where E is the energy of the hydrogen atom in its ground state. Similarly, the operators corresponding to l and m are H = L^2 * l(l+1) and H = Lz * m, where Lz is the z-component of angular momentum. (ii) The atomic electron in its ground state is, for a point-like nucleus, described by the wave function ψ100 = (1/πa03)^(1/2) * e^(-r/a0), where a0 is the Bohr radius and a0 = 4πε0h^2/(me^2). We need to show that the wave function is normalized by calculating the integral of the square of the wave function.∫ |ψ100|^2 dV = ∫ |(1/πa03)^(1/2) * e^(-r/a0)|^2 dV= (1/πa03) ∫ e^(-2r/a0) 4πr^2 dr= (1/πa03) * [(a0/2)^3 * π] = 1The expectation value of the position of the electron is = ∫ ψ* r ψ dV= (1/πa03) ∫ r^3 e^(-2r/a0) dr= (3/2) * a0and the expectation value of the position squared is = ∫ ψ* r^2 ψ dV= (1/πa03) ∫ r^4 e^(-2r/a0) dr= 3a02Ar = (∫ ψ* r^2 ψ dV - (∫ ψ* r ψ dV)^2)^(1/2) = [(3/2)a0 - (3/2)^2 a0]^(1/2) = a0/2(iii) For a finite size nucleus, we can model the nucleus as a uniformly charged hollow spherical shell of radius R. For 0 ≤ r ≤ R, the potential V(r) is constant, and for r > R, the potential is identical to the Coulomb potential generated by a point-like nucleus.

The potential is given by:V(r) = kq/r for r > R, andV(r) = kqR/r^2 for 0 ≤ r ≤ R, where q is the total charge of the nucleus and k is the Coulomb constant. We need to sketch this potential. See the attached image. The perturbation AV relative to the Coulomb potential that is generated by a point-like nucleus is given by:AV(r) = V(r) - Vcoulomb(r) = kqR * (1/r^2 - 1/R^3), for 0 ≤ r ≤ R.The shift in the ground-state energy level due to the finite size of the nucleus can be calculated using first-order perturbation theory. The shift in the energy level is given by:ΔE1 = <ψ100|AV|ψ100>where ψ100 is the wave function of the ground state of the hydrogen atom when the nucleus is point-like. Substituting the values, we get:ΔE1 = (kqR/πa03) * ∫ e^(-2r/a0) r^2 (1/r^2 - 1/R^3) e^(-r/a0) dr= (kqR/πa03) * ∫ e^(-3r/a0) (r/R^3 - r^2/a0R^2) dr= (kqR^4/πa04) * (1/9R^3 - 1/3a0R^2)Now, we know that the total charge of the nucleus is q = Ze, where Z is the atomic number and e is the charge of an electron. The expression for the ground-state energy of the hydrogen atom is given by:E = - (me^4/32π^2ε0^2h^2) * 1/n^2Substituting the values, we get:E = -13.6 eVWe can express the shift in the energy level in terms of the unperturbed ground-state energy E, and a function of the ratio R/a0 as:ΔE1 = - (2Ze^2/3a0) * (R/a0)^3 * [1/9 - (R/a0)^2/3] = - (2/3)E * (R/a0)^3 * [1/9 - (R/a0)^2/3](iv) Perturbation theory can be applied in this case because the perturbation AV is small compared to the Coulomb potential. This is evident from the fact that the potential due to a uniformly charged spherical shell is nearly the same as the Coulomb potential for r > R. Therefore, we can treat the potential due to a finite size nucleus as a perturbation to the Coulomb potential generated by a point-like nucleus.

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co = 0 when the particle density is n₀ in three dimensions. f₀ = co exp(-p²/2mkbT) / n₀. The entropy of this state in a volume V is given by the formula S = kb log(n₀).

(a) To find the value of co when the particle density is n₀ in three dimensions, we need to normalize the distribution function.

The normalization condition is given by:

∫∫∫ f(x, p) dx dy dz dpₓ dpᵧ dp_z = 1

Using the given equilibrium distribution f(x, p) = co exp(-p²/2mkbT), we can split the integral into separate integrals for position and momentum:

V ∫∫∫ co exp(-p²/2mkbT) dx dy dz dpₓ dpᵧ dp_z = 1

The position integral over the volume V gives V:

V ∫∫∫ co exp(-p²/2mkbT) dpₓ dpᵧ dp_z = 1

Now we need to perform the momentum integrals. Since the distribution function only depends on the magnitude of the momentum, we can use spherical coordinates to simplify the integration. The momentum integral becomes:

2π ∫∫∫ co exp(-p²/2mkbT) p² sin(θ) dp dp dθ = 1

Here, p is the magnitude of momentum, and θ is the angle between momentum and the z-axis.

The integral over θ gives 2π:

4π² ∫ co exp(-p²/2mkbT) p² dp = 1

To evaluate the remaining momentum integral, we can make the substitution u = p²/2mkbT:

4π² ∫ co exp(-u) du = 1

The integral over u gives ∞:

4π² co ∫ du = 1

4π² co ∞ = 1

Since the integral on the left-hand side diverges, the only way for this equation to hold is for co to be zero.

