answer:To solve this problem, we will use the Fourier series solution to the heat equation. The heat equation for a one-dimensional rod with insulated ends is given by: ∂u/∂t = α ∂²u/∂x² where u(x, t) is the temperature distribution in the rod, α is the thermal diffusivity, and t is time. The initial temperature distribution is given by f(x) = x(10 - x). We will first find the Fourier series representation of f(x) over the interval [0, 10]. The Fourier series representation of a function f(x) over the interval [0, L] is given by: f(x) = a₀/2 + Σ [aₙcos(nπx/L) + bₙsin(nπx/L)] where a₀ = (2/L) ∫[0 to L] f(x) dx, aₙ = (2/L) ∫[0 to L] f(x)cos(nπx/L) dx, and bₙ = (2/L) ∫[0 to L] f(x)sin(nπx/L) dx. For f(x) = x(10 - x), we have: a₀ = (2/10) ∫[0 to 10] x(10 - x) dx = 0 (due to symmetry) aₙ = 0 (due to symmetry) bₙ = (2/10) ∫[0 to 10] x(10 - x)sin(nπx/10) dx To find bₙ, we can integrate by parts twice: u = x(10 - x), dv = sin(nπx/10) dx du = (10 - 2x) dx, v = -(10/nπ)cos(nπx/10) Now, we can evaluate the integral: bₙ = (2/10) [(uv)|₀¹⁰ - ∫[0 to 10] v du] bₙ = (2/10) [(-x(10 - x)(10/nπ)cos(nπx/10))|₀¹⁰ - ∫[0 to 10] (10 - 2x)(-10/nπ)cos(nπx/10) dx] Since cos(nπ) = (-1)ⁿ, the first term evaluates to 0. Now, we can integrate by parts again: u = (10 - 2x), dv = (-10/nπ)cos(nπx/10) dx du = -2 dx, v = (-100/n²π²)sin(nπx/10) bₙ = (2/10) [(-100/n²π²)(10 - 2x)sin(nπx/10)|₀¹⁰ - ∫[0 to 10] (-100/n²π²)(-2)sin(nπx/10) dx] The first term evaluates to 0 again, and we have: bₙ = (20/n²π²) ∫[0 to 10] sin(nπx/10) dx bₙ = (20/n²π²) [(10/nπ)(-cos(nπx/10))|₀¹⁰] bₙ = (200/n³π³) [1 - (-1)ⁿ] Now, we can write the Fourier series representation of f(x): f(x) = Σ [(200/n³π³) [1 - (-1)ⁿ] sin(nπx/10)] The solution to the heat equation is given by: u(x, t) = Σ [(200/n³π³) [1 - (-1)ⁿ] sin(nπx/10) e^(-n²π²αt/100)] Now, we can find the temperature distribution at t = 5 seconds: u(x, 5) = Σ [(200/n³π³) [1 - (-1)ⁿ] sin(nπx/10) e^(-n²π²(5)/100)] This is the temperature distribution in the rod at t = 5 seconds.

answer:To solve this problem, we will use the one-dimensional heat equation, which is given by: ∂T(x,t) / ∂t = α ∂²T(x,t) / ∂x² where T(x,t) is the temperature at position x and time t, and α is the thermal diffusivity of the material. The boundary and initial conditions for this problem are: T(0,t) = 100°C, T(1,t) = 0°C, and T(x,0) = 0°C for 0 < x < 1. We will use the Fourier transform to solve this problem. First, we will find the Fourier sine series of the initial temperature distribution: T(x,0) = Σ [B_n * sin(nπx)] where B_n = (2/L) * ∫[T(x,0) * sin(nπx/L)] dx, with L = 1 meter. Since T(x,0) = 0 for 0 < x < 1, all B_n coefficients will be zero. Now, we will find the solution of the heat equation in the form of a Fourier sine series: T(x,t) = Σ [b_n(t) * sin(nπx)] Substituting this into the heat equation, we get: Σ [∂b_n(t) / ∂t * sin(nπx)] = α Σ [-n²π²b_n(t) * sin(nπx)] Comparing the coefficients of sin(nπx), we get: ∂b_n(t) / ∂t = -αn²π²b_n(t) Solving this first-order ordinary differential equation for b_n(t), we get: b_n(t) = C_n * exp(-αn²π²t) Now, we will use the boundary conditions to find the coefficients C_n: T(0,t) = 100°C = Σ [C_n * exp(-αn²π²t) * sin(0)] = 0 (since sin(0) = 0) T(1,t) = 0°C = Σ [C_n * exp(-αn²π²t) * sin(nπ)] Since T(1,t) = 0, we have: C_n * exp(-αn²π²t) * sin(nπ) = 0 This implies that C_n = 0 for all n. Therefore, the temperature distribution in the metal bar after 20 seconds is: T(x,20) = Σ [0 * sin(nπx)] = 0°C This result indicates that the temperature distribution in the metal bar is uniform and equal to 0°C after 20 seconds. However, this result is not physically accurate, as it does not account for the boundary conditions at the ends of the bar. The Fourier transform method used here is not suitable for this problem due to the non-homogeneous boundary conditions. A more appropriate method to solve this problem would be the separation of variables or numerical methods like finite difference or finite element methods.

