Munich Use of Geothermal Energy in the Gulf States Essay
Munich Use of Geothermal Energy in the Gulf States Essay
D
4 attachmentsSlide 1 of 4
UNFORMATTED ATTACHMENT PREVIEW
PHYSICS FORMULAS 2425 Types of Errors: Personal errors due to bias or mistakes. Systematic errors due to miscalibration of instruments, personal bias, or reaction time. Random errors are unknown or unpredictable, such as voltage or temperature fluctuations, vibration, etc. Accuracy – how close measurement comes to accepted value Precision – how consistent or repeatable measurements are operation: A = L × W 2 g = 9.8 m/s = 32 f/s gmoon = 1.62 m/s2 (for constant a) v = (for constant a) average velocity x = v 0t + 1 at 2 + 2ax 2 2 2 Mi Mf − b ± b 2 − 4ac 2a θ R = tan −1 y R R= A + B − 2 AB cos γ sin θ = 2 γ 2 R θ A u = speed of the exhaust relative the to rocket [m/s] r r r r Rr = Ar+ B +r C r Resultant = Sum of the vectors Rx = Ax + Bx + Cx x-component A x = A cos θ r r r r Ry = Ay + By + C y y-component A y = A sin θ R = Rx 2 + Ry 2 Law of Cosines: (for constant a) Addition of Multiple Vectors: Quadratic Equation: x= 2 v f − v i = u ln ∆M M × ∆V ∆D = ± + V V2 2 v 0 +v burning of fuel. M operation: D = V error: v = v 0 + at =v0 (for constant a) Rocket Science: The relationship between velocity and the error: ∆A = ± ( L × ∆W + W × ∆L ) Division: s] v Calculation of Errors: Multiplication: Formulas for Velocity: [units: v: m/s; a: m/s2; x: m; t: B Ry Rx Magnitude (length) of R or tanθ R = Ry Rx Angle of the resultant Unit Vectors: x B sin γ Cross Product or Vector Product: i× j=k j × i = −k i ×i = 0 Newton’s Laws: First Law: Law of Inertia. An object at rest will remain at rest unless acted on by an external force. An object in motion will remain in motion unless . . . Second Law: ΣF = ma , Στ = Iα The sum of external forces on a object is equal to its mass (or inertia for rotational forces) times the acceleration. Third Law: Every action has an equal and opposite reaction. Law of Gravity: F = force of attraction exerted on each body G = gravitational constant 6.67 × 10-11 m1m2 F=G 2 [N · m 2/kg] or [m3/kg · s2] r r = distance between centers [m] Positive direction: i j k Dot Product or Scalar Product: i⋅ j = 0 i⋅i =1 j i Velocity is the derivative of position with respect k to time: v= d dx dy dz ( xi + yj + zk) = i + j+ k dt dt dt dt Acceleration is the derivative of velocity with respect to time: a= dv y dv dv d ( v x i + v y j + v zk ) = x i + j+ z k dt dt dt dt Mass/Density: Drag: [kilograms] M =V × D mass = volume × density vt = Projectile Motion: v x0 = v0 cos θ 0 v y0 = v0 sin θ 0 horizontal component of velocity vertical component of velocity x = v x 0t v y = v y 0 − gt horizontal distance to find apex, let vy = 0 y = y 0 + v y 0 t − 21 gt 2 y = (tan θ0 ) x − vertical distance gx 2 2(v 0 cosθ0 ) 2 v y = v y 0 − 2 gy 2 2 Relative Motion: vertical distance v PA = v PB + v BA v PA 2mg CρA Force: v PB + v BA = 1 + v PB v BA / c 2 D = Drag Force [N] C = Coeficient of drag [ ] ρ = density [kg/m3] (air: 1.2, water: 1000) A = effective cross-sectional area [m2] v = speed [m/s] vt = Terminal Velocity [m/s] g = acceleration due to gravity [9.8 m/s2] [F is in Newtons; m is kilograms] Newtons = kg × m / s = 2 dynes = grams × cm / s 2 grams × g = dynes × 10, 000 1000 1 lb = 4. 448 N F = ma force = mass X acceleration F= force = ∆ρ ∆t J F= ∆t vertical velocity The relative velocity of object P with respect to A is equal to the velocity of P with respect to B plus the velocity of B with respect to A. For velocities approaching the speed of light, the formula changes to: D = 21 CρAv 2 force = change in momentum time interval impulse time interval conservative force – work done is independent of the path taken non-conservative force – depends on the path taken F =G m 1m 2 r2 force of gravitational attraction, where G is the constant of universal gravitation 6. 