Therefore, co = 0 when the particle density is n₀ in three dimensions.

(b) To find the value of f₀ for which our definition reproduces the equation for the absolute entropy of an ideal gas, we use the equation:

S = Nkb[log(nq/n₀) + 5/2]

We know that the equilibrium distribution function f(x, p) = co exp(-p²/2mkbT). We can compare this to the ideal gas equation:

f(x, p) = f₀ n(x, p)

Where n(x, p) is the particle density and f₀ is the value we are looking for.

Equating the two expressions:

co exp(-p²/2mkbT) = f₀ n(x, p)

Since the particle density is n₀, we can write:

n(x, p) = n₀

Therefore, we have:

co exp(-p²/2mkbT) = f₀ n₀

Solving for f₀:

f₀ = co exp(-p²/2mkbT) / n₀

(c) To calculate the entropy of this state in a volume V using the definition of entropy, which is:

S = -kb ∫∫∫ f(x, p) log(f(x, p)/f₀) dx dy dz dpₓ dpᵧ dp_z

Substituting the equilibrium distribution function and the value of f₀ we found in part (b):

S = -kb ∫∫∫ co exp(-p²/2mkbT) log(co exp(-p²/2mkbT) / (n₀ co exp(-p²/2mkbT))) dx dy dz dpₓ dpᵧ dp_z

Simplifying:

S = -kb ∫∫∫ co exp(-p²/2mkbT) log(1/n₀) dx dy dz dpₓ dpᵧ dp_z

Using properties of logarithms:

S = -kb ∫∫∫ co exp(-p²/2mkbT) (-log(n₀)) dx dy dz dpₓ dpᵧ dp_z

Pulling out the constant term (-log(n₀)):

S = kb log(n₀) ∫∫∫ co exp(-p²/2mkbT) dx dy dz dpₓ dpᵧ dp_z

The integral over position and momentum is simply the normalization integral, which we found to be 1 in part (a):

S = kb log(n₀)

Therefore, the entropy of this state in a volume V is given by the formula S = kb log(n₀).

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Yes, the potentials look different when your eyes are open or closed. They look different because of the neural noise produced by the neural activity occurring in the visual system that is present when our eyes are open.

When our eyes are closed, there is less neural noise present, which leads to cleaner and more easily discernible signals.

2. The amplitude of the potential is affected by how far you move your eyes and how quickly. When you move your eyes, the potential changes in amplitude due to changes in the orientation of the neural sources generating the signal. The amplitude will also change depending on the speed of the eye movement, with faster eye movements producing larger potentials.

Other variables that can affect the amplitude of the potential include the size and distance of the object being viewed and the intensity of the light.

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The question pertains to a long copper rod with a length of 2 m and thermal diffusivity of k. The rod has insulated lateral surfaces, with the left end maintained at 0 °C and the right end insulated. The initial temperature distribution along the rod is described by the function 100x.

The problem involves analyzing the temperature distribution in a copper rod under the given conditions. The rod has insulated lateral surfaces, meaning there is no heat exchange through the sides. The left end of the rod is held at a constant temperature of 0 °C, while the right end is insulated, preventing heat transfer to the surroundings. The initial temperature distribution along the rod is given by the function 100x, where x represents the position along the length of the rod.

To analyze the temperature distribution and the subsequent heat transfer in the rod, we would need to solve the heat conduction equation, which involves the thermal diffusivity of the material. The thermal diffusivity, denoted by k, represents the material's ability to conduct heat. By solving the heat conduction equation, we can determine how the initial temperature distribution evolves over time and obtain the temperature profile along the rod. This analysis would involve considering the boundary conditions at the ends of the rod and applying appropriate mathematical techniques to solve the heat conduction equation.

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Pouillet's law is a law in physics that expresses the relationship between current, resistance, and voltage. According to Pouillet's law, the amount of electric current carried through a conductor is directly proportional to the time and intensity of the current itself.

The expression R = ρ L / A is used to calculate the resistance of a wire, where R represents resistance, ρ represents resistivity, L represents the length of the wire, and A represents the cross-sectional area of the wire. This statement is true. Detailed Explanation :Pouillet's  law states that the amount of electric current carried through a conductor is proportional to the time and intensity of the current itself.

This means that the longer the time and the higher the intensity of the current, the more electric current that is being carried through the conductor.  Hence, this equation can be used to calculate the resistance of a wire, given its length and cross-sectional area. Therefore, the statement "Pouillet's law states that the amount of electric current carried through a conductor is proportional to the time and intensity of the current itself" is true.

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a) At what temperature does water boil at 10,000ft (3000 m) of elevation? When the elevation is increased, the atmospheric pressure decreases, and the boiling point of water decreases as well.

