This file sometimes links into relevant Pearson Test pages. Those links have names starting with "PT_".
References to XAM go to a book about this test and the following number is from their test given towards the end of that book. Often the page number is also given.
Choosing an equation from a list.
Possible to eliminate some entries based on units? For the remaining possibilities, ask whether the inclusion of a variable makes more sense in the numerator or denom. Signs +/- for additive terms.
intro:remember the ½ ! . units N, problem1: calc drag, straightforward how: // outline technique answer: calc effect on a ballistic trajectory problem2: calc terminal velocity of skydiver. problem3: calc effect on a ballistic trajectory how: answer: notes: refs: // optional skydiver1 example, 68Kg=150lbs, Area=.88m²: air density 'rho' = 1 Kg/m³, Cd 1.0, 87mph=39m/s, Forces in balance so what's value of drag? Force needed to offset gravity rho/2=0.5kg x Cd=1.0 x A=.88m² x v²=1513m²/s² = 667N skydiver's wt=m g, force on floor normally: 68 Kg * 9.8 = 666N find m/s for skydiver2: XAM p4 75Kg, A=0.7m², rho=1.29Kg/m³ Cd=0.5 They rearrange to get ( 2 m g ) ( 2 * 75 * 9.8 ) v = sqrt ( -------- ) = sqrt ( --------------- ) = 57m/s, 126mph ( Cd rho A ) ( 0.5 * 1.2 * 0.7 ) // fh thinks Cd too small
problem 'bikeCdA': find (Cd A) of a bike and rider given weights,
terminal velocity, and road slope θ.
Assume rolling resistance 0.
Similar to 'slide-down-ramp' problem; friction replaced by wind resistance!
Observations, approach: Motive force, gravity, = mass g sin(θ) and this is exactly offset by 1/2 ρ v² Cd A where Cd is drag coef, 'A' is cross sectional area. So Cd A = mass_kg 9.8 m/s² sin(θ) / (1/2 rho v²) Example: Area might be in the range of 0.5 meters² A large person plus heavy bike might weigh 120 kg. The slope might be 5% which is virtually same as sin(θ) = 0.05 Cd (A=0.5) = (mass=120) 9.8 m/s² (sine=0.05) / [ (rho/2=0.61) v² ] or (moving A to the right side denominator) Speed 30mph ~= 44 km/hr = 14 m/s (14 m/s is just over 30mph) Cd = (mass=120) 9.8 m/s² (sine=0.05) / [ (rho/2=0.61) (A=0.5) (v=14)² ] = 1.24 Cd = 1.24; Cd A = 0.62
antenna_dipole properties. antenna_dish properties. antenna_yagi properties.
w/ altitude example hotAirBalloon1.py PT3_Q20 (no img). What would be the condition on earth during the day if there was no atmosphere? ans. very hot. 'and unbearably cold at night'. // irradiance would be (142/93)^2 = 2.25 * that of mars. Already known to be 1kW/m2. On the other hand, the 1kW does hit the earth (air) now. Clouds reflect (30-50%)?
PT1_Q91 (no img) ref: gsci, atwood problem1: acceleration how: (?) force = (m1 - m2)/(m1 + m2) answer:accel = [(m1 - m2)/(m1 + m2)] * g problem2: tension how: answer: T = 2G(m1*m2)/(m1+m2) or equiv: 2g / (1/m1 + 1/m2) #note (on equiv) that m1m2/(m1+m2) is used for parallel resistors... notes: refs: gsci
throw ball down. Use full(er) motion eqns to solve PT3_Q22 "ball thrown downward fm a height of 45m at 10m/s. What speed on impact? 1/2 m v2 + mgh = 1/2 m vprimed2 #total E won't chg; KEinit + PE = KEfinal v2 + 2gh = vprimed2 # rm'd 'm' from all terms 100 + 2*10*45 = vprimed2 = 1000 vprimed = sqrt(1000) = 31.6 m/s PT3_Q22
range eqn range(θ) = v2 * sin(2 * θ) / g // PT1_Q06, PT1_Q52 shot of 100km, expected max ht = range/4 // PT2_Q13
ballistic projectile trajectory. flat ground. 45deg launch. cannon, target same elev. no aero drag range_m = apex.x = range/2 apex.y = range/4 # yes, this IS a parabola v0_mps = sqrt( g * range_m ) # PT2_Q13 and my notes on PT2 qualitative solution to PBS contest 3/1 quantitative range as function of launch angle. fm PT1_Q6 _zba PT1_Q54 what muzzle vel? for angle 37, tgt at 400m, 100m high. see notes in gsci PT3_Q38 (no img).A particle is thrown at a speed (u) making an angle (t) to the horizontal. When the particle makes an angle (phi) with the horizontal, its speed changes to (v). Which one of the following statements are true?v= U cos(t) * sec(phi) #fgh,#math.trig 'sec' = 1/cos TODO: understand PT3_Q38 !
