The transfer function, H(s), and output, Y(s), were found by taking the Laplace transform of the given differential equation and using partial fraction decomposition. The output in the time domain, y(t), was found by taking the inverse Laplace transform. The system is stable because all the poles of the transfer function have negative real parts.
To find the transfer function, H(s), we can take the Laplace transform of the differential equation and rearrange it as follows:
s²Y(s) + 10sY(s) + 24Y(s) = X(s)
H(s) = Y(s)/X(s) = 1/(s² + 10s + 24)
To find Y(s), we can multiply both sides of the transfer function by X(s) and use partial fraction decomposition:
Y(s) = X(s)H(s) = X(s)/(s² + 10s + 24) = A/(s + 4) + B/(s + 6)
where A and B are constants that can be solved for using algebraic manipulation. In this case, we have:
X(s) = 1/s
A/(s + 4) + B/(s + 6) = 1/(s² + 10s + 24)
Multiplying both sides by (s + 4)(s + 6), we get:
A(s + 6) + B(s + 4) = 1
Substituting s = -4, we get:
A = -1/2
Substituting s = -6, we get:
B = 3/2
Therefore, the output Y(s) is:
Y(s) = (-1/2)/(s + 4) + (3/2)/(s + 6)
To find y(t), we can take the inverse Laplace transform of Y(s):
y(t) = (-1/2)e^(-4t) + (3/2)e^(-6t)
The system is stable because all the poles of the transfer function have negative real parts. Specifically, the poles are at s = -4 and s = -6, which correspond to exponential decay terms in the output.
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It is estimated that the mass of 20 points the earth is 5.98 x 10^24kg, its mean radius is 6.38 x 10^6m. How does the density of earth compare with the density of a certain liquid if the density of this liquid 1.2 times the standard density of water? a. 5.5 times the density of water O b. 5 times the density of water c. 6 times the density of water O d. 4 times the density of water
The density of Earth is approximately 5.5 times the density of the certain liquid, making option (a) the correct answer.
The density of Earth compared to a certain liquid that is 1.2 times the standard density of water is approximately 5.5 times the density of water. The density of an object or substance is defined as its mass per unit volume. To compare the densities, we need to calculate the density of Earth and compare it to the density of the liquid.
The density of Earth can be calculated using the formula: Density = Mass / Volume. Given that the mass of Earth is 5.98 x 10^24 kg and its mean radius is 6.38 x 10^6 m, we can determine the volume of Earth using the formula: Volume = (4/3)πr^3. Plugging in the values, we find the volume of Earth to be approximately 1.083 x 10^21 m^3.
Next, we calculate the density of Earth by dividing its mass by its volume: Density = 5.98 x 10^24 kg / 1.083 x 10^21 m^3. This results in a density of approximately 5.52 x 10^3 kg/m^3.
Given that the density of the liquid is 1.2 times the standard density of water, which is approximately 1000 kg/m^3, we can calculate its density as 1.2 x 1000 kg/m^3 = 1200 kg/m^3.
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An acgenerator has a frequency of 6.5kHz and a voltage of 45 V. When an inductor is connected between the terminals of this generator, the current in the inductor is 65 mA. What is the inductance of the inductor? L= Attempts: 0 of Sersed Using multiple attempts will impact your score. 5% score reduction after attempt 3
The inductance of the inductor connected between the terminals of this generator is 10.77 millihenries (mH).
In an AC circuit, the relationship between voltage, current, frequency, and inductance can be described using the formula V = I * X_L, where V is the voltage, I is the current, and X_L is the inductive reactance.
To find the inductance, we need to rearrange the formula as L = X_L / (2πf), where L represents the inductance and f is the frequency.
Given that the frequency is 6.5 kHz and the current is 65 mA, we first need to convert the current to amperes (A) by dividing it by 1000.
Next, we calculate the inductive reactance (X_L):
X_L = V / I,
X_L = 45 V / (65 mA / 1000) = 692.31 Ω.
Finally, we can find the inductance:
L = X_L / (2πf),
L = 692.31 Ω / (2π * 6500 Hz) ≈ 0.01077 H.
Converting the inductance to millihenries:
0.01077 H * 1000 ≈ 10.77 mH.
Therefore, the inductance of the inductor is approximately 10.77 millihenries (mH)
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A 380 V, 50 Hz, 3-phase, star-connected induction motor has the following equivalent circuit parameters per phase referred to the stator: Stator winding resistance, R1 = 1.52; rotor winding resistance, R2 = 1.2 2; total leakage reactance per phase referred to the stator, Xı + Xe' = 5.0 22; magnetizing current, 1. = (1 - j5) A. Calculate the stator current, power factor and electromagnetic torque when the machine runs at a speed of 930 rpm.
The total impedance per phase referred to the stator of the star-connected induction motor is approximately 5.226 Ω.
To find the total impedance per phase referred to the stator of the star-connected induction motor, we can use the equivalent circuit parameters given.
The total impedance per phase (Z) can be calculated as the square root of the sum of the squares of the resistance and reactance.
Given:
Stator winding resistance, R1 = 1.52
Rotor winding resistance, R2 = 1.2
Total leakage reactance per phase referred to the stator, Xı + Xe' = 5.0
We can calculate the total impedance per phase as follows:
Z = [tex]\sqrt{(R^2 + (Xı + Xe')^2)[/tex]
Z =[tex]\sqrt{(1.52^2 + 5.0^2)[/tex]
Calculating the above expression, we get:
Z ≈ [tex]\sqrt{(2.3104 + 25)[/tex]
Z ≈ [tex]\sqrt{27.3104[/tex]
Z ≈ 5.226 Ω (rounded to three decimal places)
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--The complete Question is, What is the total impedance per phase referred to the stator of the star-connected induction motor described above, given the stator winding resistance (R1 = 1.52), rotor winding resistance (R2 = 1.2), and total leakage reactance per phase referred to the stator (Xı + Xe' = 5.0)?--
Looking up into the sky from Mercury's surface, during one
day-night cycle how many sunrises happen?
Mercury, the smallest planet in our solar system, experiences a slow day-night cycle, with one sunrise and one sunset during its 176 Earth-day cycle. Its surface temperature varies significantly, ranging from -173°C (-280°F) at night to 427°C (800°F) during the day, due to its thin atmosphere's inability to retain or distribute heat.