Since the boiling point of water decreases by approximately 1°C per 300-meter increase in elevation, the boiling point of water at 10,000ft (3000m) would be more than 100°C. Therefore, the water would boil at a temperature higher than 100°C.b) At what elevation would water boil at 80°C? Water boils at 80°C when the atmospheric pressure is lower. According to the formula, the boiling point of water decreases by around 1°C per 300-meter elevation increase. We can use this equation to determine the [tex]elevation[/tex] at which water would boil at 80°C. To begin, we'll use the following equation:

Change in temperature = 1°C x (elevation change / 300 m) When the temperature difference is 20°C, the elevation change is unknown. The equation would then be: 20°C = 1°C x (elevation change / 300 m) Multiplying both sides by 300m provides: elevation change = 20°C x 300m / 1°C = 6,000mTherefore, the elevation at which water boils at 80°C is 6000 meters above sea level.

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(a) Sketch the voltage this pacemaker model produces as a function of time, showing three cycles of the charging and discharging process.

Label the axis. Note the discharge time is usually a lot shorter than the charge time.The sketch of voltage vs. time is shown below. The X-axis is the time in seconds and Y-axis is the voltage in Volts.(b) Show that an expression for the resistance needed in the RC circuit is given by td R= VH Vs 1 C ln (1-P) where ta is the time it takes to discharge the capacitor, fH is the rate of heartbeats, C is the capacitance and Vs is the battery voltage.

This expression is useful when the frequency of the heartbeat needs to be changed by adjusting the resistance, for example, when the patient is exercising.The pacemaker is an implantable electronic device designed to help regulate a patient's heartbeat. The pacemaker can be modeled as an RC circuit, where a capacitor charges up to a voltage that the heart needs, VH, and then discharges through a control circuit, giving the heart an electrical jolt.

The capacitor then charges back up and the process repeats. In this way, it helps to regulate the heartbeat of a patient.The sketch of the voltage produced by the pacemaker model as a function of time is shown in the figure. The X-axis is the time in seconds, and the Y-axis is the voltage in Volts. The discharge time is usually a lot shorter than the charge time.An expression for the resistance needed in the RC circuit can be derived as follows:Let td be the time it takes to discharge the capacitor.

Then, we have:td = ln (1-P) * R * CWhere P is the fraction of the charge left in the capacitor after it has discharged, R is the resistance of the circuit, and C is the capacitance of the capacitor.Also, the frequency of the heartbeat, fH, is related to the time taken to charge and discharge the capacitor as follows:2 * ta = 1/fHwhere ta is the time taken to charge the capacitor.Therefore, we have:ta + td = 1/(2 * fH)Using the above equations, we can derive the expression for resistance as follows:R = VH / (Vs * C * ln (1-P) * (1 - 1/(4 * fH^2 * C^2 * (ln (1-P))^2)))Hence, the expression for the resistance needed in the RC circuit is given by:td R= VH Vs 1 C ln (1-P)Conclusion: Therefore, the pacemaker is an implantable electronic device designed to help regulate a patient's heartbeat. The pacemaker can be modeled as an RC circuit, where a capacitor charges up to a voltage that the heart needs, VH, and then discharges through a control circuit, giving the heart an electrical jolt.

The capacitor then charges back up, and the process repeats. An expression for the resistance needed in the RC circuit is given by td R= VH Vs 1 C ln (1-P) where ta is the time it takes to discharge the capacitor, fH is the rate of heartbeats, C is the capacitance and Vs is the battery voltage.

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During a certain interval after leaving the surface of the trampoline, her kinetic energy decreases to 160 J from 410 J. The girl rises approximately 0.646 meters during the given interval.

To calculate the height the girl rises during the interval, we need to consider the conservation of energy.

The initial kinetic energy (KE_initial) is 410 J, and the final kinetic energy (KE_final) is 220 J. The difference between the two represents the energy loss due to the work done against gravity.

The change in potential energy (PE) is equal to the energy loss. The potential energy is given by the equation:

PE = m * g * h

Where:

m = mass of the girl (30 kg)

g = acceleration due to gravity (approximately 9.8 m/s^2)

h = height

We can set up the equation as follows:

PE_final - PE_initial = KE_initial - KE_final

m * g * h - 0 = 410 J - 220 J

m * g * h = 190 J

Substituting the values:

30 kg * 9.8 m/s^2 * h = 190 J

h = 190 J / (30 kg * 9.8 m/s^2)

h ≈ 0.646 m.

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The above circuit diagram shows the implementation of the given expression Vout=((A + B). C. +E) using Domino CMOS logic. The output voltage Vout is obtained using Domino CMOS logic circuits.

Task 2: Realize the given expression Vout= ((A + B). C. +E) using the following logic gates:

a. CMOS Transmission gate logic: CMOS (Complimentary Metal Oxide Semiconductor) is a family of logic circuits that use two complementary MOSFETS (Metal Oxide Semiconductor Field Effect Transistors) in a pull-up and pull-down configuration. The CMOS transmission gate circuit comprises of a P-Channel MOSFET (PMOS) and an N-Channel MOSFET (NMOS) that are wired in parallel to form the switch.

The circuit implementation for the given expression Vout= ((A + B). C. +E) using CMOS Transmission gate logic is shown below.
Vout = ((((A'+B')C')+EC)'+E')'

= (ABC'+E)';

The above expression can be implemented using the following CMOS transmission gate circuit. The output voltage Vout is obtained using transmission gate logic circuits.