673 × 10−11 N ⋅m 2 kg 2 c = the speed of light = 299,792,458 m/s Atwood’s Machine: Inclined Plane: F = mg sinθ W = mg Fn = mg cosθ (the normal force) [F and W are in Newtons; m is kilograms] Fk a= m2 − m1 a= g m1 + m2 m1 F θ W Fn Fk = µk Fn tanθ Acceleration in m/s2: (force of friction, opposite the direction of movement) = µk The coefficient of friction µk is found when the angle θ is adjusted for zero acceleration of the sliding object. g sinθ (acceleration) Tension in Newtons: 2m1 m2 T = g m1 + m2 m2 Tension: [Newtons] T m θ T = m (g + a ) T = m (g − a ) W Fn (where m is accelerating upward) (where m is accelerating downward) PE s = Work: [joules or Newton-meters] T= W = F•d 2 F cos θ 1 f T = 2π W = ( F cos θ )s kx (elastic potential energy) Simple Harmonic Motion: F θ 1 2 s (work done on the object by F) a=− work = force × displacement = PE i − PE f (work done by gravity, y is (T is period in seconds; f is frequency in Hz) m k k x m T = period (s) m = mass (kg) k = spring constant (N/m) (acceleration) x is the location in meters W = KE f − KEi k 2 ( A − x2) m x = A cos( 2π f t ) The work done by a conservative force on a particle is independent of the path taken. Pendulum: see also: Energy, Spring T = 2π W g = mgy i − mgy f vertical distance in meters) Power: [watts] P= Power is the rate of work. P = Energy: KE = 1 2 dW dt W ∆t = Fv watts = mv v = fλ second v= A falling object loses potential energy as it gains kinetic energy. In an isolated system, energy can be transferred from one type to another but total energy remains the same. W net = ∆KE = − ∆PE E total = KE + PE (mechanical energy) PE i + KE i = PE f + KE f [i = initial; f = final, energy is conserved] + 12 mv i = mgy f + 12 mv f 2 2 [F is in Newtons; W is in Joules; x is in meters; k is in Newtons per meter: N/m] F = spring with a spring constant k a distance x) (average force required to compress a spring–or kx average force output from a decompression over a distance x) kx 2 W = − kx 1 2 center of mass is unaffected. In an elastic collision, kinetic energy KE = 12 mv 2 is conserved. m1v1i + m2 v 2 i = m1v1 f + m2 v 2 f m1 − m2 2m2 v1i + v m1 + m2 m1 + m2 2i m − m1 2m1 v1i + 2 v = m1 + m2 m1 + m2 2i spring (work done on a spring by an applied force) 2 (work done by a spring) (momentum) v1 f = (elastic only) v2 f (elastic only) p = mv ∑F = dp dt Linear Momentum in a system of particles: M = total mass of the system [m] P = Mv cm vcm = velocity of the center of mass [m/s] Hooke’s Law (force required to compress a 1 2 Collisions: In all collisions, momentum is conserved and the 8 F = kx W = F µ Momentum: [kg · m/s] speed light 3.00 × 10 m/s See also: Rotation and Torque 1 2 (f is frequency in Hz; λ is wavelength in m) [y = vertical distance] E is the mass energy, m is mass, c is the 2 Spring: Third Order Approximation (kinetic energy) ∆KE = KE f − KE i = 21 mv 2 − 21 mv 0 2 = Work E = mc First Order Approximation for small angles L is length in m; g is gravity (F is tension in N; µ is mass per unit length of string in kg/m) see also: Oscillation (gravitational potential energy, y is vertical distance in meters) i (x is position in m; f is frequency Hz) Waves: joules PE = mgy mgy L g (A is amplitude in m; x is position) L 1 θ 9 θ sin 4 1 + sin 2 + g 4 2 64 2 T = 2π [joules] 2 v=± in Impulse: [kg · m/s] J = ∆p = F∆ t = mv f − mv i impulse = force × duration or the change in momentum see also Force Center of Mass: The center of mass of a body or a system of bodies is the point that moves as though all of the mass were concentrated there and all external forces were applied there. xcm = distance from origin [m] 1 n x cm = mi x i M = total mass [m] M i =1 m = mass of object [m] This can be applied to x = distance of object from origin y and z axis as well. [m] ∑ Rotation and Torque: [θ is in radians] ∆θ ω= ∆t ω = ω 0 + αt θ = ω 0 t + 12 αt 2 τ = torque (vector) (positive is in the counterclockwise direction) [N · m] τ = rFt = r⊥ F = rF sin φ τ = magnitude of the torque r = radius [m] F = force [N] r⊥ = perpendicular distance between axis and an extended line running through F. φ = the angle between r and F [° or rad] ∑ τ = Iα ∑τ = the net torque acting on a body [N · m] I = Inertia [kg · m2] α = angular acceleration [rad/s2] (if constant acceleration) [rad/s] (if constant acceleration) [radians] (if constant acceleration) I = [kg · m2] Inertia: orbiting object: I = mr sphere: Is = 2 5 2 mr ∑ mr (inertia) Ir = ring: 2 2 1 2 m (r 1 + r 2 ) 2 Id = disk or cyl.: 2 1 2 mr 2 average