Final Exam Equation Sheet Physics 116

tαωω += 0

2

21 ttiif αωθθ +=−

tt ifaveif )(21

ωωωθθ +==−

)(220

2if θθαωω −+=

ωrv = αra =

Rolling condition

22

21

21

CMMvIK += ω

ωRvCM = αRaCM =

θsinBABAC

=×=

Fr

×=τ prL

×=

dtLd

ext

=∑τ

2ii rmI ∑=

Idisk=(1/2)mR2 Ihoop=mR2

Krot= ½ Iω2

ωIL =

ατ Iext =∑

ffii II ωω =

0=∑ extF

(static equilibrium) 0=∑ extτ

)32(95 −= FC TT

KTT C 15.273+=

kTKaverage 23

= Ideal gas

nRTNkTU23

23

int == Monatomic ideal gas

NkTnRTPV == KmolJR ⋅= /314.8

molemoleculesxNA /1002.6 23= KJxk /1038.1 23−=

MRT

mkTvrms

33==

)()( ifif TTncTTmcQ −=−= mLQ =

RCV 23

=

RCP 25

=

!

"U int =Q #Wdone by gas )( ifisobaric VVPW −=

)ln(i

fisothermal V

VnRTW =

ch QQW −=

h

c

h QQ

QWe −== 1

h

c

h

c

TT

QQ

= for a ideal cycle

h

cideal T

Te −=1

TdQdS r=

∫=Δf

ir

TdQS

)ln(i

fV T

TncS =Δ Isovolumetric Process

TmLS =Δ Phase change

)ln(i

fP T

TncS =Δ Isobaric Process

VM

=ρ

AFP =

1 atm = 1.01x105 Pa == vAIv constant

=++ 2

21 vghP ρρ constant

submergedfluid gVB ρ=

3310mkg

water =ρ

32.1mkg

air =ρ ( ) ( )φω += tAtx cos

ωπ2

=T

Tf 1=

mk

=ω (spring)

Lg

=ω (pendulum)

( ) ( )φωω +−= tAtv sin ( ) ( )φωω +−= tAta cos2

2

21 kAEtotal =

221 kxUspring =

Fspring=-kx ( ) ( )vtxftxy ±=,

µTv = (taut string)

( ) ( ) ( )tkxAvtxAtxy ωλπ

±=⎥⎦

⎤⎢⎣

⎡ ±= sin2sin,

kfv ωλ ==

λπ2

=k

Tπ

ω2

=

( ) ( )tkxstxs ω±= cos, max

Threshold of hearing: 2

1210mWIo

−=

⎟⎟⎠

⎞⎜⎜⎝

⎛=

0

log10II

β

24 rPI av

π=

fvvf L ⎟⎠

⎞⎜⎝

⎛+=ʹ′ 1

Listener moving toward source.

fvvf L ⎟⎠

⎞⎜⎝

⎛−=ʹ′ 1

Listener moving away from source.

f

vvfS ⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

+=ʹ′1

1

Source moving away from listener.

f

vvfS ⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

−=ʹ′1

1

Source moving toward listener.

fvvvvfS

L⎟⎟⎠

⎞⎜⎜⎝

⎛ ±=ʹ′ ffbeat Δ=

)cos()sin(),( 0 txkstxs nnnn ω=

If both ends are open or closed,

2nnL λ

= , ,...3,2,1=n

1nffn = , ,...3,2,1=n If one end is open and one is closed,

4nnL λ

= , ,...5,3,1=n

1nffn = , ,...5,3,1=n