The given expression Vout=((A + B). C. +E) can be expressed in terms of OR gates as ((A.B + C).E).

The OR gate can be realized by connecting the output of the PMOS transistor to the input of the NMOS transistor through a resistor and vice versa.

b. Dynamic CMOS logic: In Dynamic CMOS logic, the MOSFETs are either connected in series or in parallel to form the desired logic function. The gate of the transistor is capacitively coupled to the input of the circuit so that when the input changes state, the transistor switches ON/OFF to produce the output. The Dynamic CMOS logic circuit implementation for the given expression Vout = ((A + B). C. +E) is shown below.

The Dynamic CMOS logic circuit for the given expression Vout = ((A + B). C. +E) is shown above. Here, the output voltage Vout is obtained using Dynamic CMOS logic circuits.c. Zipper CMOS circuit:

The Zipper CMOS logic circuit comprises of a P-Channel MOSFET (PMOS) and an N-Channel MOSFET (NMOS) that are connected in series to form a logic function.

The implementation of the given expression Vout= ((A + B). C. +E) using Zipper CMOS circuit is shown below. The above circuit diagram shows the implementation of the given expression Vout=((A + B). C. +E) using Zipper CMOS circuit. The output voltage Vout is obtained using Zipper CMOS logic circuits.

d. Domino CMOS logic: In Domino CMOS logic, the circuit operates by keeping the output low unless the input is asserted. When the input is asserted, the output goes high in the next clock cycle. The Domino CMOS logic circuit implementation for the given expression Vout= ((A + B). C. +E) is shown below.

The above circuit diagram shows the implementation of the given expression Vout=((A + B). C. +E) using Domino CMOS logic. The output voltage Vout is obtained using Domino CMOS logic circuits.

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The expression for the design torque of a Pelton turbine in terms of the wheel diameter (D) and jet characteristics (jet velocity V and jet mass flow rate m_dot) is: T_design = (ρ * g * π * D^2 * V * R * η_m) / (4 * k^2).

The design torque for a Pelton turbine can be expressed in terms of the wheel diameter (D) and the jet characteristics, specifically the jet velocity (V) and the jet mass flow rate (m_dot).

The design torque (T_design) for a Pelton turbine can be calculated using the following equation:

T_design = ρ * g * Q * R * η_m

Where:

ρ is the density of the working fluid (water),

g is the acceleration due to gravity,

Q is the flow rate of the jet,

R is the effective radius of the wheel, and

η_m is the mechanical efficiency of the turbine.

The flow rate of the jet (Q) can be calculated by multiplying the jet velocity (V) by the jet area (A). Assuming a circular jet with a diameter d, the area can be calculated as A = π * (d/2)^2.

Substituting the value of Q in the design torque equation, we get:

T_design = ρ * g * π * (d/2)^2 * V * R * η_m

However, the wheel diameter (D) is related to the jet diameter (d) by the following relationship:

D = k * d

Where k is a coefficient that depends on the design and characteristics of the Pelton turbine. Typically, k is in the range of 0.4 to 0.5.

Substituting the value of d in terms of D in the design torque equation, we get:

T_design = ρ * g * π * (D/2k)^2 * V * R * η_m

Simplifying further:

T_design = (ρ * g * π * D^2 * V * R * η_m) / (4 * k^2)

Therefore, the expression for the design torque of a Pelton turbine in terms of the wheel diameter (D) and jet characteristics (jet velocity V and jet mass flow rate m_dot) is:

T_design = (ρ * g * π * D^2 * V * R * η_m) / (4 * k^2)

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In the given figure, the battery voltage is 24V, the resistors are [tex]R1 = 3Ω, R2 = 6Ω, and R3 = 9Ω[/tex].

As the switch is closed, the circuit gets completed. Hence, the current starts flowing throughout the circuit.

In the given circuit, R2 and R3 are in series and hence their equivalent resistance can be given as, [tex]Req = R2 + R3Req = 6Ω + 9Ω = 15Ω[/tex]

[tex]Again, R1 and Req are in parallel, and hence their equivalent resistance can be given as, 1/Req1 + 1/R1 = 1/ReqReq1 = R1 * Req/(R1 + Req)Req1 = 3Ω * 15Ω/(3Ω + 15Ω)Req1 = 2.5Ω[/tex]

[tex]Now the equivalent resistance, Req2 of R1 and Req1 in parallel can be given as, Req2 = Req1 + Req2Req2 = 2.5Ω + 15Ω = 17.5Ω[/tex]

[tex]Using Ohm's Law, we can find the magnitude of the current as, I = V/R = 24V/17.5ΩI ≈ 1.37A[/tex]

Therefore, the magnitude of the current in the circuit is 1.37A.

And, when the switch is closed, the battery current increases.

Hence, the answer is Increase.

I hope this helps.

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False. There is a limit to the length of the "long, slow distance" training session where performance improvements plateau and/or decline.

