Science Calculations

This file contains calculation methods and related details, rather than definitions as found in 'gsci'. It's also likely that many entries here become 'pseudo code' or actual python programs in /Python34/examples/.

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.

General Methods, multiple choice tests

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.

Specific algorithms, equations

aero drag
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
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)
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.
atmospheric density w/ altitude
w/ altitude

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%)?

Atwood machine
PT1_Q91 (no img)
ref: gsci, atwood
problem1: acceleration
       (?)    force =  (m1 - m2)/(m1 + m2)
answer: accel = [(m1 - m2)/(m1 + m2)] * g

problem2: tension
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...
refs:  gsci
ballistics, throw down
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
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

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 !

black body radiation
P = sigma A e T4

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 = Fb = 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)

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:
SSpecific heat of the reaction
ΔTchange in temperature, Kelvin (or Celsius)
('Q', the standard symbol for heat energy, is the requested result.) Try 'basic logic' to figure out the relationship (all multiplicative. no denominator term) PT3_Q42.
capacitance, calculating
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.
chord (of a circle)
circuits, theory, calcs, behavior
self-inductance _unit_ // same as any inductance, the Henry


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)
circuit components
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?
circuit parallel
calc'g R's (or C's) in parallel, inductors in series

can do pairs R1 || R2 = R1 * R2 / (R1 + R2)
circuit RLC 1
book example problem. Note use of Omega and its relation to freq in Hz. Note also how the L and C reactance values compete, that their difference is used.

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

CO2 generation
    bakingSoda //
    paraffin   //
    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 plus its 10mph. So it rockets up at 30mph.

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.v * b.mass; = t.v (neg) * t.mass
after the collision, -= and +=
See website "rocket balls"
1. Solving for post reaction concentrations using quadratic formula: XAM p117 input chemicals: "reactants"
c1. ethanoic acid. 80g, MW (Molecular Wt) 60g
c2. ethanol, 85g, MW 46g
outputs: "products"          this taken from XAM pg117
c3. ethyl acetate
c4. H2O

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]

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]
---------------------   =   ------------------------- = 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
    e- collected by solar panel. batt storage...
centripetal accel, 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.
PT1_Q31,     PT3_Q01.
diffraction, slit experiments
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 (?))
diffraction, Double slit
(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.
diffration grating
width of signal at various settings.
prism XAM p68 btm, rainbow
Doppler effect
surprise: movements of Obs, Src are wrt medium ! (which could move in wind etc)

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 = fs((v + vo)/(v - vs)) becomes
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%.

dimensional analysis
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
electric field
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 = 6.1m
    V = 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
exponential decay
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
exponential growth
population. XAM p119. and conversion to pH
Free Body Diagram.
    traffic light example:

there are some nice examples early in the XAM book.
"Beaker Ball Balance". From Veritasium. TODO (more)

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_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
friction, tires
    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.
fundamental frequency
see wave frequency, fundamental

    gas molecular Wt, given PV=nRT info

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)
heat, conduction
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.
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
XL (inductive reactance) = L ω = 0.05H * 377 = 188 ohms
XC (capacitive reactance) = 1/ωC = 442 ohms
Reactance Xi = XL - XC
vector Z = R + Xi, i imaginary. XAM p93

impedance amplitude Z = sqrt(R² + (XL - XC)²) ;
phase angle = arctan((XL - XC)/R) # if C larger than L, phase angle negative

determine the amplitude of a signal at dif frequencies (just change ω)
prevent reflections, a loss of signal, power
light, energy, properties
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
magnetic field
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

magnetic field
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
magnetic moment
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

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

magnetic field, particle1
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 charge
Lorentz 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
magnetic field, particle2
    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

magnetic attraction
FILLINexpand (+/-) of parallel wires, coils
magnetic Dipole Moment
    Biot-Savart, 1/r2

    Earnshaw's theorem, etc

    see find B in 1/r³ area.
magnetic induction
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.
magnetize a rod
"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".
mass vs weight
PT1_Q72 (no img)

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
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
nuclear, atomic, spectra
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??
    real vs virtual
    focal length vs radius of curvature.
    snell's law: XAM_Q36, XAM p149


    critical 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(n2 / n1), where n2 must < n1; Example: rays rising into air (n2=1.0) from water(n1=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, n1 sin(θ1) = n2 sin(θ2 == 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.

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
    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
pendulum period
    std 6degrees
    use fuller eqn for much larger angle (?)
From your study of a particular reaction and the resultant concentrations, find
pH = -log10 [H+] // (where [H+] is concentration of the H+ ion)
eg: XAM p119, [H+] is 0.0005 mol/L
pH = -log10(0.0005) = 3.3
XAM p119. For _figuring_ concentration, see concentrations
energy, photon, wavelength
    energy of photon, use hbar (?) in another example
power fm stars, radio freq
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 ;
quadratics, completing square
    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 ]

quadratics, factoring
must be other apps too. TODO
ramp, pushing up, w/ friction
    pushing, w/ friction
ramp, thing sliding
rolling a 3kg, 4"diam. ball down'
ramp, ball, rolling down
problem: ball/wheel/cylinder rolling down a slope/ramp/incline.
    angular momentum important
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.

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
        # 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) ; = Vf/r
    off parabola.  focal point
    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
resistor networks
    parallel 2 Rs
    parallel > 2 Rs
    weird. common center resistor, eg: PT1_Q36
RC time constant
    555 timer chip
roller coaster
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

sagitta of a cirle
for now, see gsci
simple harmonic motion, "SHM"
XAM p144

sine wave generation
    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
specific heat
    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

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  ?!
Find torque for an object which steadily increases its angular velocity from 0 to 6 rads/s in 200 ms. The object has a moment of inertia of 10 kg m², XAM p45, btm.

    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.

tunnel thru earth
'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

thermodynamics, Otto cycle
FILLINexpand. XAM p56
thermodynamics, carnot
FILLINexpand. XAM p56
transformer V, I
(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)
units, cgs, conversion, trivia
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

voltages induced
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
wave, beat frequency
requency. seen in a couple of tests.
_zdo  beat freq = f u / v taken fm f(1_u/v) - f


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
wave equation
wave frequency, fundamental
 of a closed pipe.
    PraxisDiscussionArea,3rd 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 types
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 (?)

wave equation
 TODO (understand!) gsci for understanding. calc here (?)
waves, standing waves
    take in k,x,omega and setup function which WOULD work if ... serway p505,eg 507
    gsci entry
Wheatstone bridge
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.
Work/Energy from Force * Distance
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
work, from Voltage and Charge
FILLINexpand_J(V, Q)  # XAM p78
    W_J = V * Q
work. ?electrostatics?
move a charge fm infinity to dist r from another charge
    work = PE of the destination = V * Q (voltage * charge being moved) XAM p78
work to move object against Force
vector, wt.  XAM_Q10, p143,
PT1_Q57 (no img) grav work total after object pushed up ramp then down ramp.  zero.