2012 Notes on Physics and Calculus: 2012

Wednesday, September 5, 2012

Significant Physics Constants

Sources

$\epsilon_0 \approx 8.854187817620... 10^{-12} $

Source = Wikipedia
$\mbox{Used in Coulomb's Law defining the electrostatic force}$

$F_c = \left(\dfrac{1}{4 \pi \epsilon_0}\right) \left(\dfrac{q_1 q_2}{r^2}\right)$

$\epsilon_0 \mbox{is defined by:}$

$\epsilon_0 = \dfrac{1}{\mu_0 c^2} $

$\mu_0 = 4 \pi 10^{-7} \approx 1.2566370614... 10^{-6}$

Source = Wikipedia
In electromagnetism, permeability is the measure of the ability of a material to support the formation of a magnetic field within itself.

$\mbox{Gravitational constant} \; G = 6.67384 10^{-11} $

Source = Wikipedia
According to the law of universal gravitation, the attractive force (F) between two bodies is proportional to the product of their masses (m₁ and m₂), and inversely proportional to the square of the distance, r, (inverse-square law) between them:

$F_g = G \dfrac{m_1 m_2}{r^2} $

Tuesday, August 7, 2012

Final 17 - Springs and Tables

Given:

$\mbox{A 1 kg mass is connected to a spring with a constant of 350 newtons/meter.} $
$\mbox{The spring is compressed 25 centimeters and let go from 2 meters.}$

Question:

$\mbox{How far away will the mass impact the floor?} $

Rationale:

$PE_{spring} = KE_{mass} $

$.5 K x^2 = .5 m V_x^2 $
$V_x^2 = \dfrac{(.5 * K * x^2 )}{(.5 * m)}$

$t_y^2 = \dfrac{2 * \Delta{y}}{g} $
$t_y = t_x $
$\Delta{x} = V_x * t_x $

$\Delta{x} = \sqrt{\dfrac{(.5 * K * x^2 )}{(.5 * m)}} * \sqrt{\dfrac{2 * \Delta{y}}{g}} $

Calculate:

$\Delta{x} = \sqrt{\dfrac{(.5 * 350 * .25^2 )}{(.5 * 1)}} * \sqrt{\dfrac{2 * 2}{10}} $

$\Delta_{x} = \sqrt{10.9375 * .5} * \sqrt{.4} = 4.6770717 * .632455 = 2.9580374$
$\Delta_{x} \approx 2.96 \; \mbox{meters}$

References:

Final 16 - Bikes and Buses

Final 15 - Gravity and Electricy

Given:

$\mbox{A mass of 8 kg and a charge of 0.1 coulombs}$

$\mbox{Is dropped from a height of 1.2 meters}$
$\mbox{In the continuing presence of an electric field of 100 newtons/coulomb.} $

Question:

$\mbox{What distance d does the mass hit the floor?} $

Rationale:

$\Delta{y} = 0.5 g t^2 $

$ t^2 = \dfrac{2 \Delta{y} }{g} $

$F_{electric} = F_x = m a_x $

$E q = m a_x $
$a_x = \dfrac{(E q)}{m} $

$t_y^2 = t_x^2 $
$\Delta{x} = .5 (\dfrac{(E q)}{m}) \dfrac{2 \Delta{y} }{g} $

Calculate:

$\Delta{x} = .5 (\dfrac{(100 * .1)}{8}) \dfrac{2 * 1.2 }{10} = 0.15 \; \mbox{meters}$

References:

Final 14 - Pushing A Car

Given:

$\mbox{A car is being pushed by three equal force with one on each side at 30 degrees and one in the rear.} $
$\mbox{The 1000 kg car is being accelerated at 0.5 m/s/s by the efforts of these three.} $

Question:

$\mbox{What force are each exerting on the car?} $

Rationale:

$F = F_1 = F_2 = F_3 $
$F_y = F \cos\alpha \; \mbox{(and there are two of these forces)} $