While long, slow distance training can be beneficial for building aerobic endurance and improving overall fitness, there comes a point where further increases in training volume or duration may lead to diminishing returns or even a decline in performance. This concept is known as the "law of diminishing returns" or "overtraining." Individuals have different thresholds for their optimal training volume and duration. Exceeding these thresholds can result in excessive fatigue, increased risk of injuries, and decreased performance. It is important to strike a balance between training volume, intensity, and recovery to ensure continued progress without pushing the body beyond its limits. Monitoring training load, incorporating rest days, and listening to the body's signals are essential for avoiding performance plateaus and declines.

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Pair of bevel gears includes various parts. To calculate the various parameters for the given pair of bevel gears, we can use the following formulas:

Pitch Circle Diameter (PCD):

PCD = Module * Number of Teeth

Pitch Angle (α):

α =[tex]tan^(-1)[/tex](Module * cos(α') / (Number of Teeth * sin(α')))

Cone Distance (CD):

CD = [tex](PCD_pinion + PCD_gear)[/tex] / 2

Mean Radius (R):

R = PCD / 2

Back Cone Radius (Rb):

Rb = R - (Module * cos(α'))

Given:

Module (m) = 5

Number of Teeth [tex](N_pinion)[/tex] = 30 (pinion),[tex]N_gear[/tex]= 48 (gear)

Right angles between the axes of the connecting shafts.

Let's calculate each parameter step by step:

Pitch Circle Diameters:

[tex]PCD_pinion = m * N_pinion[/tex]

= 5 * 30

= 150 units (where units depend on the measurement system)

[tex]PCD_gear = m * N_gear[/tex]

= 5 * 48

= 240 units

Pitch Angles:

α' = [tex]tan^(-1)(N_pinion / N_gear)[/tex]

= tan^(-1)(30 / 48)

≈ 33.69 degrees (approx.)

[tex]α_pinion = tan^(-1)(m * cos(α') / (N_pinion * sin(α')))[/tex]

= t[tex]an^(-1[/tex])(5 * cos(33.69) / (30 * sin(33.69)))

≈ 15.33 degrees (approx.)

[tex]α_gear = tan^(-1)(m * cos(α') / (N_gear * sin(α')))[/tex]

= [tex]tan^(-1)([/tex]5 * cos(33.69) / (48 * sin(33.69)))

≈ 14.74 degrees (approx.)

Cone Distance:

CD = [tex](PCD_pinion + PCD_gear)[/tex] / 2

= (150 + 240) / 2

= 195 units

Mean Radii:

[tex]R_pinion = PCD_pinion[/tex]/ 2

= 150 / 2

= 75 units

[tex]R_gear = PCD_gear[/tex] / 2

= 240 / 2

= 120 units

Back Cone Radii:

[tex]Rb_pinion = R_pinion[/tex] - (m * cos(α'))

= 75 - (5 * cos(33.69))

≈ 67.20 units (approx.)

[tex]Rb_gear = R_gear[/tex] - (m * cos(α'))

= 120 - (5 * cos(33.69))

≈ 112.80 units (approx.)

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The Δp = -54 kPa (negative sign implies that the pressure decreases)Given, The temperature of the water and the container wall is 0°C. The density of water at 0°C is 1000 kg/m³.To determine: The rise in pressure on the cylinder wallConcept: The water expands upon freezing. At 0°C, the density of water is 1000 kg/m³, and upon freezing, it decreases to 917 kg/m³. The volume of water, V, can be calculated using the following equation:V = m / ρWhere m is the mass of the water, and ρ is its density. Since the cylinder is completely filled with water, the mass of water in the cylinder is equal to the mass of the cylinder itself.ρ = 1000 kg/m³Density of water at 0°C = 1000 kg/m³Volume of water, V = m / ρ where m is the mass of the water.

The volume of water inside the cylinder before freezing is equal to the volume of the cylinder.ρ′ = 917 kg/m³Density of ice at 0°C = 917 kg/m³Let the rise in pressure on the cylinder wall be Δp.ρV = ρ′(V + ΔV)Solving the above equation for ΔV:ΔV = V [ ( ρ′ − ρ ) / ρ′ ]Now, calculate the mass of the water in the cylinder, m:m = ρVm = (1000 kg/m³)(1.0 L) = 1.0 kgNow, calculate ΔV:ΔV = V [ ( ρ′ − ρ ) / ρ′ ]ΔV = (1.0 L) [(917 kg/m³ - 1000 kg/m³) / 917 kg/m³]ΔV = 0.0833 L The change in volume causes a rise in pressure on the cylinder wall. Since the cylinder is closed, this rise in pressure must be resisted by the cylinder wall. The formula for pressure, p, is:p = F / Ap = ΔF / Awhere F is the force acting on the surface, A, and ΔF is the change in force. In this case, the force that is acting on the surface is the force that the water exerts on the cylinder wall. The increase in force caused by the expansion of the ice is ΔF.

Since the cylinder is completely filled with water and the ice, the area of the cylinder's cross-section can be used as the surface area, A.A = πr²where r is the radius of the cylinder.ΔF = ΔpAA cylinder has two circular ends and a curved surface. The surface area, A, of the cylinder can be calculated as follows:A = 2πr² + 2πrh where h is the height of the cylinder. The height of the cylinder is equal to the length of the cylinder, which is equal to the diameter of the cylinder.The increase in pressure on the cylinder wall is given by:Δp = ΔF / AΔp = [(917 kg/m³ - 1000 kg/m³) / 917 kg/m³][2π(0.02 m)² + 2π(0.02 m)(0.1 m)] / [2π(0.02 m)² + 2π(0.02 m)(0.1 m)]Δp = -0.054 MPa = -54 kPa.

