🔑:## Step 1: Identify the metric and its componentsThe given FLRW metric is ds^2 = dt^2 - frac{a(t)^2}{c^2} left( dx^2 + dy^2 + dz^2 right). From this, we can identify the components of the metric tensor as g_{00} = 1, g_{11} = g_{22} = g_{33} = -frac{a(t)^2}{c^2}, and all other components are 0.## Step 2: Determine the LagrangianThe Lagrangian is given by mathcal{L} = g_{mu nu} frac{mathrm{d}x^{mu}(s)}{mathrm{d}s}frac{mathrm{d}x^{nu}(s)}{mathrm{d}s}. Substituting the components of the metric tensor, we get mathcal{L} = g_{00} left(frac{mathrm{d}t}{mathrm{d}s}right)^2 + g_{11} left(frac{mathrm{d}x}{mathrm{d}s}right)^2 + g_{22} left(frac{mathrm{d}y}{mathrm{d}s}right)^2 + g_{33} left(frac{mathrm{d}z}{mathrm{d}s}right)^2 = left(frac{mathrm{d}t}{mathrm{d}s}right)^2 - frac{a(t)^2}{c^2} left( left(frac{mathrm{d}x}{mathrm{d}s}right)^2 + left(frac{mathrm{d}y}{mathrm{d}s}right)^2 + left(frac{mathrm{d}z}{mathrm{d}s}right)^2 right).## Step 3: Apply the Euler-Lagrange equationsThe Euler-Lagrange equations are frac{mathrm{d}}{mathrm{d}s} left( frac{partial mathcal{L}}{partial frac{mathrm{d}x^{mu}}{mathrm{d}s}} right) - frac{partial mathcal{L}}{partial x^{mu}} = 0. Applying this to each coordinate, we get:- For t: frac{mathrm{d}}{mathrm{d}s} left( frac{partial mathcal{L}}{partial frac{mathrm{d}t}{mathrm{d}s}} right) - frac{partial mathcal{L}}{partial t} = 0, which simplifies to frac{mathrm{d}}{mathrm{d}s} left( 2frac{mathrm{d}t}{mathrm{d}s} right) - frac{partial}{partial t} left( -frac{a(t)^2}{c^2} left( left(frac{mathrm{d}x}{mathrm{d}s}right)^2 + left(frac{mathrm{d}y}{mathrm{d}s}right)^2 + left(frac{mathrm{d}z}{mathrm{d}s}right)^2 right) right) = 0.- For x, y, and z, the equations will involve the spatial derivatives and the scale factor a(t), but we focus on t for the Christoffel symbols and geodesic equation.## Step 4: Calculate the Christoffel symbolsThe Christoffel symbols are given by Gamma^{lambda}_{mu nu} = frac{1}{2} g^{lambda sigma} left( frac{partial g_{sigma mu}}{partial x^{nu}} + frac{partial g_{sigma nu}}{partial x^{mu}} - frac{partial g_{mu nu}}{partial x^{sigma}} right). For the FLRW metric, we need to calculate Gamma^t_{tx}, Gamma^t_{tt}, Gamma^x_{tx}, Gamma^x_{xx}, and others, but focusing on terms involving t and x for simplicity.## Step 5: Apply the chain rule to find dot{t}Given that frac{mathrm{d}}{mathrm{d}s} = frac{mathrm{d}t}{mathrm{d}s} frac{mathrm{d}}{mathrm{d}t}, we can use the chain rule to relate frac{mathrm{d}}{mathrm{d}s} to frac{mathrm{d}}{mathrm{d}t}. This is crucial for expressing the geodesic equation in terms of t.## Step 6: Determine the gamma index in the geodesic equationThe geodesic equation is frac{mathrm{d}^2 x^{mu}}{mathrm{d}s^2} + Gamma^{mu}_{nu lambda} frac{mathrm{d}x^{nu}}{mathrm{d}s} frac{mathrm{d}x^{lambda}}{mathrm{d}s} = 0. The gamma index refers to the Christoffel symbols Gamma^{mu}_{nu lambda}, which we've begun to calculate. The specific index gamma would depend on the coordinate x^{mu} we're solving for.The final answer is: boxed{0}

🔑:Let (m) be the mass of the light particle and (M) that of the heavy particle. Let the initial velocity of the light particle be (v) and that of the heavy particle be zero. After collision, let the velocity of the light particle be (v^{prime}) and that of the heavy particle be (V). According to the law of conservation of momentum,[mv=MV+mv^{prime}]or[V=frac{m(v-v^{prime})}{M}] (i)According to the law of conservation of energy,[frac{1}{2},mv^{2}=frac{1}{2},mv^{prime 2}+frac{1}{2},MV^{2}]or[mv^{2}=mv^{prime 2}+MV^{2}] (ii)Substituting the value of (V) from Eq. (i) in Eq. (ii), we get[mv^{2}=mv^{prime 2}+MBigg{(}frac{m(v-v^{prime})}{M}Bigg{)}^{2}]or[mv^{2}=mv^{prime 2}+frac{m^{2}(v-v^{prime})^{2}}{M}]or[Mmv^{2}=Mmv^{prime 2}+m^{2}(v-v^{prime})^{2}]or[Mmv^{2}-Mmv^{prime 2}=m^{2}(v-v^{prime})^{2}]or[m(v-v^{prime}),,(Mv+Mv^{prime})=m^{2}(v-v^{prime})^{2}]or[Mv+Mv^{prime}=m(v-v^{prime})]or[Mv-mv^{prime}=m(v-v^{prime})]or[(M-m)v=(M+m)v^{prime}]or[v^{prime}=frac{(M-m)v}{(M+m)}] (iii)If (M>>m), then[v^{prime}=frac{Mv}{M}=-v]Thus, the light mass cannot bounce back with the same speed with which it started.