$F_{total} = F + (2 F \cos\alpha) $

$F_{car} = m a = F + (2 F \cos\alpha) $
$ F (1+(2 \sin\alpha) = m a $
$ F = \dfrac{( m a)}{(1+(2 \cos\alpha)} $

Calculate:

$F = \dfrac{(100 * .5)}{(1 + (2 * .866025)} = \dfrac{500}{2.73205} = 183.012756 \approx 183$

References:

Final 13 - Double Inclined Plane

Given:

$\mbox{Two masses connected by a rope and pully} $

$\mbox{mass 1 is on a plane inclined at an angle alpha} $
$\mbox{mass 2 is on a plane inclined at an angle beta} $

$a = g \frac{M_{unknown}\sin{\beta} {(operator)} M_{unknown}\sin{\alpha}}{M_1 ({operator}) M_2} $

Question:

$What are subscripts and operators? $

Rationale:

$M_1 \; \mbox{is associated with angle alpha} $
$M_2 \; \mbox{is associated with angle beta} $

$a = g \frac{M_2 \sin{\beta} {(operator)} M_1 \sin{\alpha}}{M_1 ({operator}) M_2} $

$\mbox{The denominator operator cannot be multiply or divide}$

$\mbox{Since if either mass is zero, then the equation would be indeterminate} $
$\mbox{So the operator must be + or - }$

$\mbox{But it cannot be minus, since it would indeterminate is the masses are equal}$
$\mbox{Therefore the denominator operator is plus.}$

$\mbox{If the masses are equal and the angles are equal, }$

$\mbox{Then the numerator must be zero}$
$\mbox{Thus the numerator operator must be minus.}$

Calculate:

References:

Final 12 - Water Clock

Given:

$\mbox{In one time interval, a ball rolls down the incline 0.8 meters.} $

Question:

$\mbox{What interval does the ball travel in the second and third time intervals?} $

Rationale & Calculate:

$\Delta{x} \varpropto t^2 $

$\Delta{x-1} = 1 * K $
$\Delta{x-2} = 4 * K$
$\Delta{x-3} = 9 * K $

$ K = 0.8 \; \mbox{since at t-1, x-1 is 0.8} $

$\Delta{x-2} = 3.2 $
$\Delta{x-3} = 7.2 $

$ Interval_{2-1} = 3.2 - 0.8 = 2.4 $
$ Interval_{3-2} = 7.2 - 3.2 = 4.0 $

References:

Unit 2A-8 of PH-100

Final 11 - Dead Reckoning

Given:

$\mbox{A ship travels:}$

$\mbox{East at 12 km/hour for 2 hours} $
$\mbox{Then south at 20 km/hour for 1 hour}$
$\mbox{Then east at 15 km/hour for 3 hours}$
$\mbox{Then northeast into port at 8 km/hour for 1 hour.} $

Question:

$\mbox{What is the straight line distance traveled to port ? }$

Rationale:

$\mbox{Segment-distance} = velocity * time $
$East_{distance} = \Delta{x} $
$South_{distance} = \Delta{y} $
$-\Delta{y_{northeast} = \Delta{x_{northeast}} = \sin{45^{\circ}}} * Northeast_{distance} $
$\mbox{Straight-line-distance} = \sqrt{(\Delta_{x-total})^2 + (\Delta_{y-total})^2 } $

Calculate:

$East_{distance} = ((12) (2.5)) + ((15) (3)) = 75 $
$South_{distance} = (20) (1) = 20 $
$ -\Delta{y_{northeast}} = \Delta{x_{northeast}} = (0.707106781) (8) (1) = 5.656854 $
$\Delta_{x} = 75 + 5.656854 = 80.656854 $
$\Delta_{y} = 20 - 5.656854 = 14.343146 $
$\mbox{Straight-line-distance} = \sqrt{(80.656854)^2 + (14.343146)^2} $

