Tiêu đề Sir Waqar Formula.pdf ✅

Page 1 of 5 Chapter 2 “Scalars and Vectors “ Unit Vector â = A⃗ A â = Ax i + Ay j + Az k √Ax2+Ay 2+ Az 2 Resultant of two vectors R = │A⃗ + B⃗⃗ │ = √A2 + B2 + 2ABcosθ R = │A⃗ − B⃗⃗ │ = √A2 + B2 − 2ABcosθ Resolution of a vector into x and y components Fx = F cosθ Fy = F Sinθ F = √Fy2 + Fx2 θ = tan−1 (Fy / Fx) Scalar Product A⃗ . B⃗⃗ = ABcosθ A⃗ . B⃗⃗ = AxBx + AyBy + AzBz (A⃗ . B⃗⃗ ) B = AB = (Projection of A onto B) (A⃗ . B⃗⃗ ) A = BA = (Projection of B onto A) i.j = j.i = k.i = 0, i . i = 1, j.j = 1, k. k = 1 Vector Product │A⃗ X B⃗⃗ │ = ABSinθ i x j = k, j x k = i, k x i = j j x i = −k, k x j = − i, i x k = − j i x i = 0, j x j = 0, k x k = 0 Area of Triangle Area of Δ = 2 1 |A⃗ × B⃗⃗ | Area of Parallelogram Area of llgm = │A⃗ X B⃗⃗ │ Chapter 3 “MOTION” For constant velocity S = v x t Equations of motion 1. vf = vi + at 2. S = vit + 1 2 at 2 3. 2aS = vf 2 − vi 2 Net force F = ma Motion of bodies Connected by a string: Case#1: When both bodies move vertically a = ( m1 − m2 m1 + m2 )g T = ( 2m1m2 m1 + m1 )g Case#2: When one body is moving vertically downward and other is lying on smooth Horizontal surface: a = ( m1 m1 + m2 )g T = ( m1m2 m1 + m1 )g Law of conservation of linear momentum m1u1 + m2u2 = m1v1 + m2v2 Fnet = ∆P ∆t = m ( vf − vi t ) Inclined Plane Case-1: Body is moving upward with constant acceleration ma = −Wsin θ − f In case f= 0, a = −gsin θ Case-2: Body Moving Downward with Constant Acceleration: ma = Wsin θ − f In case f= 0, a = +gsin θ Friction f = μR Fluid Friction Fd = 6πηrv
Page 2 of 5 Chapter 4 "Motion in two Dimensions" Components of velocity vx = v0x = v0cosθ vy = v0y − gt Horizontal and Vertical Motion x = vxt y = v0yt − 1 2 gt 2 Resultant velocity v = √vx 2 + vy 2 Maximum Height and Range H = v0 2sinθ2 2g R = v0 2 sin2θ g Total time of flight T = 2v0sinθ g Angular Motion S = rθ Angular velocity ω = Δθ Δt Angular Acceleration α = Δω Δt Uniform Circular Motion ac = v 2/rFc = mv 2 r vt = rω at = rα a = √at 2 + ac 2 at = gsinθ (Tarzen Numerical) Chapter 5 Torque, Equilibrium and Angular Momentum Torque τ= (r) (F sin θ) Angular Momentum L = m (r ×v⃗ ) L = mvrSinθ If Angle θ = 90o L = mvr In Ladder Numerical F = μ R where R = weight ∑τ⃗ Clock wise = ∑τ⃗ anti-clock wise Chap. 7 “Work, Power and Energy” Work done W = Fdcosθ [Where d = v x t] Electron-volt is a unit of energy. 1 eV = 1.6 x10 -19 J Kilowatt-hour is also the unit of energy. 1 Kilowatt-hour = 3. 6 x 106 Joule K.E = 1⁄2 mv2 P.E = mgh Work Energy Equation 1⁄2 mv2 + fh = mgh Power P = t W P = F v cosθ 1Hp = 746W In case when an object is lifted upward P = W t = mgh t
Page 3 of 5 Chapter 8 “Sound and Waves” Mass Spring System Spring forceF = kx P. E = kx 2 2 K.E = 1 2 k(x0 2 − x 2) ma = kx, Total Energy E = 1 2 kx0 2 T = 2π ω , where ωisangularveloctiy T = 2π √ m k ω = √ k m v = ω√x0 2 − x 2 = √ k m √x0 2 − x 2 When an object is hanging vertically with spring then spring force = weight F = w Kx = mg T = 2π √ x g ω = √ g x Simple Pendulum T = 2π√ L g Or T = 2π ω Stationary waves in a stretched string 1 st mode of vibration λ1 = 2l, f1 = v/2l 2 nd mode of vibration λ2 = l, f2 = v/l f2 = 2 x f1 3 rd mode of vibration λ3 = 2l/3, f3 = 3v/2l f3 = 3 x f1 4 th mode of vibration l = 2λ4, f4 = 2v/ l f4 = 4 x f1 Frequency for nth mode of vibration fn = n x f1 Sonometer Fundamental frequency f1 = 1 2l √ T μ where μ = mass/length Speed of transverse waves in a string v = √ T μ T = Tension Speed of sound v = √ ΥP ρ or v = √ ΥRT M Speed of sound at 0oc, v = 332 m/s v = 332 √ T 273 Where temperature ‘T’ is in Kelvin Beat Frequency fb = f1 – f2 or fb = f2 – f1 Doppler’s Effect 1. When listener move and source at listener a) Listener moves toward source fd = f(v + v0)/v b) Listener moves away from source fd = f(v - v0)/v 2. When listener at rest and source move a) Source moves towards listener fd = vf/(v – vs) b) Source moves away from listener fd = vf/(v + vs) 3. When both are moving a) Towards fd = (v + v0)f/(v – vs) b) Away fd = (v - v0)f/(v + vs) Mach number = Speed of Object/Speed of sound Mach number = vs/v
Page 4 of 5 Chapter 6 “Gravitation” Newton’s Gravitational Law F = Gm1m2 r 2 Gravity at the surface of earth g = G ME RE 2 Me = 5. 98 × 1024Kg ρ = 5. 5 × 103Kg/m3 At the surface of any other planet g = G MX RX 2 Value of g at height ‘h’ gh = G ME (RE + h) 2 Value of g at depth‘d’ gd = (1 − d RE )g Apparent Weight When elevator is moving upward T = m(g + a) Apparent weight will increase When elevator is moving downward T = m(g − a) Apparent weight will decrease Artificial Gravity f = 1 2π √ ac R If the frequency f is increased to such an extent that artificial gravity equals to gravity at the Earth surface then ac = g f = 1 2π √ g R Chapter 9 “Nature of Light” Young’s double Slit Experiment Δy = λ L d Or Path difference = dy/L Michelson’s Interferometer λ = 2x m Newton’s Rings t = r 2/2R Constructive Interference r = √(N − 1⁄2)λ R m = N – 1 2nt cos α = ( m + 1 2 ) λ Destructive Interference r = √m λ R m = N 2nt cos α = mλ Diffraction Grating d = L / N m λ = d sinθm X-ray Diffraction m λ = 2d sinθ Chapter 10 “Geometrical Optics” Power (diopters) = 1 f(meter) This lens formula q 1 p 1 f 1   Sign Conventions 1. The object distance p is positive for real object and negative for virtual object. 2. The image distance q is positive for real object and negative for virtual object. 3. The focal length f is positive for convex lens and negative for concave lens.