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)

Number of molecules = 0.25 x 6.022 x 10^23 = 1.506 x 10^23 molecules. Each CO2 molecule has 3 atoms (1 C + 2 O), so number of atoms = 3 x 1.506 x 10^23 = 4.52 x 10^23 atoms.

N = n x N_A

Divide each percentage by atomic mass: C = 24.27/12 = 2.02, H = 4.07/1 = 4.07, Cl = 71.65/35.5 = 2.02. Divide by the smallest (2.02): C = 1, H = 2, Cl = 1. Empirical formula = CH2Cl (empirical mass = 12 + 2 + 35.5 = 49.5). n = molar mass / empirical mass = 98.96/49.5 = 2. Molecular formula = C2H4Cl2.

n = molar mass / empirical formula mass

Molar mass of NaOH = 23 + 16 + 1 = 40 g/mol. Moles of NaOH = 5/40 = 0.125 mol. Volume = 450 mL = 0.450 L. Molarity = 0.125 / 0.450 = 0.278 M (approximately 0.28 M).

M = n / V(L)

Moles of H2 = 3/2 = 1.5 mol; moles of O2 = 29/32 = 0.906 mol. From the equation, 2 mol H2 need 1 mol O2. For 1.5 mol H2, O2 required = 0.75 mol, but 0.906 mol O2 is available, so O2 is in excess and H2 is the limiting reagent. Water formed = moles of H2 (since 2H2 -> 2H2O) = 1.5 mol = 1.5 x 18 = 27 g.

mass = moles x molar mass

1) The electron revolves around the nucleus only in certain fixed circular orbits of definite energy called stationary states, without radiating energy. 2) Angular momentum of the electron is quantised: mvr = nh/2pi, where n = 1, 2, 3... 3) Energy is emitted or absorbed only when an electron jumps from one orbit to another, and the energy difference equals hv (delta E = E2 - E1 = h v).

mvr = nh/2pi ; delta E = h v

For a hydrogen atom, r_n = n^2 x r_1. For n = 3: r_3 = 3^2 x 0.529 = 9 x 0.529 = 4.761 Angstrom.

r_n = n^2 x r_1

de Broglie wavelength lambda = h / (m v) = (6.626 x 10^-34) / (9.1 x 10^-31 x 2.05 x 10^7). Denominator = 1.866 x 10^-23. lambda = 3.55 x 10^-11 m = 0.355 Angstrom.

lambda = h / (m v)

1) Principal quantum number (n): gives the main energy level/shell and size of the orbital (n = 1, 2, 3...). 2) Azimuthal (angular momentum) quantum number (l): gives the subshell and shape of the orbital (l = 0 to n-1, i.e. s, p, d, f). 3) Magnetic quantum number (m_l): gives the orientation of the orbital in space (m_l = -l to +l). 4) Spin quantum number (m_s): gives the direction of electron spin (+1/2 or -1/2).

l = 0 to (n-1); m_l = -l ... +l

Cr (Z = 24): [Ar] 3d5 4s1 (not 3d4 4s2). Cu (Z = 29): [Ar] 3d10 4s1 (not 3d9 4s2). One 4s electron shifts to 3d because exactly half-filled (3d5) and completely filled (3d10) subshells have extra stability due to symmetrical distribution of electrons and greater exchange energy.

Across a period (left to right), atomic radius decreases because nuclear charge increases while electrons are added to the same shell, so the increased effective nuclear charge pulls the electron cloud closer. Down a group, atomic radius increases because a new shell is added at each step and the number of inner shielding electrons increases, outweighing the rise in nuclear charge.

Ionization enthalpy is the minimum energy required to remove the most loosely bound electron from an isolated gaseous atom in its ground state. Across a period it increases because nuclear charge increases and atomic size decreases, so the electron is held more tightly. Down a group it decreases because atomic size increases and shielding by inner electrons increases, so the outer electron is more easily removed.

M(g) -> M+(g) + e-