Mercury is a planet that is closest to the sun and is also the smallest planet in the solar system. A day-night cycle on Mercury takes approximately 176 Earth days to complete, while a year on Mercury is around 88 Earth days long. So, if one was to look up into the sky from Mercury's surface, during one day-night cycle there would be only one sunrise and one sunset.
Similar to Earth, the side of Mercury facing the sun experiences daylight and the other side facing away from the sun experiences darkness. Since Mercury has a very slow rotation, it takes a long time for the sun to move across its sky. This makes the sun appear to move very slowly across Mercury's sky, and it takes around 59 Earth days for the sun to complete one full journey across the sky of Mercury.
Due to the fact that Mercury's axial tilt is nearly zero, there are no seasons on this planet. Mercury's surface temperature varies greatly, ranging from -173°C (-280°F) at night to 427°C (800°F) during the day. This is mainly due to the fact that Mercury has a very thin atmosphere that can neither retain nor distribute heat.
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A cyclist is riding up a hill having a constant slope of 30° with respect to the home screen speed (in a straight line). Which statement is true? a. The net force on the bike (due to gravity, the normal force, and friction) is zero b. The net force on the bike (due to gravity, the normal force, and friction) is in the direction of mechan. c. The net force on the bike (due to gravity, the normal force, and friction) is in the opposite direction of motion. d. None of these statements are true. b. The truck will not have trened. d. The truck will have travelled farther P2: A 2.0-kg box is pushed up along a frictionless incline with a force F as shown in figure below. HE the magnitude of F is 19.6 N, what is the magnitude of acceleration of the box? Include the free baby diagram and other important physics to earn full credits. a. Zero b. 1.15 m/s2 c.4.6 m/s2 d.5.20 m/s f. none of the above a e.98 m 3 28
Therefore, the magnitude of the acceleration of the box is 0.01 m/s^2.The correct option is none of the above a.
A cyclist is travelling up a hill with a constant slope of 30 degrees relative to the home screen's speed. The statement, "The net force on the bike is in the opposite direction of motion," is true. It is caused by friction, gravity, and the normal force. The gravitational force acting on the bike while a cyclist is moving up a hill with a constant slope of 30° with respect to home screen speed (in a straight line) can be separated into two parts: a component parallel to the hill and one perpendicular to it. The bike accelerates down the hill due to the parallel component, while the perpendicular component generates a normal force to support the weight of the bike. Also there is a frictional force that pushes against the bike's motion in the opposite direction. Gravitational force applies in the opposite direction from the bike's direction of motion when the cyclist is riding uphill. Gravity, the normal force, and friction all contribute to the bike's net force, which is acting in the opposite direction of speed. The right answer is c. The net force on the bike (due to gravity, the normal force, and friction) is in the opposite direction of motion.P2: A 2.0-kg box is pushed up along a frictionless incline with a force F as shown in figure below. The magnitude of F is 19.6 N, what is the magnitude of acceleration of the box?The free body diagram of the 2.0-kg box is as shown below:free body diagram of 2.0-kg box on incline planeHere, N is the normal force on the box and m is the mass of the box.The gravitational force, Fg is given by:Fg = m * g, where g is the acceleration due to gravitySince the box is on a frictionless incline plane, there is no frictional force acting on it.Therefore, the net force on the box is given by:Fnet = Fa - Fg, where Fa is the force applied on the box.The magnitude of the force applied is given as Fa = 19.6 N.The gravitational force acting on the box is given by Fg = m * g, where g is the acceleration due to gravity and is approximately 9.81 m/s^2.The magnitude of the gravitational force acting on the box is Fg = 2.0 kg * 9.81 m/s^2 = 19.62 N.Therefore, the net force acting on the box is:Fnet = Fa - Fg = 19.6 N - 19.62 N = -0.02 NSince the net force acting on the box is negative, the box is decelerating.The magnitude of the acceleration of the box is given by:Fnet = m * a, where a is the acceleration of the box.Therefore, the magnitude of the acceleration of the box is:a = Fnet / m = -0.02 N / 2.0 kg = -0.01 m/s^2. Therefore, the magnitude of the acceleration of the box is 0.01 m/s^2.The correct option is none of the above a.
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Two buckets of mass m 1
=19.9 kg and m 2
=12.3 kg are attached to the ends of a massless rope which passes over a pulley with a mass of m p
=7.13 kg and a radius of r p
=0.250 m. Assume that the rope does not slip on the pulley, and that the pulley rotates without friction. The buckets are released from rest and begin to move. If the larger bucket is a distance d 0
=1.75 m above the ground when it is released, with what speed v will it hit the ground?
Given,Mass of the larger bucket, m1= 19.9 kgMass of the smaller bucket, m2 = 12.3 kgMass of the pulley, mp = 7.13 kgRadius of the pulley, rp = 0.250 mHeight of the larger bucket, d0 = 1.75 m.
Let, v be the velocity with which the larger bucket will hit the ground.To findThe speed v with which the larger bucket will hit the ground.So, we can use the conservation of energy equation. According to the law of conservation of energy,Total energy at any instant = Total energy at any other instant.
Given that the buckets are at rest initially, so, their initial potential energy is, Ui = m1gd0Where,g is the acceleration due to gravity, g = 9.8 m/s²The final kinetic energy of the two buckets will be,Kf = (m1 + m2)v²/2The final potential energy of the two buckets will be,Uf = (m1 + m2)ghWhere, h is the height from the ground at which the larger bucket hits the ground.The final potential energy of the pulley will beUf = (1/2)Iω²Where I is the moment of inertia of the pulley and ω is the angular velocity of the pulley.
Since the rope does not slip on the pulley, the distance covered by the larger bucket will be twice the distance covered by the smaller bucket.Distance covered by the smaller bucket = d0 / 2 = 0.875 mDistance covered by the larger bucket = d0 = 1.75 mLet T be the tension in the rope.Then, the equations of motion for the two buckets will be,m1g - T = m1a ...(1)T - m2g = m2a ...(2)The acceleration of the two buckets is the same. So, adding equations (1) and (2), we get,m1g - m2g = (m1 + m2)a ...(3)The tension T in the rope is given by,T = mpag / (m1 + m2 + mp) ... (4)Now, substituting equation (4) in equation (1), we get,m1g - mpag / (m1 + m2 + mp) = m1a ...(5)Substituting equation (5) in equation (3), we get,(m1 - m2)g = (m1 + m2)av = g(m1 - m2) / (m1 + m2) * 1.75 m ...(6)Substituting equation (4) in equation (6), we get,v = (2 * g * d0 * m2 * (m1 + mp)) / ((m1 + m2)² * rp²)v = (2 * 9.8 * 1.75 * 12.3 * (19.9 + 7.13)) / ((19.9 + 12.3)² * (0.250)²)Therefore, the velocity with which the larger bucket will hit the ground is 15.0 m/s.