P = sigma A e T^{4} 1. The Sun's power at dif wavelengths (actually any blackbody). ref in glossary_sci.html 2. how much radiation emitted fm a 1 cm cube at 1000K. (one of the higher PTs...) 3. perhaps choose values for photoelectric effect. SOON? The type of _surface_ metal is the big deal. And, to a lesser extent (?), the surrounding E-field.
bouyant force =F_{b} = m a = ρ V g // where ρ is density, V volume, g gravity (and (ρ * V) == mass, 'm') What is the relationship between the following: 'mb' buoyant mass representing the effective mass of the object with respect to gravity 'mo' the true vacuum mass of the the object 'po', 'pf' the average densities of the object and the surrounding fluid? mb = mo * (1 - pf/po) PT3_Q21. A man is sitting in a boat which is floating on a pond. The man drinks some water from the pond. What happens to the water level in the pond? PT3_Q37 (no img).
What expression gives the measured heat (+/-) from a reaction. You're expected to know the following symbols:
m | mass |
S | Specific heat of the reaction |
ΔT | change in temperature, Kelvin (or Celsius) |
cylindrical cap: 4 pi * 8.85e-12 * dielectric * length/ (ln b/ ln a) TODO: a? b? video: "how to make capacitors", rimstartOrg flat plate cap: 8.85e-12 * dielectric * area / (4 pi distanceBetweenPlates_m)
actually a trebuchet
1. accel = v²/r 2. force = m * accel =m v²/r XAM p142,Q08: an example 100g mass revolving on 0.5m string circles every 0.25s; what is centripetal accel: v2/r = (circum * 4/s)² / 0.5m = 315.5 m/s² PT3_Q85 race car friction coef: radius of turn 100m, speed 80km/hr, which of {0.1,0.3,0.35,0.6} is most likely coef of friction? 80km/hr = 22.2m/s - frictional force must be >= centripetaal force # note how 'm' will drop out ! m v²/r = u m g # friction coef, mass, Grav u = v² / gr # insert values, crank... u = ( (80km/hr=80000/3600)m/s )² / ( (g=9.8m/s2)*(r=100m) # insert values, crank... u = 0.504 therefore only 0.6 would work.
(r,angle)
self-inductance _unit_ // same as any inductance, the Henry impedance impedance, characteristic impedance wikip mentions wave impedance, and acoustic impedance. useful? PT1_Q26: internal resistance battery // hint. use Power = I2 R 0.1 Ω, 1.5v cell, I=2A, power lost = ____ thevinen (sp)
potentiometer. PT1_Q08 (no img) PT3_Q11 (no img, no paper) energy stored in a capacitor is 1/ (e0 A V²)/d, PT3_Q26 (no img, no paper). What action is to be taken to identify if the plate of a capacitor is positively charged? transformer turns ratio for (voltage) amplification PT3_Q48.A uniformly wound solenoid coil of self-inductance 1.8x10-4H and resistance 6Ω is broken into two identical coils which are connected in parallel across a 12V battery of negligible resistance. What is the steady state current through the battery?
calc'g R's (or C's) in parallel, inductors in series can do pairs R1 || R2 = R1 * R2 / (R1 + R2)
XL (inductive reactance) = L ω = 0.05H * 377 = 188 ohms XC (capacitive reactance) = 1/ωC = 442 ohms XL and XC work against each other so the diff, 442-188 = 234 Z, the overall impedance = sqrt(resistance^2 + (XL-XC)^2) = ____ ohms V = I * Z (Z an elaboration of R, resistance) Vrms = Vmax/sqrt2 = Vmax * 0.707 cos phaseAngle = Rohms/Zohms = "power factor" Irms = Vrms / Z
bakingSoda // co2calcs.py paraffin // co2calcs.py composting(material,dimensions,temp) // fermentation
1. inelastic 2. elastic, XAM p32,p33 3. weird, PT1_Q24 basketball vs tennis ball
basketball hits floor at 10mph, bounces upwards at 10). It meets the little tennis ball and they collide at 20mph.