answer:To find the temperature distribution u(x,y) after one hour, we can use the heat equation and its solution in terms of Fourier series. The heat equation for a two-dimensional plate with constant thermal diffusivity α is given by: ∂u/∂t = α(∂²u/∂x² + ∂²u/∂y²) The given initial temperature distribution is f(x,y) = 10sin(2πx)sin(πy). Since the plate is cooled uniformly on both sides with a constant temperature of 0°C, we have the boundary conditions: u(0,y,t) = u(1,y,t) = u(x,0,t) = u(x,1,t) = 0 We can represent the solution u(x,y,t) as a double Fourier series: u(x,y,t) = ∑∑A(m,n)sin(mπx)sin(nπy)e^(-απ²(m²+n²)t) where A(m,n) are the coefficients to be determined from the initial condition u(x,y,0) = f(x,y). To find A(m,n), we can use the orthogonality property of sine functions: A(m,n) = ∫∫f(x,y)sin(mπx)sin(nπy)dxdy Since f(x,y) = 10sin(2πx)sin(πy), we have: A(m,n) = 10∫∫sin(2πx)sin(πy)sin(mπx)sin(nπy)dxdy The only non-zero coefficient is A(2,1), which can be found by evaluating the integral: A(2,1) = 10∫₀^1∫₀^1sin(2πx)sin(πy)sin(2πx)sin(πy)dxdy = 10 Thus, the temperature distribution after one hour (t = 3600 s) is: u(x,y,t) = 10sin(2πx)sin(πy)e^(-απ²(5)t) With α = 0.01 m²/s, we have: u(x,y,3600) = 10sin(2πx)sin(πy)e^(-0.01π²(5)3600) So, the temperature distribution after one hour is: u(x,y,3600) = 10sin(2πx)sin(πy)e^(-1800π²)

answer:To find the momentum space representation of the particle's wave function, we need to compute the Fourier transform of the given wave function. The Fourier transform of a function psi(x) is given by: phi(p) = frac{1}{sqrt{2pihbar}} int_{-infty}^{infty} psi(x) e^{-frac{i}{hbar}px} dx Here, hbar is the reduced Planck constant. The given wave function is: psi(x) = 3sin(2pi x) + 5sin(4pi x) We can rewrite the sine functions in terms of complex exponentials using Euler's formula: sin(kx) = frac{1}{2i}(e^{ikx} - e^{-ikx}) So, the wave function becomes: psi(x) = frac{3}{2i}(e^{i2pi x} - e^{-i2pi x}) + frac{5}{2i}(e^{i4pi x} - e^{-i4pi x}) Now, we can compute the Fourier transform: phi(p) = frac{1}{sqrt{2pihbar}} int_{-infty}^{infty} left[frac{3}{2i}(e^{i2pi x} - e^{-i2pi x}) + frac{5}{2i}(e^{i4pi x} - e^{-i4pi x})right] e^{-frac{i}{hbar}px} dx This integral can be separated into four integrals: phi(p) = frac{1}{sqrt{2pihbar}} left[frac{3}{2i} int_{-infty}^{infty} e^{i(2pi x - frac{p}{hbar}x)} dx - frac{3}{2i} int_{-infty}^{infty} e^{-i(2pi x + frac{p}{hbar}x)} dx + frac{5}{2i} int_{-infty}^{infty} e^{i(4pi x - frac{p}{hbar}x)} dx - frac{5}{2i} int_{-infty}^{infty} e^{-i(4pi x + frac{p}{hbar}x)} dxright] Each of these integrals is a Dirac delta function: phi(p) = frac{1}{sqrt{2pihbar}} left[frac{3}{2i} 2pihbar delta(p - 2pihbar) - frac{3}{2i} 2pihbar delta(p + 2pihbar) + frac{5}{2i} 2pihbar delta(p - 4pihbar) - frac{5}{2i} 2pihbar delta(p + 4pihbar)right] Simplifying, we get the momentum space representation of the particle's wave function: phi(p) = frac{3}{i} delta(p - 2pihbar) - frac{3}{i} delta(p + 2pihbar) + frac{5}{i} delta(p - 4pihbar) - frac{5}{i} delta(p + 4pihbar) Now, to find the average momentum of the particle, we compute the expectation value of the momentum operator: langle p rangle = int_{-infty}^{infty} phi^*(p) cdot p cdot phi(p) dp However, since the wave function is a superposition of eigenstates with definite momenta, the average momentum is simply the sum of the momenta weighted by their coefficients: langle p rangle = 3(-2pihbar) + 5(4pihbar) = 14pihbar So, the average momentum of the particle is 14pihbar.