angular acceleration [rad/s2] thin rod (on side): I = tangential speed [m/s] cylinder on its side (axis ctr): I = m velocity of the center of mass [m/s] Parallel Axis Theorem: If you know the rotational inertia a t = rα at = tangential acceleration [m/s2] r = radius [m] α = angular acceleration [rad/s2] 2 vt = rω 2 r ar = radial acceleration or centripetal acceleration [m/s2] ar 1 12 ml 2 rod (axis end): I = 13 ml ( r2 4 l + 12 2 2 ) of a body about any axis that passes through its center of mass, you can find its rotational intertia about any other axis parallel to that axis with the parallel-axis theorem: I = I cm + Mh 2 (directed inward to center) v = speed [m/s] r = radius [m] ω = angular speed [rad/s] at I = Inertia [kg · m2] Icm = Inertia with axis at the center of mass [kg · m2] M = mass [kg] h = distance from the center of mass to the axis [m] Kinetic Energy: [Joules] KE r = 12 Iω 2 rotational kinetic energy KEt = 12 mv 2 translational kinetic energy KE = 12 I cmω 2 + 12 mv cm 2 rolling kinetic energy total acceleration [m/s2] a = at 2 + a r 2 Fc = ma r = m τ =r×F average angular speed [rad/s] ω 2 = ω 0 2 + 2αθ ∆ω α = ∆t v t = rω v cm = rω ar = Torque: vt 2 r Fc = centripetal force [N] ar = radial acceleration or centripetal acceleration [m/s2] (directed inward to center) 2π r 2π T= = v ω 4π 2 3 2 r = K sr 3 T = GM s T ∑ F = T − Mg = Ma ∑ τ = TR = Iα T = period [s] Kepler’s Third Law (planetary motion) Yo-yo: 0 where T = the period 2 G = 6. 673 × 10−11 N ⋅m2 kg K s = 2. 97 × 10 −19 s 2 m3 a = −α R0 a= T = tension [N] M = mass [kg] R0 = radius of axle [m] −g 1 + I / MR0 2 R0 Mg Angular Momentum: Oscillation: l = Iω rigid body on fixed axis [kg · m2/s or J · s] l = r × p = m(r × v ) l = angular momentum of a particle [J · s] r = a position vector p = linear momentum [kg · m2/s or J · s] m = mass [kg] v = linear velocity [m/s] I i ωi = I f ωf tan θ = v 2 F = force [N] A = crossectional Fl0 A∆L area [m2] V ∆P V= original vol. [m3] ∆V = chg. in vol. ∆V [m3] l0 = initial length [m] ∆L = chg. in length [m] ∆P = change in pressure [Pa or N/m2] f F = ma = −( mω 2 ) x U ( t ) = 12 kx m 2 cos 2 (ω t + φ ) Potential Energy K ( t ) = 12 kx m 2 sin 2 (ω t + φ ) Kinetic Energy y = mx + b Ax + By + C = 0 d (sin u ) = u’ cos u dx (m = − A / B ) y − y1 = m ( x − x1 ) A Ax + By = Ax 1 R = rate of flow [m3/s] A = crossectional area [m2] v = velocity [m/s] R = A1v1 = A2 v 2 + By 1 point-slope, alt. a + y b =1 (m = − b / a ) 2-point intercept a = x-intercept b = y-intercept Tom Penick tomzap@eden.com www.teicontrols/notes December 6, 1997 Bernoulli’s Equation: P1 + 12 ρ v12 + ρ gy1 = P2 + 12 ρ v 2 2 + ρ gy 2 P1 = pressure [Pa or N/m2] v = velocity [m/s] y = height [m] x slope-intercept first degree point-slope y 2 − y1 y − y1 = ( x − x1 ) x 2 − x1 Rate of Flow: For a horizontal pipe: Total Mechanical Energy Equations of a Line: F a [rad] k = spring constant [N/m] T = period [s] f = frequency [Hz] F = force [N] m = mass [kg] I = moment of inertia [kg · m2] h = distance between axis and center of mass [m] d (cos u) = − u’sin u dx P0 = atmospheric pressure if applicable [Pa or N/m2] ρ = density [kg/m3] g = gravity [m/s2] h = height [m] f = force [N] a = area [m2] F = force [N] A = area [m2] 1 f x’ = velocity of the oscillating object [m/s] x” = acceleration of the oscillating object m/s2] 1 atm = 1.01 × 105 Pa = 760 torr = 14.7 in2 f F = a A k (spring) m ω= E = U + K = 12 kx m 2 Pressure in a liquid: (due to gravity) [Pa or N/m2] P = P0 + ρ gh x = position [m] xm = amplitude [m] ω = angular frequency [rad/s] t = time [s] φ = phase angle [rad] (ω t + φ) = phase of the motion m (spring) k I T = 2π mgh Volume Elasticity: Bulk Modulus [Pa or N/m2] B =− 2π = 2π f T T = 2π rg Elasticity in Length: Young’s Modulus [Pa or N/m2] Y= ω= T= Angular momentum is conserved when torque is zero. An Optimally Banked Curve: x = x m cos(ω t + φ ) The Position Function for oscillating motion: ρ = density [kg/m3] g = gravity [m/s2] P1 + 12 ρ v12 = P2 + 12 ρ v 2 2 •1 SSM Earth is approximately a sphere of radius 6.37 × 106 m. What are (a) its circumference in kilometers, (b) its surface area in square kilometers, and (c) its volume in cubic kilometers? 