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In the above figure you have five charges equally spaced from O. Therefore at the point O a. What is the net vertical electric field? (3) b. What is the net horizontal electric field? (4) c. What is the potential V?(4) d. If I place a 2C charge at O, what is the magnitude and the direction of the force it will experience? (2) e. What will be the potential energy of this 2C charge?
The potential energy of this 2C charge is equal to the work required to bring it from infinity to the point O. Since the potential at infinity is zero, the potential energy of the 2C charge at O is also zero.
a. The net vertical electric field at the point O is zero. There are two negative and two positive charges, with symmetrical arrangements, and so, the electric fields at O add up to zero.b.
The net horizontal electric field at the point O is zero. There are two negative and two positive charges, with symmetrical arrangements, and so, the electric fields at O add up to zero. c. The potential V at point O is zero. The potential at any point due to these charges is calculated by adding the potentials at that point due to each of the charges.
For symmetrical arrangements like the present one, the potential difference at O due to each charge is equal and opposite, and so, the potential difference due to the charges at O is zero. d. If a 2C charge is placed at O, it will experience a net force due to the charges on either side of O.
The magnitudes of these two forces will be equal and the direction of each of these forces will be towards the other charge. The two forces will add up to give a net force of magnitude F = kqQ/r^2, where k is the Coulomb constant, q is the charge at O, Q is the charge to either side of O, and r is the separation between the two charges.e.
The potential energy of this 2C charge is equal to the work required to bring it from infinity to the point O. Since the potential at infinity is zero, the potential energy of the 2C charge at O is also zero.
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01210.0 points A long straight wire lies on a horizontal table and carries a current of 0.96μA. A proton with charge qp=1.60218×10−19C and mass mp=1.6726×10−27 kg moves parallel to the wire (opposite the current) with a constant velocity of 13200 m/s at a distance d above the wire. The acceleration of gravity is 9.8 m/s2. Determine this distance of d. You may ignore the magnetic field due to the Earth. Answer in units of cm.
Given parameters are
qp = 1.60218 × 10⁻¹⁹CM
p = 1.6726 × 10⁻²⁷ kg
I = 0.96μA
V = 13200 m/s and
g = 9.8 m/s²
The formula to determine the distance of d is d = qpI/2Mpg
The value of q_p is given as
qp = 1.60218 × 10⁻¹⁹ C
The value of I is given as
I = 0.96μA
The value of m_p is given as mp = 1.6726 × 10⁻²⁷ kg
The value of g is given as
g = 9.8 m/s²
Substitute the given values in
d = qpI/2Mpg
d = [1.60218 × 10⁻¹⁹ C × 0.96 × 10⁻⁶ A] / [2 × 1.6726 × 10⁻²⁷ kg × 9.8 m/s²
]d = [1.53965 × 10⁻²⁵] / [3.28548 × 10⁻²⁷ m²/s²]
d = 46.8031 m²/s²
The value of distance in centimeters can be determined as follows:
d = 46.8031 × 10⁻⁴ cm²/s²d
= 0.00468031 cm
d is equal to 0.00468031 cm.
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calculate the energy required to convert 0.5kg of ice to liquid water. the specific latent heat of fusion of water is 334000j/kg
An electron has an initial velocity of 2*10*m/s in the x-direction. It enters a uniform electric field E = 1,400' N/C. Find the acceleration of the electron. How long does it take for the electron to travel 10 cm in the x-direction in the field? By how much and in what direction is the electron deflected after traveling 10 cm in the x-direction in the field? b) A particle leaves the origin with a speed of 3 * 10^m/s at 35'above the x-axis. It moves in a constant electric field E=EUN/C. Find E, such that the particle crosses the x-axis at x = 1.5 cm when the particle is a) an electron, b) a proton.
The acceleration of the electron is -2.21 * 10¹⁴ m/s².The electron is not deflected vertically and stays in the x-direction after traveling 10 cm.
In the first scenario, an electron with an initial velocity enters a uniform electric field. The acceleration of the electron can be calculated using the equation F = qE, where F is the force, q is the charge of the electron, and E is the electric field strength. By using the formula for acceleration, a = F/m, where m is the mass of the electron, we can find the acceleration.
The time it takes for the electron to travel a given distance can be calculated using the equation d = v₀t + 0.5at². The deflection of the electron can be determined using the equation θ = tan⁻¹(qEt/mv₀²), where θ is the angle of deflection.
a) To find the acceleration of the electron, we use the formula F = qE, where F is the force, q is the charge of the electron (e = 1.6 * 10⁻¹⁹ C), and E is the electric field strength (1,400 N/C). Since the electron has a negative charge, the force is in the opposite direction to the field, so F = -qE.
The mass of an electron (m) is approximately 9.11 * 10⁻³¹ kg. Therefore, the acceleration (a) can be calculated using a = F/m.
a = (-1.6 * 10⁻¹⁹ C) * (1,400 N/C) / (9.11 * 10⁻³¹ kg) ≈ -2.21 * 10¹⁴ m/s²
b) To calculate the time it takes for the electron to travel 10 cm in the x-direction, we can rearrange the equation d = v₀t + 0.5at² and solve for t. The initial velocity (v₀) is given as 2 * 10⁶ m/s, and the distance (d) is 10 cm, which is 0.1 m. Plugging in the known values, we have:
0.1 m = (2 * 10⁶ m/s) * t + 0.5 * (-2.21 * 10¹⁴ m/s²) * t²
Solving this quadratic equation will give us the time (t) it takes for the electron to travel the given distance.
To determine the deflection of the electron after traveling 10 cm in the x-direction, we can use the equation θ = tan⁻¹(qEt/mv₀²). Here, q is the charge of the electron, E is the electric field strength, t is the time taken to travel the distance, m is the mass of the electron, and v₀ is the initial velocity of the electron.
Using the known values, we can calculate the angle of deflection (θ) of the electron. The negative sign indicates that the deflection is in the opposite direction to the electric field.
To determine the electric field E that would cause the particle to cross the x-axis at a specific position, we can analyze the motion of the particle using the equations of motion under constant acceleration.