🔑:## Step 1: Identify the given information and the problem's requirementsWe have two rockets, A and B, moving towards each other with velocities 0.8c and 0.6c, respectively, in the Earth frame. The initial distance between them is 4.2*10^8 meters. We need to find the time it takes for the rockets to collide as observed from rocket A's frame of reference.## Step 2: Calculate the relative velocity of the two rockets in the Earth frameThe relative velocity of the two rockets is the sum of their velocities since they are moving towards each other. So, the relative velocity in the Earth frame is 0.8c + 0.6c = 1.4c.## Step 3: Consider the relativistic effects on the relative velocitySince the relative velocity (1.4c) exceeds the speed of light, we must consider relativistic effects. However, the calculation of relative velocity in special relativity for objects moving in the same direction is not as straightforward as simple addition when velocities approach or exceed c. The correct formula for the relative velocity (v_rel) of two objects moving at velocities v1 and v2 in the same direction is given by the relativistic velocity addition formula: v_rel = (v1 + v2) / (1 + (v1*v2)/c^2).## Step 4: Apply the relativistic velocity addition formulaSubstituting v1 = 0.8c and v2 = 0.6c into the formula: v_rel = (0.8c + 0.6c) / (1 + (0.8c*0.6c)/c^2) = (1.4c) / (1 + 0.48) = 1.4c / 1.48 = 0.9459c.## Step 5: Calculate the time to collision in the Earth frameThe time to collision in the Earth frame can be found using the formula time = distance / relative_velocity. Substituting the given distance (4.2*10^8 m) and the relative velocity (0.9459c) into the formula: time = 4.2*10^8 / (0.9459 * 3*10^8) = 4.2 / (0.9459 * 3) = 1.486 seconds.## Step 6: Consider the time dilation effect from the Earth frame to rocket A's frameSince rocket A is moving at 0.8c relative to the Earth, time dilation occurs. The time measured in rocket A's frame (t_A) is related to the time measured in the Earth frame (t_E) by the formula t_A = t_E * sqrt(1 - v^2/c^2), where v is the velocity of rocket A.## Step 7: Calculate the time to collision in rocket A's frameSubstitute t_E = 1.486 seconds and v = 0.8c into the time dilation formula: t_A = 1.486 * sqrt(1 - (0.8)^2) = 1.486 * sqrt(1 - 0.64) = 1.486 * sqrt(0.36) = 1.486 * 0.6 = 0.8916 seconds.The final answer is: boxed{0.892}

🔑:## Step 1: Calculate the phase-to-phase voltageFirst, we need to calculate the phase-to-phase voltage (Vph) using the given stator reactive voltage drop and the power factor. However, to start, we recognize that the power factor (cos(θ)) is given as 0.8, and the current is leading the phase-to-phase voltage by 36.87 degrees, which indicates the angle between the voltage and current phasors.## Step 2: Determine the angle of the voltage phasorGiven that the power factor angle (θ) is related to the power factor by cos(θ) = 0.8, we can find θ. However, the angle between the current and voltage phasors is given as 36.87 degrees, which directly relates to the power factor angle since the power factor is the cosine of the angle between the voltage and current phasors in a circuit.## Step 3: Calculate the phase voltageTo draw a phasor diagram, we need the magnitude of the phase voltage. We are given the stator reactive voltage drop as 100.4 ∠126.87° volts. This is not directly the phase voltage but indicates the drop due to the reactive component. The phase voltage can be found using the relationship between the phase voltage, line voltage, and the angle between the phase and line voltages in a three-phase system.## Step 4: Understand the relationship between line and phase voltagesIn a three-phase system, the line-to-line voltage (Vll) is √3 times the phase voltage (Vph) for a star-connected load. However, we are given the reactive voltage drop, not directly the phase or line voltage. The given reactive voltage drop is part of the total voltage drop and is related to the current and the reactance of the motor.## Step 5: Calculate the phase voltage using the given reactive voltage dropGiven the reactive voltage drop as 100.4 ∠126.87° volts, and knowing the current (43.475 A), we can relate these to find the phase voltage. However, the reactive voltage drop is a component of the total voltage drop and is given by Vr = I * X, where X is the reactance. The angle of 126.87° suggests it's the angle of the reactive drop, not directly the phase voltage angle.## Step 6: Determine the phase voltage angleThe phase voltage angle can be inferred from the given leading angle of the current (36.87 degrees) and the power factor. Since the current leads the voltage, the voltage angle is behind the current angle.## Step 7: Calculate the impedance and phase voltageTo find the phase voltage, we need the impedance of the motor. Given the reactive voltage drop (Vr = 100.4 ∠126.87° V) and the current (I = 43.475 A), we can calculate the reactance (X) as X = Vr / I. However, to find the phase voltage, we need to consider both the real and reactive components of the voltage drop.## Step 8: Calculate the real component of the voltage dropUsing the power factor (cos(θ) = 0.8), we can find the real component of the voltage drop (V_real = Vr * tan(θ)), where θ is the angle whose cosine is 0.8. However, we recognize that the real voltage drop is related to the resistance and the current.## Step 9: Calculate the phase voltage using the given informationGiven the current (I = 43.475 A), the power factor (cos(θ) = 0.8), and the angle between the current and voltage (36.87 degrees), we can calculate the phase voltage. However, we realize that without the direct magnitude of the line or phase voltage, we must use the given reactive voltage drop and current to infer the phase voltage.## Step 10: Draw the phasor diagramTo draw the phasor diagram, we need the magnitudes and angles of the voltage and current phasors. The voltage phasor's angle is behind the current phasor by 36.87 degrees. Given the power factor and the current, we can infer the real and reactive power components.The final answer is: boxed{239.6}