2-The micrometer (1 μm) is often called the micron. (a) How many microns make up 1.0 km? (b) What fraction of a centimeter equals 1.0 μm? (c) How many microns are in 1.0 yd? 3. The fastest growing plant on record is a Hesperoyucca whipplei that grew 3.7 m in 14 days. What was its growth rate in micrometers per second? 4. Earth has a mass of 5.98 × 1024 kg. The average mass of the atoms that make up Earth is 40 u. How many atoms are there in Earth? 5. Iron has a density of 7.87 g/cm3, and the mass of an iron atom is 9.27 × 10−26 kg. If the atoms are spherical and tightly packed, (a) what is the volume of an iron atom and (b) what is the distance between the centers of adjacent atoms? Part 2: 1. While driving a car at 90 km/h, how far do you move while your eyes shut for 0.50 s during a hard sneeze? 2. An automobile travels on a straight road for 40 km at 30 km/h. It then continues in the same direction for another 40 km at 60 km/h. (a) What is the average velocity of the car during the full 80 km trip? (Assume that it moves in the positive x direction.) (b) What is the average speed? (c) Graph x versus t and indicate how the average velocity is found on the graph. 3. The position of an object moving along an x axis is given by x = 3t − 4t2 + t3, where x is in meters and t in seconds. Find the position of the object at the following values of t: (a) 1 s, (b) 2 s, (c) 3 s, and (d) 4 s. (e) What is the object’s displacement between t = 0 and t = 4 s? (f) What is its average velocity for the time interval from t = 2 s to t = 4 s? (g) Graph x versus t for 0 ≤ t ≤ 4 s and indicate how the answer for (f) can be found on the graph. 4. You are to drive 300 km to an interview. The interview is at 11:15 a.m. You plan to drive at 100 km/h, so you leave at 8:00 a.m. to allow some extra time. You drive at that speed for the first 100 km, but then construction work forces you to slow to 40 km/h for 40 km. What would be the least speed needed for the rest of the trip to arrive in time for the interview? 5. At a certain time a particle had a speed of 18 m/s in the positive x direction, and 2.4 s later its speed was 30 m/s in the opposite direction. What is the average acceleration of the particle during this 2.4 s interval? 6. SSM An electron with an initial velocity v0 = 1.50 × 105 m/s enters a region of length L = 1.00 cm where it is electrically accelerated (Fig. 2-26). It emerges with v = 5.70 × 106 m/s. What is its acceleration, assumed constant? 7. An electric vehicle starts from rest and accelerates at a rate of 2.0 m/s2 in a straight line until it reaches a speed of 20 m/s. The vehicle then slows at a constant rate of 1.0 m/s2 until it stops. (a) How much time elapses from start to stop? (b) How far does the vehicle travel from start to stop? 8. A car traveling 56.0 km/h is 24.0 m from a barrier when the driver slams on the brakes. The car hits the barrier 2.00 s later. (a) What is the magnitude of the car’s constant acceleration before impact? (b) How fast is the car traveling at impact? 9. Figure 2-29 depicts the motion of a particle moving along an x axis with a constant acceleration. The figure’s vertical scaling is set by xs = 6.0 m. What are the (a) magnitude and (b) direction of the particle’s acceleration? 10. With what speed must a ball be thrown vertically from ground level to rise to a maximum height of 50 m? (b) How long will it be in the air? (c) Sketch graphs of y, v, and a versus t for the ball. On the first two graphs, indicate the time at which 50 m is reached. 11. At a construction site a pipe wrench struck the ground with a speed of 24 m/s. (a) From what height was it inadvertently dropped? (b) How long was it falling? (c) Sketch graphs of y, v, and a versus t for the wrench. 12. As a runaway scientific balloon ascends at 19.6 m/s, one of its instrument packages breaks free of a harness and free-falls. Figure 2-34 gives the vertical velocity of the package versus time, from before it breaks free to when it reaches the ground. (a) What maximum height above the break-free point does it rise? (b) How high is the break-free point above the ground? x (m) x x 2 1 (s) Figure 2-29 Problem 37. 아 -1 (s) 2 6 8 Figure 2-34 Problem 51.
