Roads are banked at curves so that the horizontal component of the normal reaction provides the centripetal force needed to turn, reducing reliance on friction and the risk of skidding. On a frictionless banked road, the ideal (safe) speed is v = sqrt(r g tan(theta)).

v = sqrt(r g tan(theta))

v_t = (2 r^2 (rho - sigma) g) / (9 eta), where r is the radius of the sphere, rho its density, sigma the density of the fluid, g the acceleration due to gravity and eta the coefficient of viscosity. It is obtained by balancing the weight against the buoyant force and the viscous (Stokes) drag.

v_t = 2 r^2 (rho - sigma) g / (9 eta)

A refrigerator is a heat engine working in reverse: it uses external work W to extract heat Q_c from a cold body and reject a larger heat Q_h to the hot surroundings, so Q_h = Q_c + W. Its coefficient of performance is beta = Q_c / W = Q_c / (Q_h - Q_c). A good refrigerator has a high beta.

beta = Q_c / (Q_h - Q_c)

Velocity v = dx/dt = A omega cos(omega t), with magnitude v = omega sqrt(A^2 - x^2); it is maximum (A omega) at the mean position (x = 0) and zero at the extremes. Acceleration a = dv/dt = - A omega^2 sin(omega t) = - omega^2 x; its magnitude is maximum (A omega^2) at the extreme positions (x = +/- A) and zero at the mean position.

v = omega sqrt(A^2 - x^2) ; a = - omega^2 x

Newton assumed sound propagation in a gas is isothermal, giving v = sqrt(P / rho), where P is pressure and rho density; this gave a value about 15% lower than the measured value. Laplace corrected this by treating the process as adiabatic (compressions and rarefactions are too rapid for heat exchange), giving v = sqrt(gamma P / rho), where gamma is the ratio of specific heats; this matches experiment.

v = sqrt(gamma P / rho)

When the lift accelerates upward, the apparent weight (normal reaction) is N = m (g + a) = 50 (10 + 2) = 50 * 12 = 600 N. (At rest it would be m g = 500 N, so the person feels heavier.)

N = m (g + a)

At the highest point, gravity alone must provide the centripetal force, so m g = m v_top^2 / r, giving the minimum speed at the top v_top = sqrt(g r). Using energy conservation between the lowest and highest points, the minimum speed at the bottom is v_bottom = sqrt(5 g r).

v_top = sqrt(g r) ; v_bottom = sqrt(5 g r)

In an open pipe (open at both ends) both even and odd harmonics are present; the fundamental is f1 = v/(2L) and the harmonics are f1, 2 f1, 3 f1, ... In a closed pipe (closed at one end) only odd harmonics are present; the fundamental is f1 = v/(4L) and the harmonics are f1, 3 f1, 5 f1, ... So an open pipe of the same length gives a higher fundamental and a richer set of harmonics.

Open: fn = n v/2L ; Closed: fn = (2n-1) v/4L

Distance in the nth second: s_n = u + (a/2)(2n - 1). With u = 0, a = 2, n = 5: s_5 = 0 + (2/2)(2*5 - 1) = 1 * 9 = 9 m.

s_n = u + (a/2)(2n - 1)

Given q1 = 2 x 10^-6 C, q2 = 3 x 10^-6 C, r = 30 cm = 0.3 m. Using F = k q1 q2 / r^2 = (9 x 10^9)(2 x 10^-6)(3 x 10^-6)/(0.3)^2 = (9 x 10^9 x 6 x 10^-12)/0.09 = 0.054/0.09 = 0.6 N. The force is repulsive since both charges are positive.

F = k q1 q2 / r^2

Given Q = 5 x 10^-6 C, r = 0.3 m. The field due to a point charge is E = kQ/r^2 = (9 x 10^9)(5 x 10^-6)/(0.3)^2 = (45000)/0.09 = 5 x 10^5 N/C. The field is directed radially outward since the charge is positive.