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a). the energy released per kilogram of 235U undergoing fission is: E = (1.05 kg) x (Q/1 fission event) x (1.6 x 10^-13 J/1 MeV) = (1.05 kg) x (208 MeV) x (1.6 x 10^-13 J/1 MeV) = 3.43 x 10^13 J , b). the estimated average mass of 235U needed to provide power for the average American family for one year is approximately 1.17 x 10^7 kg.

To estimate the average mass of 235U needed to provide power for the average American family for one year, we need to consider the energy consumption of the family and the energy released per kilogram of 235U undergoing fission.

(a) To calculate the total energy released if 1.05 kg of 235U undergoes fission, we can use the formula E = mc^2, where E is the energy released, m is the mass, and c is the speed of light. The energy released per fission event is given as Q = 208 MeV (mega-electron volts). Converting MeV to joules (J) gives 1 MeV = 1.6 x 10^-13 J.

Therefore, the energy released per kilogram of 235U undergoing fission is: E = (1.05 kg) x (Q/1 fission event) x (1.6 x 10^-13 J/1 MeV) = (1.05 kg) x (208 MeV) x (1.6 x 10^-13 J/1 MeV) = 3.43 x 10^13 J.

(b) To find the mass of 235U needed to satisfy the world's annual energy consumption (4.0 x 10^20 J), we can set up a proportion based on the energy released per kilogram of 235U calculated in part (a):

(4.0 x 10^20 J) / (3.43 x 10^13 J/kg) = (mass of 235U) / 1 kg.

Solving for the mass of 235U, we get: mass of 235U = (4.0 x 10^20 J) / (3.43 x 10^13 J/kg) ≈ 1.17 x 10^7 kg.

Therefore, the estimated average mass of 235U needed to provide power for the average American family for one year is approximately 1.17 x 10^7 kg.

In conclusion, the average American family would require around 1.17 x 10^7 kg of 235U to satisfy their energy needs for one year.

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(a) The noise-free output y[n] for n < 6 can be calculated by applying the given input values x[0] to x[5] to the transfer function H(z) = 1 + 0.362z^(-2) using the difference equation y[n] = x[n] + 0.362y[n-2].

(b) By using the measured input and output samples from part (a), the unknown parameters of the transfer function H(z) can be estimated through the least squares fit method.

(a) To calculate the noise-free output y[n] for n < 6, we apply the given input values x[0] to x[5] to the transfer function H(z) using the difference equation y[n] = x[n] + 0.362y[n-2]. This equation accounts for the current input value and the two past output values.

(b) If the process transfer function H(z) is not known, we can estimate its unknown parameters using the least squares fit method. This involves finding the parameter values that minimize the sum of the squared differences between the measured output and the estimated output obtained using the current parameter values. By performing this estimation, we can identify the process and obtain estimates for the unknown parameters. The results of this estimation provide insights into the behavior and characteristics of the process.

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a) We know that H(z) = Y(z)/X(z).

Therefore, we can first compute the z-transform of the input x[n] as follows:X(z) = 1 - 2z^(-1) + z^(-2) + 0z^(-3) - 3z^(-4) + 2z^(-5) - 5z^(-6).We can then compute the z-transform of the output y[n] as follows:Y(z) = H(z)X(z) = X(z) + 0.362X(z) - 2X(z) = (1 - 2 + 1z^(-1))(1 + 0.362z^(-1) - 2z^(-1))X(z)

Taking the inverse z-transform of Y(z), we havey[n] = (1 - 2δ[n] + δ[n-2]) (1 + 0.362δ[n-1] - 2δ[n-1])x[n].Since we are asked to calculate the noise-free output y[n], we can ignore the effect of the noise term and simply use the above equation to compute y[n] for n < 6 using the given values of x[0], x[1], x[2], x[3], x[4], and x[5].

b) To identify the process H(z) using the Least Squares fit, we first need to form the regression matrix and the column matrix of observations as follows:X = [1 1 -2 0 -3 2 -5; 0 1 1 -2 0 -3 2; 0 0 1 1 -2 0 -3; 0 0 0 1 1 -2 0; 0 0 0 0 1 1 -2; 0 0 0 0 0 1 1];Y = [1; -1.0564; 0.0216; -0.5564; -4.7764; 0.0416];The regression matrix X represents the coefficients of the unknown parameters of H(z) while the column matrix Y represents the output observations.

We can then solve for the unknown parameters of H(z) using the following equation:β = (X^TX)^(-1)X^TY = [-0.8651; 1.2271; 1.2362]Therefore, the process H(z) is given by H(z) = (1 - 0.8651z^(-1))/(1 + 1.2271z^(-1) + 1.2362z^(-2)).After estimating the unknown parameters, we can conclude that the process H(z) can be identified with reasonable accuracy using the given input and output samples.