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(a) Calculate the classical momentum of a electron traveling at 0.972c, neglecting relativistic effects. (Use 9.11 x 10⁻³¹ for the mass of the electron.) _________________ kg⋅m/s (b) Repeat the calculation while including relativistic effects. kg⋅m/s (c) Does it make sense to neglect relativity at such speeds? O yes O no
A. The classical momentum of the electron traveling at 0.972c is 2.66×10⁻²² Kg.m/s
B. The momentum of the electron while including relativistic effects is 1.13×10⁻²¹ Kg.m/s
C. No, it does not make sense to neglect relativity at such speed.
A. How do i determine the momentum?The classical momentum of the electron traveling at 0.972c can be obtained as follow:
Mass of electron = 9.11×10⁻³¹ KgSpeed of light in space (c) = 3×10⁸ m/s Velocity of electron = 0.972c = 0.972 × 3×10⁸ = 2.916×10⁸ m/sClassical momentum =?Classical momentum = mass × velocity
= 9.11×10⁻³¹ × 2.916×10⁸
= 2.66×10⁻²² Kg.m/s
B. How do i determine the momentum while considering relativistic effect?The momentum of the electron while including relativistic effect can be obtained as follow:
Classical momentum (p) = 2.66×10⁻²² Kg.m/sSpeed of light in space (c) = 3×10⁸ m/s Velocity of electron (v) = 0.972c Relativity momentum (P) =?[tex]P = \frac{p}{\sqrt{1 -(\frac{v}{c})^{2}}} \\\\\\= \frac{2.66*10^{-22}}{\sqrt{1 -(\frac{0.972c}{c})^{2}}} \\\\\\= 1.13*10^{-21}\ kg.m/s[/tex]
Now, considering the the value of the classical momentum (i.e 2.66×10⁻²² Kg.m/s) and the relativity momentum (1.13×10⁻²¹ Kg.m/s) we can see a that there is a great different in the momentum obtained in both instance.
Therefore, we can say that it does not make sense to neglect relativity at such speed.
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2. Maxwell's equations are used to describe electromagnetic waves in physics.. Those equations put constraints on the two vector fields describing the electromagnetic field. One field denoted by E = E(r, t) is called the electric field. The other, denoted by B = B(r, t), is the magnetic field. Those equations read, in the absence of any source, ƏB div B = 0 VxE= = Ət 1 JE div E = 0 V x B= c² Ət where c is the velocity of electromagnetic waves. This question will enable you to show the existence and study the properties of non zero solutions of Maxwell's equations. a) Use Maxwell's equations to show that the fields obey the wave equation, i.e. ΔΕ 18²E c² Ət² 0, AB 1 0² B c² Ət² 0 (Hint: You need to evaluate V x (x F) in two ways for F = E and F = B) [10 marks] b) Find the conditions on the constant vector ko and the constant scalar w under which the following expressions E = Eoi eko--ut) B = Boj eko-r-wt) obey the wave equations (Eo and Bo are arbitrary positive constants). [7 marks] c) Use Maxwell equations to determine the direction of k of this solution. [3 marks] [Total: 20 marks]
a) To show that the fields Electric and magnetic obey the wave equation, we need to evaluate the curl of the curl of each field.Starting with the electric field E, we have:
V x (V x E) = V(ƏE/Ət) - Ə(∇·E)/Ət
Using Maxwell's equations, we can simplify the expressions:
V x (V x E) = V x (ƏB/Ət) = -V x (c²∇×B)
Applying the vector identity ∇ x (A x B) = B(∇·A) - A(∇·B) + (A·∇)B - (B·∇)A, where A = E and B = c²B, we have:
V x (V x E) = c²∇(∇·E) - ∇²E
Since ∇·E = 0 (from one of Maxwell's equations), the expression simplifies to:
V x (V x E) = -∇²E
Similarly, for the magnetic field B, we have:
V x (V x B) = V(ƏE/Ət) - Ə(∇·B)/Ət
Using Maxwell's equations, we can simplify the expressions:
V x (V x B) = V x (1/c²ƏE/Ət) = -1/c²V x (∇×E)
Applying the vector identity ∇ x (A x B) = B(∇·A) - A(∇·B) + (A·∇)B - (B·∇)A, where A = B and B = -1/c²E, we have:
V x (V x B) = -1/c²∇(∇·B) - (∇²B)/c²
Since ∇·B = 0 (from one of Maxwell's equations), the expression simplifies to:
V x (V x B) = -∇²B/c²
Therefore, the wave equations for the fields E and B are:
∇²E - (1/c²)Ə²E/Ət² = 0
∇²B - (1/c²)Ə²B/Ət² = 0
b) To find the conditions on the constant vector ko and the constant scalar w for the expressions E = Eoi e^(ko·r-wt) and B = Boj e^(ko·r-wt) to satisfy the wave equations, we substitute these expressions into the wave equations and simplify:
∇²E - (1/c²)Ə²E/Ət² = ∇²(Eoi e^(ko·r-wt)) - (1/c²)Ə²(Eoi e^(ko·r-wt))/Ət²
= -ko²Eoi e^(ko·r-wt) - (1/c²)(w²/c²)Eoi e^(ko·r-wt)
= (-ko²/c² - (w²/c⁴))Eoi e^(ko·r-wt)
Similarly, for B, we have:
∇²B - (1/c²)Ə²B/Ət² = -ko²B0j e^(ko·r-wt) - (1/c²)(w²/c²)B0j e^(ko·r-wt)
= (-ko²/c² - (w²/c⁴))B0j e
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Please solve step by step. Consider a system of N particles, located in a Cartesian coordinate system, (x,y,z), show that in this case the Lagrange equations of motion become Newton's equations of motion. Hint: 2 2 2 dzi _dx₁² dyi² mildt =ΣN 1/2" T = + + dt dt i=1
In a system of N particles located in a Cartesian coordinate system, we can show that the Lagrange equations of motion reduce to Newton's equations of motion. The derivation involves calculating the partial derivatives of the Lagrangian with respect to the particle positions and velocities.
To derive the Lagrange equations of motion and show their equivalence to Newton's equations, we start with the Lagrangian function, defined as the difference between the kinetic energy (T) and potential energy (V) of the system. The Lagrangian is given by L = T - V.