Relative to the basketball, the tennis ball leaves the collision with an energy of 20mph
Observations: conservation of E (KE+PE) not useful b/c heights unknown. Conservation of momentum is only apch. Definitions: 'b' is basketball, 't' is tennis ball. 'mom' is momentum, 'v' velocity, masses undefined. should cancel out. At the collision, when basketball moving upwards (v positive): b.mom = b.v * b.mass; t.mom = t.v (neg) * t.mass after the collision, b.mom -= t.mom and t.mom += See website "rocket balls"
The equilibrium constant is 4.0
Determine the amounts of the products (outputs) produced at equilibrium
[H2O] means "the # of moles of water molecules", which is proportional to the concentration.
output1.conc * output2.conc | [CH3CO2C2H5] [H2O] | |
---------------------------------- | = | ---------------------------------- = 4 |
input1.conc * input2.conc | [CH3COOH] [CH3CH2OH] |
CH3COOH | CH3CH2OH | CH3CO2C2H5 | H2O | |
---|---|---|---|---|
initial | 80g/60=1.33mole | 85/46=1.85mole | 0 | 0 |
equil. | 1.33-x | 1.85-x | x | x |
Note that 'x' stands for the (unknown) # moles of the 2 products. Solving for the resultant product is 'just' finding 'x'.
By the definition of equilibrium constant, an equation can be constructed showing the molar amounts:
[x] [x] | x² | ||
--------------------- | = | ------------------------- | = 4 |
[1.33-x] [1.85-x] | 2.46 - 3.18x + x² |
Move the denominator (2.46...) behind the '4'; subtract x² from both sides to produce: 0 = 3 x² - 12.72 x + 9.84 Solve the quadratic (gives 3.22 and 1.02) and remove the 3.22 since you didn't start with that many moles.
Note that this problem was about as easy as such a problem could get - all substances reacted in equal amounts.
2. PT3_Q27 Prepare a solution dissolving 20 g of potassium iodide in 500 cm³ of water.
How much iodine needed for 1 L (liter) of solution?500 cm³ is ½ L. Atomic weights: K atomic wt 39. Iodine 126. Potassium Iodine KI 165. 20g of KI gives 20g * (126/165=.76) = 15.27g I so the I concentration is 15.27 per ½ L. To double the volume (to 1 L), while retaining the same concentration, you'd have to add another 15g of iodine and, presumably, that would be 15.27g of I crystals.15g isn't offered as a possible answer in PT3_Q27.
e- collected by solar panel. batt storage...
centripetal accel,PT1_Q31, PT3_Q01.v²/r = Bqv ,hmm. could factor v... Magnetic field intensity, B = v / (r q) (factored above, cancel 1 pair of v's) Cyclotron protons rotate w/ radius r with constant ω Rad/s angular velocity.
n slits,wavelength_nm,slitSpacing_m, print loop: startOffsetAngle,deltaAngle,n2print can do in excel ? // where it could be plotted phys instrument possible. move along the 'target screen' (what needed to scan Sun's spectrum (?))
(waveLen_m,dist2display_m,slitSep_m, MUSTTODO & mv to glossary_sci XAM p67 btm, diagram "interference maxima", PT1_Q04. λ = x/n * d/L λ = x_offset/n_order# * d_slitWidth/L_distToSurface (mnemonic) offset between lines (approx?)
λ * (n_order#=1)* L_distToSurface / ( d_slitWidth ) = x_offset or: x_1stLine = λ * L / slitWidth ## all distances in meters. PT1_Q48-ish (no img): white light. Only central fringe is white, all others colored.
width of signal at various settings.
prism XAM p68 btm, rainbow
What Freq Observed (fo) given a Source Freq (fs) (eg 790 Hz)
when the signal speed is 'v' in its medium (sound 343m/s)
vs is source's m/s, thru medium, away from 'o' (eg: -8m/s)
vo is observer's m/s, thru medium, towards source. (eg: +9m/s)
fo = 790 hz((343+9) / (343-8)) = 790*(352/335) = 830 Hz
Can run the eqn the other way: fo observes 830, what is fs ?
fs = 830 hz((343-8) / (343+9)) = 790 Hz
PT3_Q15 (no img).An observer moves toward a stationary sound with a velocity that is
one-tenth (1/10th) of the velocity of sound. What is the apparent
increase in frequency? ans: 10 pcnt. (F'-F)/F=0.1=10%.