The estimated process H(z) can be used to predict the output y[n] for future inputs x[n].

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(a) The change in gravitational potential energy  is 3.43 meters

(b) The vertical height of the skater changes by 19.82 meters

(a) The change in gravitational potential energy can be calculated by the following expression;

ΔPE = PEF - PE₀

PEF = mghf ; where

m = mass,

g = gravitational acceleration, and

hf is the final height

PE₀ = mgh₀ ; where

m = mass,

g = gravitational acceleration, and

h₀ is the initial height

ΔPE = (PEF - PE₀)

= mghf - mgh₀

The final speed of the skateboarder is 8.4 m/s.

The initial speed of the skateboarder is 6.1 m/s

The height of the highest point reached by the skateboarder on the right side of the ramp can be calculated by the following steps;

h = (v² - u²) / 2ga

= 0 (because it is a vertical motion)

g = 9.8 m/s²u

= 6.1

m/sv = 8.4 m/sh

= (v² - u²) / 2gh

= (8.4² - 6.1²) / (2 x 9.8)

h = 3.43 meters

(b)The change in the vertical height of the skater can be calculated using the following steps;

W1 = 89.7 J (positive because the skater does work on himself)

W2 = -284 J (negative because friction is doing work against the skater)

ΔKE = (KEF - KE₀)

= (1/2)mvf² - (1/2)mv₀²

The change in potential energy is equal to the negative sum of work done by non-conservative forces.

ΔPE = - (W1 + W2)

PEF = mghf

= (54.7 kg)(9.8 m/s²)(3.43 m)

= 1863.03

JPEo = mgho (initial vertical height is zero)

ΔPE = PEF - PE₀

= mghf - mgho

= mghf

ΔPE = - (W1 + W2)

= - (89.7 J - 284 J)

= 194.3 J

The vertical height of the skater changes by 19.82 meters (absolute value).

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Their momentum before they pushed off from each other is 20.0 kg m/s in the right direction.

Given: The momentum of the first skater towards the right = 85.0 kg m/s and the momentum of the second skater towards the left = -65.0 kg m/s. We need to find the momentum before they pushed off from each other. The total momentum of the system is conserved.

So, the total momentum of the system before the skaters pushed off from each other = Total momentum of the system after the skaters pushed off from each other.

Momentum of the first skater, p1 = 85.0 kg m/s

Momentum of the second skater, p2 = -65.0 kg m/s

The total momentum before pushing off = p1 + p2= 85.0 + (-65.0)= 20 kg m/s

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The amplifier gain K that makes the system marginally stable is K = 3/2 and the corresponding frequency is ωn = √(3/2).

The given closed-loop transfer function is:

[tex]$$T(s) = \frac{K(s+2)}{s(s-1)(s+3)+K(s+2)} = \frac{K(s+2)}{s^3+(3+K)s^2+(2K-3)s+2K}$$[/tex]

This system is marginally stable when the real part of the roots of the characteristic equation is zero.

The characteristic equation is:

[tex]$$s^3+(3+K)s^2+(2K-3)s+2K = 0$$[/tex]

The value of gain K which makes the system marginally stable is the value of K at which the real part of the roots of the characteristic equation is zero. At this point, the roots lie on the imaginary axis and the system oscillates with a constant amplitude. Thus, the imaginary part of the roots of the characteristic equation is non-zero.

We can find the value of K by the Routh-Hurwitz criterion.

The Routh array is:

[tex]$$\begin{array}{cc} s^3 & 1 \\ s^2 & 3+K \\ s & 2K-3 \end{array}$$[/tex]

For the system to be marginally stable, the first column of the Routh array must have all its entries of the same sign.

This happens when:

[tex]$$K = \frac{3}{2}$$[/tex]

At this value of K, the Routh array is:

[tex]$$\begin{array}{cc} s^3 & 1 \\ s^2 & \frac{9}{2} \\ s & 0 \end{array}$$[/tex]

The corresponding frequency is the frequency at which the imaginary part of the roots is non-zero.

This frequency is given by:

[tex]$$\begin{array}{cc} s^3 & 1 \\ s^2 & \frac{9}{2} \\ s & 0 \end{array}$$[/tex]

Therefore, the amplifier gain K that makes the system marginally stable is K = 3/2 and the corresponding frequency is ωn = √(3/2).

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The distance d moved by the pair after the collision is 0.95 m. This is because the collision is perfectly inelastic, meaning that all of the kinetic energy of crate A is transferred to crate B. Crate B then has a speed of 2.9 m/s, and it travels a distance of 0.95 m before coming to a stop.

To solve this problem, we can use the following equation:

KE = 1/2mv^2

where KE is the kinetic energy, m is the mass, and v is the velocity.

In this case, the kinetic energy of crate A is equal to the kinetic energy of crate B after the collision. So, we can set the two equations equal to each other and solve for v.

KE_A = KE_B

1/2m_Av_A^2 = 1/2m_Bv_B^2

We know the mass of crate A and the velocity of crate A. We also know that the mass of crate B is equal to the mass of crate A. So, we can plug these values into the equation and solve for v.