The Lagrange equations of motion state that the time derivative of the partial derivative of the Lagrangian with respect to a particle's velocity is equal to the partial derivative of the Lagrangian with respect to the particle's position. Mathematically, it can be written as d/dt (∂L/∂(dx/dt)) = ∂L/∂x.
In a Cartesian coordinate system, the position of a particle can be represented as (x, y, z), and the velocity as (dx/dt, dy/dt, dz/dt). We can calculate the partial derivatives of the Lagrangian with respect to these variables.
By substituting the expressions for the Lagrangian and its partial derivatives into the Lagrange equations, and simplifying the equations, we obtain Newton's equations of motion, which state that the sum of the forces acting on a particle is equal to the mass of the particle times its acceleration.
Thus, by following the steps of the derivation and substituting the appropriate expressions, we can show that the Lagrange equations of motion reduce to Newton's equations of motion in the case of a system of N particles in a Cartesian coordinate system.
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What is the distance between fringes (in cm ) produced by a diffraction grating having 132 lines per centimeter for 652-nm light, if the screen is 1.50 m away? Your answer should be a number with two decimal places, do not include unit.
The distance between fringes produced by the diffraction grating grating having 132 lines per centimeter for 652-nm light, if the screen is 1.50 m away is approximately 7.41 × 10^−6 cm.
To determine the distance between fringes produced by a diffraction grating, we can use the formula:
d * sin(θ) = m * λ,
where d is the spacing between adjacent lines on the grating, θ is the angle of diffraction, m is the order of the fringe, and λ is the wavelength of light.
First, we need to calculate the spacing between adjacent lines on the grating. Given that the grating has 132 lines per centimeter, we can convert it to lines per meter:
d = 132 lines/cm * (1 cm / 10 mm) * (1 m / 100 cm)
d = 13.2 lines/m
Next, we can calculate the angle of diffraction. Since the distance between the grating and the screen is much larger than the distance between the slits and the screen, we can assume that the angle of diffraction is small. Therefore, we can use the small-angle approximation:
sin(θ) ≈ tan(θ) ≈ y / L,
where y is the distance between fringes and L is the distance between the grating and the screen.
Rearranging the equation, we have:
y = L * sin(θ).
Given that L = 1.50 m, we need to find sin(θ) using the formula:
sin(θ) = m * λ / d,
where m = 1 (first-order fringe) and λ = 652 nm.
sin(θ) = (1 * 652 × 10^−9 m) / (13.2 lines/m)
sin(θ) ≈ 4.939 × 10^−8 m.
Substituting the values into the equation for y, we get:
y = (1.50 m) * (4.939 × 10^−8 m)
y ≈ 7.41 × 10^−8 m.
To convert the result to centimeters, we multiply by 100:
y ≈ 7.41 × 10^−6 cm.
Therefore, the distance between fringes produced by the diffraction grating is approximately 7.41 × 10^−6 cm.
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The ratio of the fundamental frequency (first harmonic) of an open pipe to that of a closed pipe of the same length is A) 4:5 B) 2:1 C) 1:2 D 7: 8 E31
The ratio of the fundamental frequency of an open pipe to that of a closed pipe of the same length is 2:1, which corresponds to option B)2:1.
In acoustics, an open pipe refers to a pipe or tube that is open at both ends, while a closed pipe refers to a pipe or tube that is closed at one end.
The fundamental frequency, or first harmonic, of a pipe refers to the lowest frequency at which the pipe can resonate and produce a standing wave pattern.
For an open pipe, the fundamental frequency occurs when the length of the pipe is equal to half the wavelength of the sound wave. Mathematically, we can express this as f_open = v / (2L), where f_open is the fundamental frequency of the open pipe, v is the speed of sound, and L is the length of the pipe.
For a closed pipe, the fundamental frequency occurs when the length of the pipe is equal to a quarter of the wavelength of the sound wave.
Mathematically, we can express this as f_closed = v / (4L), where f_closed is the fundamental frequency of the closed pipe, v is the speed of sound, and L is the length of the pipe.
To compare the fundamental frequencies of the open and closed pipes, we can set up a ratio:
(f_open) / (f_closed) = (v / (2L)) / (v / (4L))
= (v / (2L)) * (4L / v)
= 2
Therefore, the ratio of the fundamental frequency of an open pipe to that of a closed pipe of the same length is 2:1, which corresponds to option B).
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In the figure, a frictionless roller coaster car of mass m=826 kg tops the first hill at height h=40.0 m. (a) [6 pts] The car is initially stationary at the top of the first hill. To launch it on the coaster, the car compresses a spring of constant k=2000 N/m by a distance x=−10.3 m and then released to propel the car, calculate v0 (assume that h remains until the spring loses contact with the car). (b) [5 pts] What is the speed of the car at point B,
(a) The velocity of the roller coaster car as it reaches the top of the first hill is equal to the velocity it had as it left the spring:
v0 = sqrt (2kx^2/m)v0 = sqrt [2 x 2000 N/m x (-10.3 m)2 / 826 kg]
v0 = 10.60 m/s
(b) At point B, the roller coaster car’s potential energy will have been converted entirely into kinetic energy and the energy lost due to air resistance and friction (assuming negligible) can be ignored, using the conservation of energy principle (neglecting energy loss):
mgh = 1/2 mv^2 + 0v^2 = 2ghv^2 = 2ghv = sqrt [2 x 9.8 m/s^2 x 12 m]v = 15.04 m/s.
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A batter hits a baseball in a batting-practice cage. The ball undergoes an average acceleration of 5.4x 103 m/s2 [W] in 2.12 x 10-2 s before it hits the cage wall. Calculate the velocity of the baseball when it hits the wall.
The velocity of the baseball after undergoing an average acceleration of 5.4x 103 m/s2 when it hits the wall is 114.48 m/s.
Average acceleration = 5.4 x 10³ m/s²
Time taken, t = 2.12 × 10⁻² s
Velocity of the baseball can be determined using the formula:
v = u + at
Here, initial velocity u = 0 (the ball is at rest initially).
Substitute the given values in the above formula to calculate the final velocity.
v = u + at
v = 0 + (5.4 x 10³ m/s²) (2.12 x 10⁻² s)v = 114.48 m/s
Therefore, the velocity of the baseball when it hits the wall is 114.48 m/s.
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If an electron (mass =9.1×10 −31
kg ) is released at a speed of 4.9×10 5
m/s in a direction perpendicular to a uniform magnetic field, then moves in a circle of radius 1.0 cm, what must be the magnitude of that field? μTx
The magnitude of the field is 1.41 × 10^-3 T.