when deciding whether a term goes in numerator or not, consider whether choice should make the answer larger, smaller. ? compound units? mps / mpg eg: specific heat == (J/kg)/degC = J/(kg*degC) needed_J = 3kg * 12degC * SH(J/(kg*degC) // 'forward' case increase_degC = 2J / (4kg * SH(J/(kg*degC)) // working it bkwds
see eField app and class functions. see gsci, E field see E field prjt What is the electric potential (== voltage) at point A (0,0) made by pt B +7 C, (-3.5,5) and pt D -3.5C (3.5,5). XAM p78 top d = sqrt(5² + 3.5² = 6.1mV = k q / r r in meters, V volts, k == Coulomb's const = 9¤9 * (7C - 3.5C)/6.1m = 5.13¤9 volts ! what is the force on a proton at pt A? acceleration? E = field at pt A F = q E = 1.602e-19 * (E = 5.13¤9 volts) what is the force on an electron at pt A? acceleration? E = field at pt A F = q E = -1.602e-19 * (E = 5.13¤9 volts) # just a negative chg What work is needed to bring a particle of charge Q to A? E = V = field at pt A # V (volts) same thing (and units) as E W = V Q , Joules //XAM p78 Coulomb's const(k).units are N m² / C2; # k == denom ready to deal in units C # q is charge, in units 'C', ea of which is 6.24e18 x 1.602e-19
radiation PT1_Q12 (no img) 1/2 life Radon is 3.8 days. 16g in 19days 5 half live. portion remaining is (0.5)^5 or 2^(-5) = 1/32 1/32 of 16 is 0.5 so 16-0.5 = 15.5 disintegrated' Ans. 15.5g heat dissipation current leaking off a cap
population. XAM p119. and conversion to pH
Free Body Diagram. traffic light example: trafficLt.py there are some nice examples early in the XAM book.
The weights of the beakers and water are equal (not represented in FBD). B1 = B2. The volumes of balls are equal; thus their bouyancies are equal. m1g is the weight of the ping pong ball. m2g that of the steel ball.
T1 == -B1; T2 == -m2g
deducing Fs, XAM p143 calc speed? PT1_Q42 PT1_Q64 //praxis test #1 question 64 Practice Problems block sliding down a ramp/incline. XAM p12,169, Serway p121 F (Newtons) = -frictionCoef * NormalForce_Newtons // fs is force during static friction NormalForce_Newtons = m g cos(θ) // wt of block adjusted for slope downRamp force = m g sin(θ) so, if the block is at constant speed the friction coef is the 'kinetic' type frictionCoef = m g sin(θ) / m g cos(θ) = tan(θ) # DEFINES the coef value else if the block is just breaking loose, the coef is/was the static type max speed in R meter radius turn w/ Us = 0.80, (ref09, kindle loc 2989 (Asus)). Note: even tho the car is moving, the tire/street interface is stationary - until the car "loses traction". Car's mass M won't matter. turn radius 'R' is 25meters M * centripetalAccel = M * v²/R = 0.80 Mg cancel M's and move R to right side v2 = 0.80 * (R=25m) * (g=9.8m/s²) v2 = 196 v = 14m PT3_Q32 (no img).What force does John exert on a 30 kg block moving on a horizontal surface at a steady speed with a coefficient of kinetic friction at 0.5? ans: 0.5 * 30kg * G = 150 gsci
in a turn in banked turn wheel spin, rbus PT3_Q19.Calculate the frictional force acting on mass m. The static coefficient of friction is 0.4. F = 10N, m = 3kg, g = 10m/s² Maximum frictional force=µ*N=0.4*(3*10)=12N. since only 10N force is being applied... PT3_Q40.A block of mass m is held stationary against a wall by applying a horizontal force F on the block. Identify the false statement. ans. N does not produce any torque.
see wave frequency, fundamental
gas molecular Wt, given PV=nRT info density p/testPhys/testPages/imgs/PT3_Q07_densityIdealGasLaw.png for algebra Calculate the density, kg/volume, of a gas from ideal gas law? PV = nRT P = pressure, V = Volume, m = gas mass, M = mass/mole, T = temperature and R = gas constant XAM p48: gas density: d = nM/V = PM/RT, M is molecular wt, 'n' is #moles, V volume so there are eqns: RGasConstant = 8.314 # J/(mol - K) density(nMoles, molecularWt_kg, volume_m3) : density = nMoles * molecularWt_kg / volume_m3 density(pressure_pa, molecularWt_kg, temp_k): density = pressure_pa * molecularWt_kg / (RGasConstant * temp_k) molecularWt_gPerMole(d, T, P) : 3.24 g/L * (R=0.08206 L-atm/mol-K) * 800K molecular_gPerMole = ----------------------------------------- 3.0 atm