E = kQ/r^2

An electric dipole is a pair of equal and opposite charges (+q and -q) separated by a small distance 2a. The dipole moment p = q x 2a, directed from the negative to the positive charge. On the axial line at distance r from the centre, the field due to +q is k q/(r-a)^2 (away) and due to -q is k q/(r+a)^2 (towards). Net E = kq[1/(r-a)^2 - 1/(r+a)^2] = kq[(4ar)/((r^2-a^2)^2)]. For a short dipole (r much greater than a), E = k(2p)/r^3, directed along the dipole moment.

E(axial) = 2kp/r^3

Gauss's law states that the total electric flux through any closed surface is equal to 1/epsilon0 times the total charge enclosed by that surface. Mathematically, the surface integral of E over the closed surface = q(enclosed)/epsilon0. The closed surface over which the flux is calculated is called a Gaussian surface. The law is useful for finding the electric field of symmetric charge distributions.

flux = q(enclosed)/epsilon0

Consider an infinitely long straight wire with linear charge density lambda. Choose a coaxial cylindrical Gaussian surface of radius r and length L. By symmetry, the field E is radial and uniform over the curved surface. The flux through the curved surface = E x (2 pi r L); the flat ends contribute no flux. Charge enclosed = lambda L. By Gauss's law, E x 2 pi r L = lambda L / epsilon0, so E = lambda/(2 pi epsilon0 r). The field is directed radially and varies as 1/r.

E = lambda/(2 pi epsilon0 r)

Given Q = 4 x 10^-6 C, r = 9 cm = 0.09 m. The potential due to a point charge is V = kQ/r = (9 x 10^9)(4 x 10^-6)/0.09 = (36000)/0.09 = 4 x 10^5 V. Thus the potential at the point is 4 x 10^5 volt.

V = kQ/r

Let a parallel plate capacitor have plate area A, separation d, and charge Q with surface charge density sigma = Q/A. The uniform field between the plates is E = sigma/epsilon0 = Q/(A epsilon0). The potential difference between the plates is V = E d = Q d/(A epsilon0). Hence capacitance C = Q/V = Q/(Q d/(A epsilon0)) = epsilon0 A/d. So the capacitance increases with plate area and decreases with plate separation.

C = epsilon0 A/d

In series: 1/Cs = 1/2 + 1/3 + 1/6 = 3/6 + 2/6 + 1/6 = 6/6 = 1, so Cs = 1 microfarad. In parallel: Cp = 2 + 3 + 6 = 11 microfarad. Thus the series combination gives 1 microfarad and the parallel combination gives 11 microfarad.

series: 1/Cs = sum(1/Ci); parallel: Cp = sum(Ci)

Drift velocity is the small average velocity with which free electrons move in a conductor under an applied electric field, opposite to the field. Consider a conductor of area A with n free electrons per unit volume moving with drift velocity vd. In time t, electrons in a length vd t cross a section, so charge crossing = (n)(A vd t)(e). Current I = charge/time = n A vd e t / t = n A e vd. Hence I = n A e vd.

I = n A e vd

In parallel: 1/Rp = 1/2 + 1/3 + 1/6 = 3/6 + 2/6 + 1/6 = 6/6 = 1, so Rp = 1 ohm. In series: Rs = 2 + 3 + 6 = 11 ohm. Thus the parallel combination gives 1 ohm (less than the smallest resistance) and the series combination gives 11 ohm.

parallel: 1/Rp = sum(1/Ri); series: Rs = sum(Ri)

A Wheatstone bridge has four resistances P, Q, R and S arranged in a quadrilateral, with a galvanometer between the junctions of P-Q and R-S, and a cell across the other diagonal. When the bridge is balanced, no current flows through the galvanometer, so the junction connected by the galvanometer are at the same potential. Applying Kirchhoff's laws, the potential drops give P/Q = R/S. This is the balance condition; if three resistances are known, the fourth can be found.

P/Q = R/S