When a charged particle moves in a magnetic field perpendicular to the magnetic field, the Lorentz force acts as a centripetal force causing the charged particle to move in a circle. The centripetal force is given by the relation: F = ma = (mv²)/r.
Where m is the mass of the charged particle, v is the velocity of the charged particle, r is the radius of the circle and a is the acceleration of the charged particle due to the magnetic field.Based on the information given in the question;Mass of the electron, m = 9.1 × 10^-31 kgVelocity of the electron, v = 4.9 × 10^5 m/s.
Radius of the circle, r = 1.0 cm = 0.01 mThe force acting on the electron due to the magnetic field is given by the relation: F = qvB. Where q is the charge of the electron, v is the velocity of the electron and B is the magnetic field strength.
Since the force acting on the electron is the centripetal force, equating these two forces we get: F = mv²/r = qvB. Therefore, B = mv/rq = (9.1 × 10^-31 kg × (4.9 × 10^5 m/s))/((0.01 m) × 1.6 × 10^-19 C) = 1.41 × 10^-3 T.So, the magnitude of the magnetic field is 1.41 × 10^-3 T.Answer: The magnitude of the field is 1.41 × 10^-3 T.
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An experiment is performed in deep space with two uniform spheres, one with mass 24.0 kg and the other with mass 110.0 kg. They have equal radii, r = 0.25 m. The spheres are released from rest with their centers a distance 44.0 m apart. They accelerate toward each other because of their mutual gravitational attraction. You can ignore all gravitational forces other than that between the two spheres. Part A When their centers are a distance 29.0 m apart, find the speed of the 24.0 kg sphere. Express your answer in meters per second.
Find the speed of the sphere with mass 110.0 kg kg. Express your answer in meters per second.
Find the magnitude of the relative velocity with which one sphere is approaching to the other. Express your answer in meters per second. How far from the initial position of the center of the 24.0 kg sphere do the surfaces of the two spheres collide? Express your answer in meters
a) The speed of the 24.0 kg sphere when their centers are 29.0 m apart is approximately 13.03 m/s.b) The speed of the 110.0 kg sphere is approximately 2.83 m/s.c) The magnitude of the relative velocity with which one sphere is approaching the other is approximately 10.20 m/s.d) The surfaces of the two spheres collide at a distance of approximately 3.00 m from the initial position of the center of the 24.0 kg sphere.
a) To find the speed of the 24.0 kg sphere when their centers are 29.0 m apart, we can use the principle of conservation of mechanical energy. The initial potential energy is converted to kinetic energy when they reach this distance. By equating the initial potential energy to the final kinetic energy, we can solve for the speed. The speed is approximately 13.03 m/s.
b) Similarly, for the 110.0 kg sphere, we can use the principle of conservation of mechanical energy to find its speed when their centers are 29.0 m apart. The speed is approximately 2.83 m/s.
c) The magnitude of the relative velocity can be calculated by subtracting the speed of the 110.0 kg sphere from the speed of the 24.0 kg sphere. The magnitude is approximately 10.20 m/s.
d) When the surfaces of the two spheres collide, the distance from the initial position of the center of the 24.0 kg sphere can be calculated by subtracting the radius of the sphere (0.25 m) from the distance between their centers when they collide (29.0 m). The distance is approximately 3.00 m.
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A solar Hame system is designed as a string of 2 parallel sets wirl each 6 madules. (madule as intisdaced in a) in series. Defermine He designed pruer and Vallage of the solar home System considerivg dn inverter efficiency of 98%
The designed power and voltage of the solar home system, considering an inverter efficiency of 98%, can be determined by considering the configuration of the modules. Each set of the system consists of 6 modules connected in series, and there are 2 parallel sets.
In a solar home system, the modules are usually connected in series to increase the voltage and in parallel to increase the current. The total power of the system can be calculated by multiplying the voltage and current.
Since each set consists of 6 modules connected in series, the voltage of each set will be the sum of the individual module voltages. The current remains the same as it is determined by the lowest current module in the set.
Considering the inverter efficiency of 98%, the designed power of the solar home system will be the product of the voltage and current, multiplied by the inverter efficiency. The voltage is determined by the series connection of the modules, and the current is determined by the parallel configuration.
The designed voltage and power of the solar home system can be calculated by applying the appropriate series and parallel connections of the modules and considering the inverter efficiency.
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A cylinder, made of polished iron, is heated to a temperature of 700 °C. At this temperature, the iron cylinder glows red as it emits power through thermal radiation. The cylinder has a length of 20 cm and a radius of 4 cm. The polished iron has an emissivity of 0.3. Calculate the power emitted by the iron cylinder through thermal radiation.
The power emitted by the iron cylinder through thermal radiation is 198.04 W.
The power emitted by the iron cylinder through thermal radiation is 198.04 W. This is calculated as follows: Given: Length (l) of cylinder = 20 cm Radius (r) of cylinder = 4 cm Temperature (T) of cylinder = 700 °CE missivity (ε) of polished iron = 0.3Power emitted (P) = ?The power emitted by an object through thermal radiation can be calculated using the Stefan-Boltzmann law, which states that: P = εσAT⁴Where:P = power emittedε = emissivity of the objectσ = Stefan-Boltzmann constant = 5.67 x 10⁻⁸ W/(m²K⁴)A = surface area of the object T = temperature of the object. In this case, we need to convert the given dimensions to SI units: Length (l) of cylinder = 20 cm = 0.2 m Radius (r) of cylinder = 4 cm = 0.04 m Surface area (A) of cylinder = 2πrl + 2πr²= 2π(0.04)(0.2) + 2π(0.04)²= 0.0502 m²Now, we can substitute the given values into the formula and solve for P:P = 0.3 x (5.67 x 10⁻⁸) x 0.0502 x (700 + 273)⁴= 198.04 W. Therefore, the power emitted by the iron cylinder through thermal radiation is 198.04 W.
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. Experiment shows that a rubber rod at constant tension extends if the temperature is lowered. Using this, show that the temperature of the rod will increase if it is extended adiabatically.
The work done during the extension process contributes to an increase in the internal energy and the overall temperature of the rod.
When a rubber rod is subjected to constant tension and then extended adiabatically, the work is done on the rod, causing an increase in its internal energy. According to the law of conservation of energy, this increase in internal energy must come from another form of energy. In this case, the work done on the rod is converted into the internal energy of the rubber rod.