o you similar assignment and would want someone to complete it for you? Click on the ORDER NOW option to get instant services at LindasHelp.com
Munich Use of Geothermal Energy in the Gulf States Essay
D
4 attachmentsSlide 1 of 4
UNFORMATTED ATTACHMENT PREVIEW
PHYSICS FORMULAS 2425 Types of Errors: Personal errors due to bias or mistakes. Systematic errors due to miscalibration of instruments, personal bias, or reaction time. Random errors are unknown or unpredictable, such as voltage or temperature fluctuations, vibration, etc. Accuracy – how close measurement comes to accepted value Precision – how consistent or repeatable measurements are operation: A = L × W 2 g = 9.8 m/s = 32 f/s gmoon = 1.62 m/s2 (for constant a) v = (for constant a) average velocity x = v 0t + 1 at 2 + 2ax 2 2 2 Mi Mf − b ± b 2 − 4ac 2a θ R = tan −1 y R R= A + B − 2 AB cos γ sin θ = 2 γ 2 R θ A u = speed of the exhaust relative the to rocket [m/s] r r r r Rr = Ar+ B +r C r Resultant = Sum of the vectors Rx = Ax + Bx + Cx x-component A x = A cos θ r r r r Ry = Ay + By + C y y-component A y = A sin θ R = Rx 2 + Ry 2 Law of Cosines: (for constant a) Addition of Multiple Vectors: Quadratic Equation: x= 2 v f − v i = u ln ∆M M × ∆V ∆D = ± + V V2 2 v 0 +v burning of fuel. M operation: D = V error: v = v 0 + at =v0 (for constant a) Rocket Science: The relationship between velocity and the error: ∆A = ± ( L × ∆W + W × ∆L ) Division: s] v Calculation of Errors: Multiplication: Formulas for Velocity: [units: v: m/s; a: m/s2; x: m; t: B Ry Rx Magnitude (length) of R or tanθ R = Ry Rx Angle of the resultant Unit Vectors: x B sin γ Cross Product or Vector Product: i× j=k j × i = −k i ×i = 0 Newton’s Laws: First Law: Law of Inertia. An object at rest will remain at rest unless acted on by an external force. An object in motion will remain in motion unless . . . Second Law: ΣF = ma , Στ = Iα The sum of external forces on a object is equal to its mass (or inertia for rotational forces) times the acceleration. Third Law: Every action has an equal and opposite reaction. Law of Gravity: F = force of attraction exerted on each body G = gravitational constant 6.67 × 10-11 m1m2 F=G 2 [N · m 2/kg] or [m3/kg · s2] r r = distance between centers [m] Positive direction: i j k Dot Product or Scalar Product: i⋅ j = 0 i⋅i =1 j i Velocity is the derivative of position with respect k to time: v= d dx dy dz ( xi + yj + zk) = i + j+ k dt dt dt dt Acceleration is the derivative of velocity with respect to time: a= dv y dv dv d ( v x i + v y j + v zk ) = x i + j+ z k dt dt dt dt Mass/Density: Drag: [kilograms] M =V × D mass = volume × density vt = Projectile Motion: v x0 = v0 cos θ 0 v y0 = v0 sin θ 0 horizontal component of velocity vertical component of velocity x = v x 0t v y = v y 0 − gt horizontal distance to find apex, let vy = 0 y = y 0 + v y 0 t − 21 gt 2 y = (tan θ0 ) x − vertical distance gx 2 2(v 0 cosθ0 ) 2 v y = v y 0 − 2 gy 2 2 Relative Motion: vertical distance v PA = v PB + v BA v PA 2mg CρA Force: v PB + v BA = 1 + v PB v BA / c 2 D = Drag Force [N] C = Coeficient of drag [ ] ρ = density [kg/m3] (air: 1.2, water: 1000) A = effective cross-sectional area [m2] v = speed [m/s] vt = Terminal Velocity [m/s] g = acceleration due to gravity [9.8 m/s2] [F is in Newtons; m is kilograms] Newtons = kg × m / s = 2 dynes = grams × cm / s 2 grams × g = dynes × 10, 000 1000 1 lb = 4. 448 N F = ma force = mass X acceleration F= force = ∆ρ ∆t J F= ∆t vertical velocity The relative velocity of object P with respect to A is equal to the velocity of P with respect to B plus the velocity of B with respect to A. For velocities approaching the speed of light, the formula changes to: D = 21 CρAv 2 force = change in momentum time interval impulse time interval conservative force – work done is independent of the path taken non-conservative force – depends on the path taken F =G m 1m 2 r2 force of gravitational attraction, where G is the constant of universal gravitation 6. 