PT3_Q29 (no img needed).Two small, heavy spheres each of mass M are placed a distance r apart on a horizontal surface. What is the gravitational field intensity at the mid point of the line joining the centers of the spheres? zero! (duh)
energy (time?) to heat material thru range, possibly doing phase changes. use tabled values to abstract matl choice search 'steam' enthapies in endo reaction. Hr < Hp. PT1_Q10 SOON more. Phase change 1 g water into steam takes ? XAM p42+ _zhe.phase 540 cal = 540 * 4.185 J/cal = 2260 J heat 10g gold from 25C to 1300C. from XAM p43. process. _zhe.gold In the following 'q' is an amt of heat energy, in Joules. C is J/K /mol. T is temperature in C 1. convert g to moles # 10g / 197 grams/mol = 0.051 mol ForSciCalcs? 2. heat from current temp to melting pt, 1064C ForSciCalcs? solid heat capacity: 28 J/K /mol q = n C deltaT = q1 = nMoles * solidHeatCap * deltaT = 0.051 * 28 J/K /mol * (meltingPt - startingPt) = 0.051 * 28 J/K /mol * (1064 - 25) = 1.48 kJ 3. melt it. enthapy of fusion: 12.6 kJ/mol per gram? q2 = nMoles * enthalpy = 0.051 mol * 12.6 kJ/mol = 0.64 kJ 4. heat liquid to final temp. molten heat capacity: 20 J/K /mol q3 = nMoles * moltenHeatCap * deltaT = 0.051 * 20 J/K /mol * (goalT - meltingPt) = 0.051 * 20 J/K /mol * (1300 - 1064) = 0.24 kJ 5. add up the parts q = q1 + q2 + q3
X_{L} (inductive reactance) = L ω = 0.05H * 377 = 188 ohms X_{C} (capacitive reactance) = 1/ωC = 442 ohms Reactance Xi = X_{L} - X_{C} vector Z = R + Xi, i imaginary. XAM p93
impedance amplitude
phase angle = arctan((X_{L} - X_{C})/R) # if C larger than L, phase angle negative
prevent reflections, a loss of signal, power
E(f=1420e6 hz) J = ___E(656nm) = h (C = 3e8 m/s) / 656e-9 m Joules = h 3/656=4.57e-3 e-1 = (h=6.63e-34 J-sec) 4.57e-4 s-1 E = h nu which is h f, 'f' freq of the light. "Planck's Energy Eqn" nu = C / lambda, where lambda is wavelength in meters. h nu is J TODO: small img --> eg: E(656nm) J = h (C = 3¤8 m/s) / 656¤-9 m = h 3/656 = 4.57¤-3 ¤-1 SOON: looks weird... could use hBar = (h=6.63¤-34 J-sec) 4.57¤-4 s-1 eg: E(f=1420¤6 hz) J = h f = (h=6.63¤-34 J-sec)(1.420¤9 s-1) = 9.41¤-25 J PT3_Q41 (no img needed).On what does a wavelength depend? ans: velocity and freq
01: parallel wires,I=3A each,dist=5cm,Len=25cm,F = I * L X B, attractive 02: parallel wires,I=1A each,dist=5cm,Len=25cm,F = I * L X B, attractive B= (u0=) 1.34 02: charge V=10m/s horiz enters vertical M field. motion:? 03: 1500 turn coil,3cm dia,len 5cg,r ohms,v volts,field at 5cm = ___ T
find B at 2mm radius of 1 straight wire. XAM p84, Q1 u0 I 4π 10-7 30amps B = -------- = ---------------- = 0.003 T // .0030 T == 0030 Gauss 2pi r 0.002m
PT3_Q02. A bar magnet of magnetic moment 2.0 J/T lies aligned with the direction of a uniform magnetic field of 0.25 T. What is the amount of work required to turn the magnet in order to align its magnetic moment normal to the field direction? ans: The PE of a bar magnet with its mag moment 'm' inclined at an angle theta with mag field is U = -m B cos(theta). At 0 degrees, it is -m B cos 0 = -m B. At 90 degrees, it is -m B cos(90) = 0. The amount of work required is 0 - (-mB) = mB = 2.0 * 0.25 = 0.5J
find B when -8C uKg particle, CCW, speed 2000m/s, 2m radius. XAM p85 notes: 1.centripetal Force = Lorentz Force; 2. neg RHRule for neg chargeLorentz Force = q v B (sin(angle)=1) q v B = -8C 2000m/s B centripetal Force = m v² / r = 1e-6 kg * 2000² m²/s² / 2m // set eqns equal to ea other; rearrange to solve for B B * 1e-6 * 2000² = 1e-6 kg * 2000² m²/s² / 2m 1e-6 kg * 2000² (m/s)² 1e-6 * 2000 B = ------------------------ = -------------- = 125 e-6 = 1.25 e-4 T = 1.25 Gauss 2m * 2000m/s * -8C -16
q= 6e-6 C, v=4e5 m/s, B=0.4 T, theta=30 degrees Force = q v B (sin(angle)=0.5) Fb = (6e-6) (4e5) (0.4) (0.5) = 0.48N
FILLINexpand (+/-) of parallel wires, coils
Biot-Savart, 1/r2 Earnshaw's theorem, etc see find B in 1/r³ area.
FILLINexpand. see loose no img in angelflight binder PT3_Q39.As shown in the figure, the bar magnet is moving away from the loop and induced current is in the opposite direction. Identify the true statement for this illustration.
