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Andrzej Czajkowski PHY1321/1331 exam Page 7 out of 10 Mechanics 



dx vx = dt

dr v = dt

dv ax = x dt





dv a= dt    1 2 r f = ro + vo t + a t 2

at = 

dv dt

ac =



v2 r



F = ma



F o = b v 1 R = DAv 2 2 FB = lV .g

f = N





F = k x





W =  F .d s

k= 

mv 2 2

U g = mgh





rCM

1 2 kx 2



dp F = dt

P = mv 

Ue =

m =

 i

ri

M

4 V = r 3 3

rCM =

A = 4r 2

 rdm M

A = r 2

C = 2r

Fluid Mechanics:

p = po + gh

A1v1 = A2v2

1 po + gy + v 2 = const 2

Andrzej Czajkowski PHY1321/1331 exam Page 8 out of 10 Rotational motion About a Fixed Axis Angular speed  = d

dt Angular acceleration  = dw

Net torque

 = I

dt

 f =  i + t  1  If = const .  f =  i +  i t +  t 2 2  2 2  f =  i + 2 ( f   i) f

Work W =   d i

Rotational kinetic energy K R =

1 2 I 2

Power P =   Angular momentum L = I Net torque  = dL dt

I CM = MR2 1 Hollow cylinder I CM = M ( R12 + R22 ) 2 where R1: inner radius, R2: outer radius 1 Solid cylinder or disc I CM = MR2 2 1 Thin Rectangle I CM = M ( a 2 + b 2 ) 12 1 Long thin rod with rotational axis through center I CM = ML2 12

Circular Hoop

Long thin rod with rotational axis through edge 2 MR2 5 2 2 = MR 3

Solid sphere

I CM =

Thin spherical shell

I CM

I CM =

1 2 ML 3

Andrzej Czajkowski PHY1321/1331 exam Page 9 out of 10 THERMODYNAMICS Probability of finding the speed of a particle in the range (v;v+dv )is: 3

2

mv  1 m  2 2  2kT P (v )dv = 4   v e dv  2 kT  1

v MP

1

 2kT  2 =   m 

vrms

1

 3kT  2 =   m 

vavg

1 p =  v 2  3

 8kT  2 =   m 

Nm V

=

Integrals: +

 e ax dx = 2

0

1  2 a

+

ax xe dx = 2

0 +

x

4  ax 2

e

dx =

0

Eint = Q + W

+

1 2a

2  ax  x e dx = 2

0 +

3  8 a5

e

x x

0

3

+

1  4 a3



x e

3 ax

2

1 2a 2

dx =

0

4

dx = 15 1 dQ T Q

S

S = 

pV=nRT

Change

Eint

W

P = const

nCv ΔT

-p(Vf-Vi)

nCp ΔT

nC p ln

V = const

nCv ΔT

0

nCv ΔT

nCV ln

T = const

0

Q=0

nCv ΔT

pV  = const.

CRN =

 =

 nRT ln

Vi

nRT ln

1 ( p V  piVi )  1 f f

CP CV

Vf Vi

0

nRln

Tf Ti

Vf Vi

0

what we want what we pay for it S = ST COP =

V = VT dT dx

P = e  A T4;  =5.67x 10-8W/(K4m2)

P = kA

Q = mcT Q = Lm c(water) = 4186 J/(kg C);

c(steam) = 2010J/(kg C)

L(melting) = 3.33x105 J/kg

Ti

C p  Cv = R

T W QH  Q L = =1  C Q QH TH

L = αLT

Vf

Tf

c(ice) = 2090 J/(kg C);

L (vaporization) = 2.26x106 J/kg...