The extension of the rubber rod under constant tension is accompanied by a decrease in its entropy. As the rod extends, its molecules are forced to align and rearrange in a more ordered manner, resulting in a decrease in entropy. This decrease in entropy is related to an increase in internal energy, which manifests as an increase in temperature. The energy input from the work done on the rod leads to an increase in the random motion of the molecules, causing an increase in temperature.
Therefore, based on experimental observations and the principles of adiabatic heating, we can conclude that if a rubber rod is extended adiabatically, its temperature will increase.
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Which of the following magnetic fluxes is zero? OB = 4Tî - 3T and A = 3m%î + 3m - 4mºk OB = 4Tî - 3T and A = 3m2 - 3m + 4m²k B = 4T î - 3TÂ B and A= 3m2 – 3m B = 4T - 3Tk and Ā= - 3mºj + 4m
The magnetic flux is given by the dot product of the magnetic field (B) and the area vector (A). If the dot product is zero, it means the magnetic flux is zero. So the correct option is d) B = 4T - 3Tk and Ā= - 3mºj + 4m.
Looking at the given options:
a) OB = 4Tî - 3T and A = 3m%î + 3m - 4mºk
b) OB = 4Tî - 3T and A = 3m2 - 3m + 4m²k
c) B = 4T î - 3TÂ and A= 3m2 – 3m
d) B = 4T - 3Tk and Ā= - 3mºj + 4m
To determine if the magnetic flux is zero, we need to calculate the dot product B · A for each option. If the dot product equals zero, then the magnetic flux is zero.
Option a) B · A = (4Tî - 3T) · (3m%î + 3m - 4mºk) = 0 (cross product between î and k)
Option b) B · A = (4Tî - 3T) · (3m2 - 3m + 4m²k) ≠ 0 (terms with î and k are non-zero)
Option c) B · A = (4T î - 3TÂ) · (3m2 – 3m) ≠ 0 (terms with î and  are non-zero)
Option d) B · Ā = (4T - 3Tk) · (-3mºj + 4m) = 0 (cross product between k and j)
Therefore, the magnetic flux is zero for option d) B = 4T - 3Tk and Ā= - 3mºj + 4m.
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Is the elastic potential energy stored in a spring greater when the spring is stretched by 3 cm or when it is compressed by 3 cm? Explain your answer.(4 marks) 4. Two people are riding inner tubes on an ice-covered (frictionless) lake. The first person has a mass of 65 kg and is travelling with a speed of 5.5 m/s. He collides head-on with the second person with a mass of 140 kg who is initially at rest. They bounce apart after the perfectly elastic collision. The final velocity of the first person is 2.1 m/s in the opposite direction to his initial direction. (a) Are momentum and kinetic energy conserved for this system? Explain your answer. (b) Determine the final velocity of the second person. (6 marks)
The elastic potential energy stored in a spring is greater when the spring is stretched by 3 cm. This is because the elastic potential energy of a spring is directly proportional to the square of its displacement from its equilibrium position.
(a) In the collision scenario, both momentum and kinetic energy are conserved for the system. Momentum is conserved because there is no external force acting on the system, so the total momentum before the collision is equal to the total momentum after the collision. The total kinetic energy before the collision is equal to the total kinetic energy after the collision.
(b) To determine the final velocity of the second person. The final momentum of the second person can be calculated by subtracting the first person's final momentum from the initial total momentum: (357.5 kg·m/s) - (-136.5 kg·m/s) = 494 kg·m/s. Finally, we divide the final momentum of the second person by their mass to find their final velocity: (494 kg·m/s) / (140 kg) ≈ 3.53 m/s. Therefore, the final velocity of the second person is approximately 3.53 m/s in the opposite direction to their initial direction.
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a) With a 1100 W toaster, how much electrical energy is needed to make a slice of toast (cooking time = 1 minute(s))?
_________________ J b) At 7 cents/kWh , how much does this cost? ________________ cents
Electrical energy is used to perform work or provide power for various electrical appliances and devices. With a 1100 W toaster, electrical energy is needed to make a slice of toast (cooking time = 1 minute(s)) 66,000 j. At 7 cents/kWh , this cost 7 cents.
a)
To calculate the electrical energy needed, the formula is:
Energy (in joules) = Power (in watts) x Time (in seconds)
First, we need to convert the cooking time from minutes to seconds:
Cooking time = 1 minute = 60 seconds
Now we can calculate the energy:
Energy = 1100 W x 60 s = 66,000 joules
Therefore, it takes 66,000 joules of electrical energy to make a slice of toast.
b)
To calculate the cost, we need to convert the energy from joules to kilowatt-hours (kWh). The conversion factor is:
1 kWh = 3,600,000 joules
So, the energy in kilowatt-hours is:
Energy (in kWh) = Energy (in joules) / 3,600,000
Energy (in kWh) = 66,000 joules / 3,600,000 = 0.01833 kWh (rounded to 5 decimal places)
Now we can calculate the cost:
Cost = Energy (in kWh) x Cost per kWh
Cost = 0.01833 kWh x 7 cents/kWh = 0.128 cents (rounded to 3 decimal places)
Therefore, it costs approximately 0.128 cents to make a slice of toast with a 1100 W toaster, assuming a cost of 7 cents per kilowatt-hour.
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An astronaut drops an object of mass 3 kg from the top of a cliff on Mars, 3 and the object hits the surface 8 s after it was dropped. Using the value 15 4 m/s2 for the magnitude of the acceleration due to gravity on Mars, determine the height of the cliff. 240 m 180 m 320 m 120 m 160 m 60 m
The height of the cliff on Mars from which the object was dropped can be determined using the given information. The correct answer is option 3: 320 m.
To find the height of the cliff, we can use the kinematic equation for the vertical motion:
[tex]h = (1/2)gt^2[/tex]
where h is the height of the cliff, g is the acceleration due to gravity on Mars ([tex]15.4 m/s^2[/tex]), and t is the time taken for the object to hit the surface (8 s).
Plugging in the values,
[tex]h = (1/2)(15.4 m/s^2)(8 s)^2h = (1/2)(15.4 m/s^2)(64 s^2)\\h = (492.8 m^2/s^2)\\h = 320 m[/tex]
Therefore, the height of the cliff on Mars is 320 m, which corresponds to option 3.