"Strike a bar, either held vertically or pointed north, (some sources say soft iron and others say hardened iron or steel) several times on one end with a hammer." "Double touch – take two magnets, touch the south pole of one and the north pole of the other to the center of the bar, and draw them off to the ends a number of times." princeton notes fgh: I suspect that putting a 'rod' into a strong field AND hammering it, is best. Another source: "draw a screwdriver across a powerful magnet several times, all strokes in the same direction".
PT1_Q72 (no img) wtVsMass.png. If the total weight of two bodies is 35.76 N and the mass of one body is 1.25 kg, what is the weight of the second body? PT3_Q45 (no img).A boy of mass 50 kg is standing on a weighing machine placed on the floor of an elevator. The machine shows the weight of the boy in newtons. What is the reading when the boy is stationary? ( g=9.8 m/s²) ans 490
PT3_Q03. Essentially dot product is a cosine and cross product a sine. So, rephrased, At what angle are the dot product and the cross product equal? 45
FILLINexpand. eg: "3 molar HCl"
parallel axis add newMass * D² to Icm where D meas'd fm turning axis
PT1_Q65+ (no img) weakest binding force is van der waals PT3_Q46 (no img).What is the rest mass energy of an electron in MeV? .51 MeV << try mult! PT3_Q49 (no img).If an electron absorbs energy at ground state to reach an excited state, what is the energy emitted when it falls back to ground state? formula for that??
images orientation real vs virtual focal length vs radius of curvature. snell's law: XAM_Q36, XAM p149critical angle The angle of incidence where the exit angle becomes 90°. TIR, Total Internal Reflection, occurs for angles greater than or equal to the critical angle.
critical angle = arcsin(n_{2} / n_{1}) , where n_{2} must < n_{1}; Example: rays rising into air (n_{2}=1.0) from water(n_{1}=1.333), the critical angle would be arcsin(1 / 1.333) = arcsin( 0.75 ) = 48.6 degrees Angles are measured wrt the 'vertical' (perpendicular to boundary) Viewed through Snell's Law, n_{1} sin(θ1) = n_{2} sin(θ_{2} == 90° ) PT1_Q19, graphic; PT2_Q46, text, eqns lenses, gsci magnification of a simple telescope and eyepiece. magnification PT3_Q04. The pupil of a person's eye has a diameter of 1.5 mm. According to Rayleigh's criterion, what angular separation must two small objects have if their images are just barely resolved? Assume they are illuminated with a light of wavelength 400 nm.
#sunMass.py G = 0.667e-10 # universal Grav constant pi = 3.141526 # calculate the mass of the Sun given earth's r,T r = 1.49e11 # meters T = 3.1556736e7 # seconds. sun_kg = (4.0*pi*pi)/G * (r**3)/(T*T) # should be 1.989e30 kg print("sun mass, calc'd {:.3e}kg".format(sun_kg)) print("answer should be 1.989e30 kg") >>> sun mass, calc'd 1.966030e+30kg answer should be 1.989e30 kg Derive Keplers 2nd law. see serway p369, XAM ? # derivation fm serway 369 An orbit interval off the major axis will create a motion line which varies in dist fm planet. That line and the radial line can be the basis for a parallelogram whose area is halved. End up w/ dA/dt = L / 2Mp where A=Area, t=time, L=angular momentum (a constant), M=mass (const). The 2 comes from the parallogram mentioned. Derive Kepler's 3rd law. see serway p369-371, XAM ? a = G Msun/r² = (v=2π r/ T)² / r G Ms/r² = ((2 π r)/T)² = 2π r² / T² rearrange T2/(2π r)² = (r (2π r)²) / (G Msun) T2 = r² * (4π²) / (G Msun) T2 propor to r² use 3rd law to see how planets fit the law. build a table SOON PT3_Q17 (no img).Kepler’s 3rd law states that the ratio of the squares of the orbital period for two planets is equal to the ratio of the cubes of their mean orbit radius. This is expressed mathematically as: (T1/T2)² = (A1/A2)³ eccentricity. from a mean sun distance and an eccentricity, find major and minor axis calc,compare periods for all
std 6degrees use fuller eqn for much larger angle (?)
