c
=
3.00 x 108 m s‐1
e
=
1.60 x 10‐19 C
h
=
6.63 x 10‐34 J s
μo
=
4π x 10‐7 H m‐1
G
=
6.67 x 10‐11 N m2 kg‐2
u
=
1.66 x 10‐27 kg
ε0
=
8.85 x 10‐12 F m‐1
g
=
9.81 m s‐2
me
=
9.11 x 10‐31 kg
www.MrDaveSim.com Measurement and Uncertainty
Case 1 Y = 6 A + 3B ΔY = 6( ΔA) + 3( ΔB )
C a se 2 Y = 6 ( A )( B ) ΔY ΔA ΔB = + Y A B
Forces Static Equilibrium Translational Equilibrium Fx = 0, Fy = 0
Rotational Equilibrium
•
mAu A + mB uB = mAv A + mB vB Elastic head on collision
•
u A − u B = vB − v A
∑τ = 0
Upthrust weight of fluid displaced. Spring 1 Fs = kx , E = kx 2 2
Kinematics No s
v = u + at
No v
s = ut + ½ at2
No t
v2 = u2 + 2as
No a
s = (u+v)t/2
dp Δp = ∫ Fdt dt Δp F = Impulse t Change in p − p1 momentum F = 2 t Area under F‐t m v 2 − m v1 graph F = t Conservation of Momentum Draw diagram before an after collision. Apply PCM F =
•
• •
=
1.67 x 10‐27 kg
R
=
8.31 J K‐1 mol‐1
NA
=
6.02 x 1023 mol‐1
www.MrDaveSim.com Dynamics Newton 2nd Law
Random errors ‐ scatter of readings about mean Systematic error ‐ ALL above or ALL below the true value. Precision ‐ deviation of the reading from the mean value. Accuracy ‐ deviation of the reading from the actual true value.
mp
Elastic head on between 2 same masses exchange velocity Perfectly inelastic collision
Gravitation
Measurement and Uncertainty Newton Law of Gravitation Randomoferrors ‐ scatterbetween of readingstwo aboutgiven mean The force attraction Systematic error ‐ ALL above or ALL below point masses is directly proportional to thethe product of their masses and inversely true value. proportional to the ofsquare of from theirthe Precision ‐ deviation the reading distance apart. mean value. Gravitation Field Strength, g Accuracy ‐ deviation of the reading from the gravitational force per unit mass acting on true value. mass.actual Case C a se 2 1 Gravitation Potential Energy, U Y = 6 ( A )( B ) Y = 6 A + 3B ΔY ΔA ΔB External work done to move mass from ΔY = 6( ΔA) + 3( ΔB ) = + Y A B infinity to point concerned Gravitation Potential , Ø External work done per unit mass to move mass from infinity to point concerned
FG =
GMm r2
F GM g= = 2 m r
E t
v
P = Fv
KE1 + PE1 + External work done = KE 2 + PE 2
Fs = kx , E =
1 2 kx 2 GM
φ=
U =− m r
3rd
FG = −
dU dr
dφ g =− dr
Law
1 mω 2 ( x o2 − x 2 ) 2 1 TE = mω 2 x o2 2 1 PE = mω 2 x 2 2
v = ± ω xo2 − x 2
KE =
vo = xoω a = −ω 2 x
Forced Oscillation system forced to oscillate by an external period force (driver). Driven oscillator will oscillate at the driver frequency Reasonance driver frequency = natural frequency ‐ energy and amplitude of oscillating system becomes maximum. Wave Motion
v = fλ φ
=
2π
Δx
λ
I =
=
Δt T
I = k ( A m p litu d e )
Double Slit constructive
Δ N = nλ , n = 0,1, 2
Circular Motion dθ ω = dt 2π ω = T
v
2
Stationary Wave
a sin θ
s = rθ dθ v =r dt v = rω
P P = 4π r 2 A rea
Wave Interference
T 2 ∝ R3
Work Energy Power P =
GMm r
Kepler’s
mAu A + mB uB = ( mA + mB ) v
Work Done = F × distance moved
U =−
Simple Harmonic Motion acceleration is directly proportional to its displacement from equilibrium and and is always directed towards equilibrium