673 × 10−11 N ⋅m 2 kg 2 c = the speed of light = 299,792,458 m/s Atwood’s Machine: Inclined Plane: F = mg sinθ W = mg Fn = mg cosθ (the normal force) [F and W are in Newtons; m is kilograms] Fk a= m2 − m1 a= g m1 + m2 m1 F θ W Fn Fk = µk Fn tanθ Acceleration in m/s2: (force of friction, opposite the direction of movement) = µk The coefficient of friction µk is found when the angle θ is adjusted for zero acceleration of the sliding object. g sinθ (acceleration) Tension in Newtons: 2m1 m2 T = g m1 + m2 m2 Tension: [Newtons] T m θ T = m (g + a ) T = m (g − a ) W Fn (where m is accelerating upward) (where m is accelerating downward) PE s = Work: [joules or Newton-meters] T= W = F•d 2 F cos θ 1 f T = 2π W = ( F cos θ )s kx (elastic potential energy) Simple Harmonic Motion: F θ 1 2 s (work done on the object by F) a=− work = force × displacement = PE i − PE f (work done by gravity, y is (T is period in seconds; f is frequency in Hz) m k k x m T = period (s) m = mass (kg) k = spring constant (N/m) (acceleration) x is the location in meters W = KE f − KEi k 2 ( A − x2) m x = A cos( 2π f t ) The work done by a conservative force on a particle is independent of the path taken. Pendulum: see also: Energy, Spring T = 2π W g = mgy i − mgy f vertical distance in meters) Power: [watts] P= Power is the rate of work. P = Energy: KE = 1 2 dW dt W ∆t = Fv watts = mv v = fλ second v= A falling object loses potential energy as it gains kinetic energy. In an isolated system, energy can be transferred from one type to another but total energy remains the same. W net = ∆KE = − ∆PE E total = KE + PE (mechanical energy) PE i + KE i = PE f + KE f [i = initial; f = final, energy is conserved] + 12 mv i = mgy f + 12 mv f 2 2 [F is in Newtons; W is in Joules; x is in meters; k is in Newtons per meter: N/m] F = spring with a spring constant k a distance x) (average force required to compress a spring–or kx average force output from a decompression over a distance x) kx 2 W = − kx 1 2 center of mass is unaffected. In an elastic collision, kinetic energy KE = 12 mv 2 is conserved. m1v1i + m2 v 2 i = m1v1 f + m2 v 2 f m1 − m2 2m2 v1i + v m1 + m2 m1 + m2 2i m − m1 2m1 v1i + 2 v = m1 + m2 m1 + m2 2i spring (work done on a spring by an applied force) 2 (work done by a spring) (momentum) v1 f = (elastic only) v2 f (elastic only) p = mv ∑F = dp dt Linear Momentum in a system of particles: M = total mass of the system [m] P = Mv cm vcm = velocity of the center of mass [m/s] Hooke’s Law (force required to compress a 1 2 Collisions: In all collisions, momentum is conserved and the 8 F = kx W = F µ Momentum: [kg · m/s] speed light 3.00 × 10 m/s See also: Rotation and Torque 1 2 (f is frequency in Hz; λ is wavelength in m) [y = vertical distance] E is the mass energy, m is mass, c is the 2 Spring: Third Order Approximation (kinetic energy) ∆KE = KE f − KE i = 21 mv 2 − 21 mv 0 2 = Work E = mc First Order Approximation for small angles L is length in m; g is gravity (F is tension in N; µ is mass per unit length of string in kg/m) see also: Oscillation (gravitational potential energy, y is vertical distance in meters) i (x is position in m; f is frequency Hz) Waves: joules PE = mgy mgy L g (A is amplitude in m; x is position) L 1 θ 9 θ sin 4 1 + sin 2 + g 4 2 64 2 T = 2π [joules] 2 v=± in Impulse: [kg · m/s] J = ∆p = F∆ t = mv f − mv i impulse = force × duration or the change in momentum see also Force Center of Mass: The center of mass of a body or a system of bodies is the point that moves as though all of the mass were concentrated there and all external forces were applied there. xcm = distance from origin [m] 1 n x cm = mi x i M = total mass [m] M i =1 m = mass of object [m] This can be applied to x = distance of object from origin y and z axis as well. [m] ∑ Rotation and Torque: [θ is in radians] ∆θ ω= ∆t ω = ω 0 + αt θ = ω 0 t + 12 αt 2 τ = torque (vector) (positive is in the counterclockwise direction) [N · m] τ = rFt = r⊥ F = rF sin φ τ = magnitude of the torque r = radius [m] F = force [N] r⊥ = perpendicular distance between axis and an extended line running through F. φ = the angle between r and F [° or rad] ∑ τ = Iα ∑τ = the net torque acting on a body [N · m] I = Inertia [kg · m2] α = angular acceleration [rad/s2] (if constant acceleration) [rad/s] (if constant acceleration) [radians] (if constant acceleration) I = [kg · m2] Inertia: orbiting object: I = mr sphere: Is = 2 5 2 mr ∑ mr (inertia) Ir = ring: 2 2 1 2 m (r 1 + r 2 ) 2 Id = disk or cyl.: 2 1 2 mr 2 average angular acceleration [rad/s2] thin rod (on side): I = tangential speed [m/s] cylinder on its side (axis ctr): I = m velocity of the center of mass [m/s] Parallel Axis Theorem: If you know the rotational inertia a t = rα at = tangential acceleration [m/s2] r = radius [m] α = angular acceleration [rad/s2] 2 vt = rω 2 r ar = radial acceleration or centripetal acceleration [m/s2] ar 1 12 ml 2 rod (axis end): I = 13 ml ( r2 4 l + 12 2 2 ) of a body about any axis that passes through its center of mass, you can find its rotational intertia about any other axis parallel to that axis with the parallel-axis theorem: I = I cm + Mh 2 (directed inward to center) v = speed [m/s] r = radius [m] ω = angular speed [rad/s] at I = Inertia [kg · m2] Icm = Inertia with axis at the center of mass [kg · m2] M = mass [kg] h = distance from the center of mass to the axis [m] Kinetic Energy: [Joules] KE r = 12 Iω 2 rotational kinetic energy KEt = 12 mv 2 translational kinetic energy KE = 12 I cmω 2 + 12 mv cm 2 rolling kinetic energy total acceleration [m/s2] a = at 2 + a r 2 Fc = ma r = m τ =r×F average angular speed [rad/s] ω 2 = ω 0 2 + 2αθ ∆ω α = ∆t v t = rω v cm = rω ar = Torque: vt 2 r Fc = centripetal force [N] ar = radial acceleration or centripetal acceleration [m/s2] (directed inward to center) 2π r 2π T= = v ω 4π 2 3 2 r = K sr 3 T = GM s T ∑ F = T − Mg = Ma ∑ τ = TR = Iα T = period [s] Kepler’s Third Law (planetary motion) Yo-yo: 0 where T = the period 2 G = 6. 673 × 10−11 N ⋅m2 kg K s = 2. 97 × 10 −19 s 2 m3 a = −α R0 a= T = tension [N] M = mass [kg] R0 = radius of axle [m] −g 1 + I / MR0 2 R0 Mg Angular Momentum: Oscillation: l = Iω rigid body on fixed axis [kg · m2/s or J · s] l = r × p = m(r × v ) l = angular momentum of a particle [J · s] r = a position vector p = linear momentum [kg · m2/s or J · s] m = mass [kg] v = linear velocity [m/s] I i ωi = I f ωf tan θ = v 2 F = force [N] A = crossectional Fl0 A∆L area [m2] V ∆P V= original vol. [m3] ∆V = chg. in vol. ∆V [m3] l0 = initial length [m] ∆L = chg. in length [m] ∆P = change in pressure [Pa or N/m2] f F = ma = −( mω 2 ) x U ( t ) = 12 kx m 2 cos 2 (ω t + φ ) Potential Energy K ( t ) = 12 kx m 2 sin 2 (ω t + φ ) Kinetic Energy y = mx + b Ax + By + C = 0 d (sin u ) = u’ cos u dx (m = − A / B ) y − y1 = m ( x − x1 ) A Ax + By = Ax 1 R = rate of flow [m3/s] A = crossectional area [m2] v = velocity [m/s] R = A1v1 = A2 v 2 + By 1 point-slope, alt. a + y b =1 (m = − b / a ) 2-point intercept a = x-intercept b = y-intercept Tom Penick tomzap@eden.com www.teicontrols/notes December 6, 1997 Bernoulli’s Equation: P1 + 12 ρ v12 + ρ gy1 = P2 + 12 ρ v 2 2 + ρ gy 2 P1 = pressure [Pa or N/m2] v = velocity [m/s] y = height [m] x slope-intercept first degree point-slope y 2 − y1 y − y1 = ( x − x1 ) x 2 − x1 Rate of Flow: For a horizontal pipe: Total Mechanical Energy Equations of a Line: F a [rad] k = spring constant [N/m] T = period [s] f = frequency [Hz] F = force [N] m = mass [kg] I = moment of inertia [kg · m2] h = distance between axis and center of mass [m] d (cos u) = − u’sin u dx P0 = atmospheric pressure if applicable [Pa or N/m2] ρ = density [kg/m3] g = gravity [m/s2] h = height [m] f = force [N] a = area [m2] F = force [N] A = area [m2] 1 f x’ = velocity of the oscillating object [m/s] x” = acceleration of the oscillating object m/s2] 1 atm = 1.01 × 105 Pa = 760 torr = 14.7 in2 f F = a A k (spring) m ω= E = U + K = 12 kx m 2 Pressure in a liquid: (due to gravity) [Pa or N/m2] P = P0 + ρ gh x = position [m] xm = amplitude [m] ω = angular frequency [rad/s] t = time [s] φ = phase angle [rad] (ω t + φ) = phase of the motion m (spring) k I T = 2π mgh Volume Elasticity: Bulk Modulus [Pa or N/m2] B =− 2π = 2π f T T = 2π rg Elasticity in Length: Young’s Modulus [Pa or N/m2] Y= ω= T= Angular momentum is conserved when torque is zero. An Optimally Banked Curve: x = x m cos(ω t + φ ) The Position Function for oscillating motion: ρ = density [kg/m3] g = gravity [m/s2] P1 + 12 ρ v12 = P2 + 12 ρ v 2 2 •1 SSM Earth is approximately a sphere of radius 6.37 × 106 m. What are (a) its circumference in kilometers, (b) its surface area in square kilometers, and (c) its volume in cubic kilometers? 