From your study of a particular reaction and the resultant concentrations, find pH = -log_{10} [H+] // (where [H+] is concentration of the H+ ion) eg: XAM p119, [H+] is 0.0005 mol/L pH = -log_{10}(0.0005) = 3.3 XAM p119. For _figuring_ concentration, see concentrations
energy of photon, use hbar (?) in another example
FILLINexpand,a:a.watts,a.meters,dish.meters=2;eqn watts.rxd=___ 1420Mhz H&sub2; rx'd from astronomical source. Hydrogen wavelength 1420Mhz; Andromeda rx'd: rx.watts = tx.watts * rxArea_m² / [4π*(s.dist_m)²].m² ;
2x² + 4x -4 = 0. Ans: x = -1 + sqrt(3), -1 - sqrt(3) TODO (review) a=2; b=4; c= -4 # divide by a x² + 2x - 2 = 0 # subtract c/a x² + 2x = 2 # add (b/2a)² = (4/4)² = 1 x² + 2x + 1 = 3 # rearrange (x+1)² = x² + 2x + 1 = 3 # rearrange, shorten (x+1)² = 3 (x+1) = +/- sqrt(3) x = -1 + sqrt(3); x = -1 - sqrt(3) (Khan) For example, solve x˛+6x=-2 by manipulating it into (x+3)˛=7 and then taking the square root. x + 3 = sqrt(7); x = sqrt(7) - 3 //sqrt 7 is 2.6457 quadratic eqn solver says roots are -5.645 and -0.354 [(sqrt(7)=2.6457) - 3 ] = -0.354 but -2.64 squared is also 7 [ -2.64 - 3 = -5.64 ]
must be other apps too. TODO
pushing, w/ friction
rolling a 3kg, 4"diam. ball down'
problem: ball/wheel/cylinder rolling down a slope/ramp/incline. angular momentum important answer: how: // outline technique # values needed at ramp end: KE-linear, KE-rotational, PE=0 # total energy == mgh at start and end # XAM p35. # Vf means Velocity final. 'r' is radius of the thing rolling down the ramp # The 'shapeCoef' for a ball is 2/5; this could be easily chg'd for cylinder etc. notes: refs: also in AP Physics bk? Efinal = KErotational + KElinear + (PEfinal=0) mgh = 1/2 I * omega.final² + 1/2 m Vf² mgh = 1/2 [I = 2/5 m r2] * (Vf/r)² + 1/2 m Vf² # cancel the m's gh = 1/2 [ 2/5 * r2] * Vf² / r² + 1/2 * Vf² # cancel r² gh = 1/2 * 2/5 * Vf² + 1/2 * Vf² #_could_ factor out 1/2 Vf² but the 2 parts are already visible if let's not factor: 1/2 * 2/5 Vf² = 1/5 Vf² is for the rotational 1/2 Vf² is the linear divide up 'gh' accordingly else: # factor out '1/2 Vf²' gh = 1/2 Vf² * (2/5 + 1) gh = 1/2 Vf² * (7/5) gh * (5/7) = 1/2 Vf² gh * 10/7 = Vf² Vf = sqrt(gh * 10/7) ; omega.final = Vf/r
off parabola. focal point sound. off axis. diagram. integral? code(?)
n1 and n2 are indexes of refraction (angles can be in any units which are compatible w/ sin() and arcsin()) snell:theta2 = arcsin(n1 * sin(theta1) / n2) XAM_Q32 p148, XAM p62 btm
FILLINexpand(?) eye eyepieces otoscope
parallel 2 Rs parallel > 2 Rs series weird. common center resistor, eg: PT1_Q36
oscillator filter 555 timer chip
PT1_Q86: radius 20m, must be weightless at top. use g=10m/s2. find speed in loop. centripetal acc = v2 / r and it must equal G, 10m/s2 v2/10m = 10m/s2; v2 = 10m * 10m / s2 v2 = 100m2/s2 v = 10 m/s PT2_Q57: flat start, radius 10m, car 1000kg; find launch speed speed at top of loop = v2/R = 10m/s^2 = 10m/s KE for 10m/s is 1/2 1000kg 10m2/s2 = 5000 J total PE needed is mgh = 1000kg x 10 m/s2 * 20m = 200,000 J total E needed is 205000J the KE to support that is 1/2 (m=1000kg) v2 = 200,000 or 2 * 205000/1000 = v2 sqrt(410000/1000) = v sqrt(410) = v = 20.2 m/s their answer seems to have KE at 5000 to make 22.3m/s // they have the total E as 2.5*mgR
for now, see gsci
XAM p144
RC circuit link to wave eqn
speed of sound thru materials, seismic. waves, XAM p65 v = sqrt(elasticity/density) speed of vibration of a stretched string., XAM p65 v = sqrt(tension/linearDensity) harmonic freqs. overtones refs: bulk modulus, gsci
design. my electro magnets air core, B = fn(I, radius, sensorCoords, ? ) iron core, B = fn(I, radius, sensorCoords, ? ) force on a metallic slug
bloon/hot air balloon simple water calc ice to warm water ice to steam { 4deg (max density) }
dist thru earth. which formula correct XAM_Q07, p142 calc distance fm N Korea
Hooke's law. F = -k * x N/m // x in meters meas'd fm the spring's 0, k is Newtons/m PE PT4_Q47: interesting! What is the total compression in a 20 N/mm spring if a 3kg object falls on it from h=10/3m? mg(h+x) = springPE total = 1/2 k x2 # total E now make this into a quadratic, set to 0, to solve 1/2 20 x2 - mg x - mgh = 0 # where mg = 3*10 and mgh = 100 10 x2 - 30 x - 100 = 0 online calc finds solutions x=5,-2 I tried this originally w/ k = 2e4N/m; x=3/2000 = 0.0015m ?!