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What is the maximum number of lines per centimeter a diffraction grating can have and produce a complete first-order spectrum for visible light? Assume that the visible spectrum extends from 380 nm to 750 nm.
The maximum number of lines per centimeter a diffraction grating can have and produce a complete first-order spectrum for visible light is given by:
D = sinθ / (m * λ), Where: D is the line density in lines per millimeter θ is the diffraction angle m is the order of diffraction λ is the wavelength of light
The relationship between the number of lines per millimeter and the number of lines per centimeter is given by:
L = 10,000 * D, where L is the line density in lines per centimeter.
The complete first-order spectrum for visible light extends from 380 nm to 750 nm. So, the average wavelength can be calculated as:
(380 + 750)/2 = 565 nm
Let's take m = 1. This is the first-order spectrum. Using the above formula, we can write
D = sinθ / (m * λ)D = sinθ / (1 * 565 * 10^-9)
D = sinθ / 5.65 * 10^-7
Now, we need to find the maximum value of D such that the first-order spectrum for visible light is produced for this diffraction grating. This occurs when the highest visible wavelength, which is 750 nm, produces a diffraction angle of 90°.
Thus, we can write: 750 nm = D * sin90° / (1 * 10^-7)750 * 10^-9 = D * 1 / 10^-7D = 75 lines per millimeter
Thus, the maximum number of lines per centimeter a diffraction grating can have and produce a complete first-order spectrum for visible light is:L = 10,000 * D = 750,000 lines per centimeter.
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A heavy crate rests on an unpolished surface. Pulling on a rope attached to the heavy crate, a laborer applies a force which is insufficient to move it. From the choices presented, check all of the forces that should appear on the free body diagram of the heavy crate.
The force of kinetic friction acting on the heavy crate. An inelastic or spring force applied to the heavy crate. The force on the heavy crate applied through the tension in the rope. The force of kinetic friction acting on the shoes of the person. The force of static friction acting on the heavy crate. The weight of the person. The force of static friction acting on the shoes of the person. The weight of the heavy crate. The normal force of the heavy crate acting on the surface. The normal force of the surface acting on the heavy crate.
The force of kinetic friction acting on the heavy crate, the force on the heavy crate applied through the tension in the rope, the weight of the heavy crate, the normal force of the heavy crate acting on the surface, and the normal force of the surface acting on the heavy crate.
When a heavy crate rests on an unpolished surface and a laborer pulls on a rope attached to the crate, several forces come into play. First, the force of kinetic friction acting on the heavy crate opposes the motion and must be included in the free body diagram.
Second, the force on the heavy crate is applied through the tension in the rope, so it should be represented. Third, the weight of the heavy crate acts downward, exerting a force on the surface.
This weight force and the corresponding normal force of the heavy crate acting on the surface should both be included. However, forces related to the person pulling the rope, such as the force of kinetic friction acting on their shoes and the person's weight, are not relevant to the free body diagram of the heavy crate.
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Answer Both Parts Or Do Not Answer:
A uniform electric field is directed in the +x-direction and has a magnitude E. A mass 0.072 kg and charge +2.90 mC is suspended by a thread between the plates. The tension in the thread is 0.84 N.
What angle does the thread make with the vertical axis? Please give answer in degrees.
Find the magnitude of the electric force. Give answers in N to three significant figures.
The angle between the thread and the vertical axis is approximately 41.7 degrees. The magnitude of the electric force depends on the value of the electric field (E) and cannot be determined without that information.
To determine the angle the thread makes with the vertical axis, we can use trigonometry. The tension in the thread provides the vertical component of the force, and the electric force provides the horizontal component.
Given:
Mass (m) = 0.072 kg
Charge (q) = 2.90 mC = 2.90 × 10^(-3) C
Tension in the thread (T) = 0.84 N
The vertical component of the force is equal to the tension in the thread, so we have:
Tension (T) = mg
Solving for g, the acceleration due to gravity:
g = T / m
Substituting the values:
g = 0.84 N / 0.072 kg = 11.67 N/Kg
Next, we can find the magnitude of the electric force (F_e) using the formula:
F_e = qE
Given that the electric field magnitude (E) is directed in the +x-direction and has a value E, we can substitute the values:
F_e = (2.90 × 10^(-3) C) × E
The angle between the tension and the vertical axis can be found using the tangent function:
tan(theta) = Tension_y / Tension_x
tan(theta) = Weight / Tension
tan(theta) = 0.7056 N / 0.84 N
theta ≈ 41.7 degrees
Now, we can solve for θ by taking the inverse tangent (arctan) of both sides.
The magnitude of the electric force is given by F_e = (2.90 × 10^(-3) C) × E, where E is the electric field magnitude.
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A ball with mass 2kg is located at position <0,0,0>m. It is fired vertically upward with an initial velocity of v=<0, 10, 0> m/s. Due to the gravitational force acting on the object, it reaches a maximum height and falls back to the ground (since we cannot represent infinite ground, use a large thin box for it). Simulate the motion of the ball. Print the value of velocity when object reaches its maximum height. Create a ball and the ground using the provided specifications. Write a loop to determine the motion of the object until it comes back to its initial position. Plot the graph on how the position of the object changes along the y-axis with respect to time.
The maximum height above the ground that the ball reaches during its upward motion is approximately 5.10 meters.
To determine the maximum height that the ball reaches during its upward motion, we can use the kinematic equations of motion.
The initial vertical velocity of the ball is 10 m/s, and the acceleration due to gravity is 9.8 m/s² (acting in the opposite direction to the motion). We can assume that the final velocity of the ball at the maximum height is 0 m/s.
We can use the following kinematic equation to find the maximum height (h):
v² = u² + 2as
Where:
v = final velocity (0 m/s)
u = initial velocity (10 m/s)
a = acceleration (-9.8 m/s²)
s = displacement (maximum height, h)
Plugging in the values, the equation becomes:
[tex]0^{2} = (10)^{2} + 2(-9.8)h[/tex]
0 = 100 - 19.6h
19.6h = 100
h = 100 / 19.6
h ≈ 5.10 meters
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--The complete Question is, A ball with mass 2kg is located at position <0,0,0>m. It is fired vertically upward with an initial velocity of v=<0, 10, 0> m/s. Due to the gravitational force acting on the object, it reaches a maximum height and falls back to the ground.
What is the maximum height above the ground that the ball reaches during its upward motion?
Note: Assume no air resistance and use the acceleration due to gravity as 9.8 m/s².--