$\mbox{Straight-line-distance} = \sqrt{(6505.5280 + 205.7258 )} = \sqrt{6711.2538} $
$\mbox{Straight-line-distance} =81.9222 \approx 81.9 $

References:

Final 10 - Falling Water

Given:

$\mbox{Water is flowing down a wall of 1.5 meters and reaches a velocity of 4 meters/second at the bottom.} $

Question:

$\mbox{What is the work done by friction of a 1 gram water drop to retard its falling freely?} $

Rationale:

$V^2_{water-at-bottom} = 2 a_{retarded-from-normal-g} {height} $

$a_{retarded-from-normal-g} = \dfrac{V^2_{water-at-bottom}}{(2 height)} $

$a_{friction} = g - a_{retarded-from-normal-g} $
$W_{friction} = (m) (a_{friction}) {(height)} $

Calculate:

$a_{retarded-from-normal-g} = \dfrac{(4)^2}{(2 * 1.5)} = 5.33333 $
$a_{friction} = 10 - 5.33333 = 4.66667 $
$ W_{friction} = (1 * 10^{-3})(4.66667)(1.5) = 7 * 10^{-3} = .007 $

References:

Final 9 - Strange Planet

Given:

$\mbox{Length of pendulum is 12 meters and it took 45 seconds for it to swing back and forth.} $

Question:

$\mbox{What is the force of gravity on this planet?} $

Rationale:

$T = 2 \pi \sqrt{\dfrac{l}{g}} $

Calculate:

$\dfrac{45}{10} = (2) \dfrac{22}{7} \sqrt{\dfrac{12}{g}} $

$ (4.5) \dfrac{7}{44} = \sqrt{\dfrac{12}{g}} $
$ 0.71590 = \sqrt{\dfrac{12}{g}} $
$ 0.51251281 = \dfrac{12}{g} $
$ g = \dfrac{12}{ 0.51251281} = 23.4140489 \approx 23.4 $

References:

Final 8 - Cable Car Power

Given:

$ \mbox{A 2,000 kg cable car on an alpha 10 degree slope.} $

Question:

$\mbox{How much power does it take to move the trolley at 25 km/hour?} $

Rationale:

$25 \dfrac{km}{hour} = 25 \dfrac{1000 \mbox{meters-per-km}}{3600 \mbox{seconds-per-hour}} $

$25 \dfrac{km}{hour} =6.944444 \; \mbox{meters per second} $

$ P = F_{parallel} V = m g \sin\alpha V$

Calculate:

$P = (2000) (10) (.173648178) (6.9444) = 24,117.648146 $

$ P = \approx 24,117 $

References:

Final 7 - Positive Latitude

Given:

$\mbox{A 1 m rod casts a 1 m shadow at noon on the summer solstice.} $
$ \mbox{At the equator, the sun's rays strike the earth vertically at 23.5 degrees north.}$

Question:

$\mbox{What is the latitude at the measurement point?}$

Rationale:

$\mbox{The triangle with 1 m on both sides is a 45 degree triangle.}$
$\mbox{90 - The angle at the measurement point} + \mbox{The angle at the equator} = latitude$

Calculate:

$ latitude = 45 + 23.5 = 68.5 $

References:

Calculating Latitude

Final 6 - Balloon Balance

Given:

$\mbox{In an electrical field of 10,000 N/C, a mass of 30 grams is charged.} $

Question:

$\mbox{How large must that charge be to suspend the mass neither moving up nor down?} $

Rationale:

$F_{gravity} = F{electrical} $

$m a = E q $
$q = \dfrac{m a}{E} $

Calculate:

$q = \dfrac{(30 * 10^{-3}) (10)}{10^4} = 3 * 10^{-5} $

References:

Final 5 - Trolley

Given:

$\mbox{A 1,000 kg trollley on a slope of 10 degrees.} $

Question:

$\mbox{What force is needed to cause the trolley remain stationary ?} $

Rationale:

$ F_{parallel} = m g \sin\alpha $

Calculate:

$ F_{parallel} = (1000) (10) \sin{10_{circ}} = (1000) (10) (.173648) $
$ F_{parallel} = 1,736.48 \; \approx \; 1,736.5 $

References:

Final 4 - Climbing Stairs

Given:

$\mbox{A 65 kg person climbs some 6 meter stairs.}$

Question:

$\mbox{How much work is done?} $

Rationale:

$ W = m g h $

Calculate:

$ W = (65) (10) (6) = 3,900 \; \mbox{joules} $

References:

Final 3 - Reducing the Period

Given:

$\mbox{Mass m attached to a spring with spring constant K.} $

Question:

$\mbox{What size mass should you attach to the spring to double the period of oscillation?} $

Rationale:

$ T_{period} = 2 \pi \sqrt{\dfrac{m}{k}}$

$ 2 T_{period} = 2 \pi 2 \sqrt{\dfrac{m}{k}}$
$ 2 T_{period} = 2 \pi \sqrt{\dfrac{4 m}{k}}$

Calculate:

$ \mbox{Answer is it takes 4m to double the period.} $

References:

Final 2 - Height of the Church

Given:

$\mbox{Observer #1 is at ground level} $

$\mbox{Sees the top of head of Observer #2 and top of church in line.} $

$\mbox{Observer #2} = 1 \; \mbox{meter tall} $

$\mbox{Meters from Observer #1} = 2 $
$ \mbox{Meters from church} = 90 $

Question:

$\mbox{How tall is the church?} $

Rationale:

$\mbox{We have two similar triangles one within the other.} $

$ \mbox{A small one: 1 meters vertical by 2 meters horizontal describing an angle} = \theta $
$ \mbox{A large one: the height of church vertical by 92 meters horizontal describing an angle} = \theta $

$ \dfrac{vertical_{small}}{horizontal_{small}} = \tan\theta $
$ \dfrac{vertical_{church}}{horizontal_{church}} = \tan\theta $

Calculate:

$\tan\theta = \dfrac{1}{2} \; \mbox{using the small triangle} $
$ \dfrac{vertical_{church}}{horizontal_{church}} = \tan\theta $

$\dfrac{vertical_{church}}{92} = \tan\theta = 0.5 $
$ vertical_{church} = 0.5 * 92 = 46 \; \mbox{meters} $

References:

Similar Triangles

Rosa's Summary of Class Formulae

for uniform motion (a=0):

v=v0 and
Δs=v0×Δt

for uniformly accelerated motion (

a=constant≠0):

v=v0+aΔt,
Δs=v0×Δt+12a×Δt2 and
v2=v20+2×a×Δs

for the gravitational force (weight - always in the vertical down direction):

P=m×g

for the elastic force (always opposite elastic displacement):

Felas=−kΔx, k is the elastic constant

for the electric force (attractive for opposite sign charges and repulsive for equal sign charges):

Fe=keq1q2r2, ke is the electric constant

for the second law of Newton, in the x direction (for example):

Funbalanced,x=m×ax

for work done by a constant force (careful,

Fparallel may be negative and its the force component in the displacement direction):

W=Fparallel×displacement

for average power:

P=WΔt

for instantaneous power

P=F⃗ ⋅v⃗

for the conservation of energy:

Etotalinitial=Etotalfinal

for the kinetic, potential and total energy:

K=12mv2
Ugravitacional,approximate=mgh
Ugravitacional,exact=−Gm.Mr
Uelast=12kΔx2
Uelectric=keq1q2r1,2+keq1q3r1,3+keq2q3r2,3+...
Etotal=K+Ugravitacional+Uelast+Uelectric

for simple harmonic oscilation:

a=−ω2x
ω=2πT=ΔθΔt
pendulum: ω2=km
spring-mass: ω2=gl

for the gravitic field created by a mass M, acting on a mass m:

g=GMr2
weight=FG=m×g

for the elecric field E acting on a charge q:

Fe=q×E