destructive 2
v = rω 2 ac = r
1⎞ ⎛ Δ = ⎜ n − ⎟ λ , n = 1, 2 N 2⎠ ⎝ a sin θ y =
Dλ a
Diffraction Grating
2 nodes/2 antinodes L =
nλ , n = 1, 2 2
1 node & 1 antinode L =
nλ , n = 1, 3 , 5 4
dN s in θ = n λ , n = 0 , 1 , 2 1/ N
c
=
3.00 x 108 m s‐1
e
=
1.60 x 10‐19 C
h
=
6.63 x 10‐34 J s
μo
=
4π x 10‐7 H m‐1
G
=
6.67 x 10‐11 N m2 kg‐2
u
=
1.66 x 10‐27 kg
ε0
=
8.85 x 10‐12 F m‐1
g
=
9.81 m s‐2
me
=
9.11 x 10‐31 kg
www.ThePhysicsCafe.com Thermal Physics Ideal Gases
pV = nRT
1 p = ρ c2 3
pV = NkT
3 KE = kT 2
Basics of Thermodynamics
Pt
V R
Direct Current
v
First Law of Thermodynamics
Δu = q + w Electric Field Electric Force and Electric Field Strength 1
Qq FE = ( ) 4πε 0 r 2
F 1 Q E= =( ) q 4πε 0 r 2
Potential Energy and Electric Potential 1 Qq 4πε 0 r
⎛ 1 ⎞Q V =⎜ ⎟ ⎝ 4πε 0 ⎠ r
Parallel Plate ‐ Uniform Field FE = −
V =
2
Q N = mL
W Q
dW dr
E = −
dV dr
Current of Electricity Ohm’s Law dQ ρl I = R = dt A
⎛ R ⎞ V1 = ⎜ 1 ⎟ E ⎝ R1 + R2 ⎠
1.67 x 10‐27 kg
R
=
8.31 J K‐1 mol‐1
NA
=
6.02 x 1023 mol‐1
⎛ R3 ⎞ I2 = ⎜ ⎟ I1 ⎝ R2 + R3 ⎠
Electromagnetism (EM) Magnetic Flux Density due to Current μ I μ NI μ NI B = o B = o B = o 2π r L 2r Magnetic Force on Conductor
F B = N B IL s in θ Magnetic Force on charged particles
Lenz’sLaw Direction of the induced current in a loop is such that its magnetic effect will always oppose the magnetic flux change that causes it Faraday’s Law Magnitude of the induced EMF will be proportional to the rate of change of magnetic flux linkage with the coil. coil dΦ E in d u c e d = − dt d (B A c o s θ ) E in d u c e d = − N dt
v
Alternating Current I = I 0 sin ω t
2 VRMS = IRMSVRMS R
IRMS =
Transformer
Io 2
N s V s Ip = = N p V p Is
eVs = hf − Φ
f ↑ , E ↑ , K E o ↑ ,V s ↑ ⎛ dNe ⎞ ⎛ dN p ⎞ In te n s it y ↑ , ⎜ ⎟ ↑ , ⎜ d t ⎟ ↑ , C u rr e n t ↑ d t ⎝ ⎠ ⎝ ⎠
Line Spectra Emission – Bright lines on dark background Absorption – p Dark lines on bright background g g
X‐ray production eV = h
c
λ
De‐Broglie wavelength h p
λ =
Uncertainty Principle
( Δ p x )( Δ x ) ≥
= = 5 .2 7 6 × 1 0 -3 5 2
E =
Root Mean Square RMS
2 Pac = IRMS R=
K E m ax = h f − Φ
c
λ
V = V0 sin ω t
RMS value of an AC (Irms ) is defined as the value of the steady direct current which would dissipate heat at the same rate in a given resistance as the alternating current.
Direction of magnetic force ‐ left hand Like Current Attract, Unlike Current Repel
E = hf = h
E in d u c e d = B L v
FB = B q v s in θ Magnetic Force between Parallel Current
Quantum Physics Photoelectric Effect
Electromagnetic Induction (EMI) Magnetic Flux
φ = B A cosθ
P = I R
Pt
W=
Internal Resistance
C − − +273.15 − − > K
Q N = mcΔT
=
www.ThePhysicscafe.com
2
P =
⎛ X − XL ⎞ Tθ = TL + ⎜ θ ⎟ (TH − TL ) ⎝ XH − XL ⎠ 0
Electrical Power en erg y P = tim e P = VI
mp
1 p2 2 m
Quantum Tunneling 8 π 2 m (U − E ) h2
T = e x p ( − 2 k d ), k =
Nuclear Physics
Energy released = (m x − my ) c 2 Energy released = (BE )Y − (BE ) X
Radioactive Decay dN A=− dt A = λN
⎛ N ⎜ ⎝ No
t
⎞ ⎡ 1 ⎤ t1/ 2 ⎟=⎢ ⎥ ⎠ ⎣2⎦
N = No e − λ t λ=
ln 2 t1/ 2