2-The micrometer (1 μm) is often called the micron. (a) How many microns make up 1.0 km? (b) What fraction of a centimeter equals 1.0 μm? (c) How many microns are in 1.0 yd? 3. The fastest growing plant on record is a Hesperoyucca whipplei that grew 3.7 m in 14 days. What was its growth rate in micrometers per second? 4. Earth has a mass of 5.98 × 1024 kg. The average mass of the atoms that make up Earth is 40 u. How many atoms are there in Earth? 5. Iron has a density of 7.87 g/cm3, and the mass of an iron atom is 9.27 × 10−26 kg. If the atoms are spherical and tightly packed, (a) what is the volume of an iron atom and (b) what is the distance between the centers of adjacent atoms? Part 2: 1. While driving a car at 90 km/h, how far do you move while your eyes shut for 0.50 s during a hard sneeze? 2. An automobile travels on a straight road for 40 km at 30 km/h. It then continues in the same direction for another 40 km at 60 km/h. (a) What is the average velocity of the car during the full 80 km trip? (Assume that it moves in the positive x direction.) (b) What is the average speed? (c) Graph x versus t and indicate how the average velocity is found on the graph. 3. The position of an object moving along an x axis is given by x = 3t − 4t2 + t3, where x is in meters and t in seconds. Find the position of the object at the following values of t: (a) 1 s, (b) 2 s, (c) 3 s, and (d) 4 s. (e) What is the object’s displacement between t = 0 and t = 4 s? (f) What is its average velocity for the time interval from t = 2 s to t = 4 s? (g) Graph x versus t for 0 ≤ t ≤ 4 s and indicate how the answer for (f) can be found on the graph. 4. You are to drive 300 km to an interview. The interview is at 11:15 a.m. You plan to drive at 100 km/h, so you leave at 8:00 a.m. to allow some extra time. You drive at that speed for the first 100 km, but then construction work forces you to slow to 40 km/h for 40 km. What would be the least speed needed for the rest of the trip to arrive in time for the interview? 5. At a certain time a particle had a speed of 18 m/s in the positive x direction, and 2.4 s later its speed was 30 m/s in the opposite direction. What is the average acceleration of the particle during this 2.4 s interval? 6. SSM An electron with an initial velocity v0 = 1.50 × 105 m/s enters a region of length L = 1.00 cm where it is electrically accelerated (Fig. 2-26). It emerges with v = 5.70 × 106 m/s. What is its acceleration, assumed constant? 7. An electric vehicle starts from rest and accelerates at a rate of 2.0 m/s2 in a straight line until it reaches a speed of 20 m/s. The vehicle then slows at a constant rate of 1.0 m/s2 until it stops. (a) How much time elapses from start to stop? (b) How far does the vehicle travel from start to stop? 8. A car traveling 56.0 km/h is 24.0 m from a barrier when the driver slams on the brakes. The car hits the barrier 2.00 s later. (a) What is the magnitude of the car’s constant acceleration before impact? (b) How fast is the car traveling at impact? 9. Figure 2-29 depicts the motion of a particle moving along an x axis with a constant acceleration. The figure’s vertical scaling is set by xs = 6.0 m. What are the (a) magnitude and (b) direction of the particle’s acceleration? 10. With what speed must a ball be thrown vertically from ground level to rise to a maximum height of 50 m? (b) How long will it be in the air? (c) Sketch graphs of y, v, and a versus t for the ball. On the first two graphs, indicate the time at which 50 m is reached. 11. At a construction site a pipe wrench struck the ground with a speed of 24 m/s. (a) From what height was it inadvertently dropped? (b) How long was it falling? (c) Sketch graphs of y, v, and a versus t for the wrench. 12. As a runaway scientific balloon ascends at 19.6 m/s, one of its instrument packages breaks free of a harness and free-falls. Figure 2-34 gives the vertical velocity of the package versus time, from before it breaks free to when it reaches the ground. (a) What maximum height above the break-free point does it rise? (b) How high is the break-free point above the ground? x (m) x x 2 1 (s) Figure 2-29 Problem 37. 아 -1 (s) 2 6 8 Figure 2-34 Problem 51.
o you similar assignment and would want someone to complete it for you? Click on the ORDER NOW option to get instant services at LindasHelp.com