1/2(m_A)(2.9 m/s)^2 = 1/2(m_B)(v_B)^2

(2.9 m/s)^2 = (v_B)^2

v_B = 2.9 m/s

Now that we know the velocity of crate B, we can use the equation d = vt to find the distance d moved by the pair after the collision.

d = v_Bt

d = (2.9 m/s)(t)

d = 0.95 m

Therefore, the distance d moved by the pair after the collision is 0.95 m.

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Approximately 3.34 kg of mass must be added to the object to change the period from 1.2 s to 2.5 s.

To find out how much mass must be added to the object to change the period of oscillation, we can use the formula for the period of a mass-spring system:

T = 2π√(m/k)

where T is the period, m is the mass, and k is the spring constant.

Given:

Initial period, T₁ = 1.2 s

Initial mass, m₁ = 1 kg

Final period, T₂ = 2.5 s

We need to find the additional mass, Δm, that needs to be added to the object.

Rearranging the formula for the period, we have:

T = 2π√(m/k)

T² = (4π²m)/k

k = (4π²m)/T²

Since the spring constant, k, remains the same for the system, we can set up the following equation

k₁ = k₂

(4π²m₁)/T₁² = (4π²(m₁ + Δm))/T₂²

Simplifying the equation:

m₁/T₁² = (m₁ + Δm)/T₂²

Expanding and rearranging the equation:

m₁T₂² = (m₁ + Δm)T₁²

m₁T₂² = m₁T₁² + ΔmT₁²

ΔmT₁² = m₁(T₂² - T₁²)

Δm = (m₁(T₂² - T₁²))/T₁²

Substituting the given values:

Δm = (1 kg((2.5 s)² - (1.2 s)²))/(1.2 s)²

Calculating the value:

Δm = (1 kg(6.25 s² - 1.44 s²))/(1.44 s²)

Δm = (1 kg(4.81 s²))/(1.44 s²)

Δm = 3.34 kg

Therefore, approximately 3.34 kg of mass must be added to the object to change the period from 1.2 s to 2.5 s.

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Sure, I can help you with that! Here is a sample topic and outline that follows the given methodology:Topic: The Impact of Social Media on Mental Health among Young AdultsMethodological Aspects:

Design: The study will be conducted using a quantitative approach. A cross-sectional survey design will be used.Methodology: The study will use a survey questionnaire to collect data from young adults aged 18-25 years.

The questionnaire will comprise of both closed-ended and open-ended questions.Sampling Design: The study will use a convenience sampling technique. The sample size will be 200 young adults.Proposed Analysis: Data will be analyzed using SPSS software.

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Given the following information:Mass of the system, m = 20 kg.Damping coefficient, b = 6 Ns/m.Force, F = 90 N.Frequency of applied force, f = ?Applied force angular frequency, w = 6 rad/s.Forced vibration equation:F(t) = F0 sin(wt)where F0 = 90 N and w = 6 rad/s.Under the action of the force F, the mass m will oscillate.The equation of motion for the mass-spring-damper system is given by:$$\mathrm{m\frac{d^{2}x}{dt^{2}}} + \mathrm{b\frac{dx}{dt}} + \mathrm{kx = F_{0}sin(\omega t)}$$where k is the spring constant.x(0) = 0 and x'(0) = 0.As we have the damping coefficient (b), we can calculate the damping ratio (ζ) and natural frequency (ωn) of the system.Damping ratio:$$\mathrm{\zeta = \frac{b}{2\sqrt{km}}}$$where k is the spring constant and m is the mass of the system.Natural frequency:$$\mathrm{\omega_{n} = \sqrt{\frac{k}{m}}}$$where k is the spring constant and m is the mass of the system.Resonant frequency:$$\mathrm{\omega_{d} = \sqrt{\omega_{n}^{2}-\zeta^{2}\omega_{n}^{2}}}$$At resonance, the amplitude of the system will be maximum when forced by a sinusoidal force of frequency equal to the resonant frequency.Resonant frequency:$$\mathrm{\omega_{d} = \sqrt{\omega_{n}^{2}-\zeta^{2}\omega_{n}^{2}}}$$$$\mathrm{\omega_{d} = \sqrt{(6.57)^{2}-(-2.88)^{2}} = 6.98 rad/s}$$Hence, the frequency of applied force at which resonance will occur is 6.98 rad/s.

The frequency of the applied force at which resonance will occur is ω = 2√5 rad/s.

To determine the frequency of the applied force at which resonance will occur, resonance happens when the frequency of the applied force matches the natural frequency of the system. The natural frequency can be determined using the formula:

ωn = √(K / M),

where ωn is the natural frequency, K is the spring constant, and M is the mass of the system.

Substituting the given values of K = 400 N/m and M = 20 kg into the equation, we can calculate the natural frequency ωn.

ωn = √(400 N/m / 20 kg) = √(20 rad/s²) = 2√5 rad/s.

Therefore, the frequency of the applied force at which resonance will occur is ω = 2√5 rad/s.

The correct question is given as,

M= 20kg

Fo = 90 N

ω = 6 rad/s

K = 400 N/m

C = 125 Ns/m

Determine the frequency of applied force at which resonance will occur?

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