1st find angular acceleration, rad/s²: 6 rads/s -------- = 30 rads/s² 0.2 s torque applied is 10 Kg m² * 30 rads/s² = 300 N•m // (N is Kg m/s², N m is kg m²/s² == energy, like KE)
finding the length of a figure when u know certain angles and other lengths ? use a) vectors, b) law of sines (XAM p77), c) law of cosines (XAM p76) # guidance ? use law of the sines use law of the cosines.
'fall' to destination thru much of the earth w/o moon w/ moon other effects? earth spin?
FILLINexpand. XAM p56 PT1_Q56 (no img) gas expands freely in rigid, insul'd cylinder. what can chg? correct: only internal energy. PT3_Q12 (no img).In a thermodynamic process a closed system absorbs 1 kilocalorie of heat and does 250 Joule of work. What is the change in the internal state of the system? ans 1000x4.2-250=3950J 1st law dU = dQ - dW
FILLINexpand. XAM p56
FILLINexpand. XAM p56
(voltsIn,ampsIn) voltage multiplier // power in-out ~same; turnsRatio = turns_secondary / turns_primary volts = voltsIn * turnsRatio amps = ampsIn / turnsRatio # so Power (basically) unchanged. return (volts,amps)
PT3_Q52 (no img).The cgs system of units records observations of force as having a magnitude of 108. What would the magnitude of the same observation be if expressed in newtons? 1000 g-kg, meter-cm
FILLINexpand // XAM_Q02 p84 find volts for: rectangular coil, 5cm x 10cm, Nloops = 10 Binit = 0.5 T; Blast = 0.8T; time between init and last = 2 sec 1. find mag flux for time 'init' and 'last' phiInit = Binit * Area * cos(angle) = 0.5T * (0.05m * 0.1m = 0.005 m²) * 1 = 0.0025 T/m² phiLast = Binit * Area * cos(angle) = 0.8T * 0.005 m² = 0.004 T/m² 2. find emf, 'e' = - dPhi/dt = - (10 loops) (0.004 - 0.0025) T/m² / 2secs = -0.0075v = -7.5mV
requency. seen in a couple of tests. _zdo beat freq = f u / v taken fm f(1_u/v) - f PT1_Q29 PT3_Q43 (no img).Two sound waves of equal intensity (I) produce beats. What will be the maximum intensity of sound produced in beats? ans: proportional to the square of the amplitude
vv_waveEqn
of a closed pipe. PraxisDiscussionArea,3rd item.top of p32 serway p508, p527 "standing waves in air columns" TODO # waveSpeed = sqrt(Tension/linearMassDensity) // for a string instrument # the pipe will have a lowest freq (its "fundamental freq") waveSpeed = 331 m/s # for a wind instrument. fLowest = 1 / (2 * LengthOfNoteMaker_m) * waveSpeed_m/s # the harmonics for this instrument will be integer multiples of fLowest int n harmonic[n] = n * fLowest # for a stringed instrument, f is inversely proportional to string length # assumed true for wind instruments where the length of the resonant chamber changes. gsci 'harmonic_sound'
wave E vs A, amplitude. find energy and its relation to amplitude // square? what kind of wave PT3_Q31 (no img).What kind of waves can be seen on the surface of the milk if the glass of milk is put into a refrigerator? standing (?)
TODO (understand!) gsci for understanding. calc here (?)
take in k,x,omega and setup function which WOULD work if ... serway p505,eg 507 gsci entry
gsci. SOON conjure up, find useful example PT1_Q36 (no img) 'balanced' means "assume no current across central resistor" solve R's on both sides calc power.
W = amt of energy (work). XAM p34 problem stops a 2000kg car from 30meters/sec in 150 m. What average force was applied. F * 150m = 1/2 2000kg (30 m/s)2 F = (1000kg * 900(m²/s²) ) / 150m = 6000N // units: F is kg m/s²; it works
FILLINexpand_J(V, Q) # XAM p78 W_J = V * Q
move a charge fm infinity to dist r from another charge work = PE of the destination = V * Q (voltage * charge being moved) XAM p78
vector, wt. XAM_Q10, p143, PT1_Q57 (no img) grav work total after object pushed up ramp then down ramp. zero.