Semester One Examinations, 2022
PHYS1001
PHYS1001: Useful formulae and numerical values
Mathematics
Quadratic eqn:
A< # + C< + D = 0 Solution: < = −C ± √% ! &'() #( Trigonometry DEF # - + FG># – = 1
sin(J ± K) = sin J cos K ± cos J sin K
cos(J ± K) = cos J cos K ∓ sin J sin K
J±K
J∓K
sin J ± sin K = 2 sin M
P
N cos O
2
2
J+K
J−K
cos J + cos K = 2 cos M
N cos M
N
2
2
J+K
J−K
cos J − cos K = 2 sin M
N sin M
N
2
2
Binomial expansion
(1 + < + >(> − 1) #
< +⋯ 2! Calculus d n x = n x n -1 dx x n +1 n x dx = ò n +1 1 ò x dx = ln ( x ) d sin ( ax ) = a cos ( ax ) dx d cos ( ax ) = - a sin ( ax ) dx Vectors U⃗= JK cos V J⃗∙K U⃗ = JK sin V >U⃗
J⃗×K
U⃗.
where >U⃗ is a unit vector normal to both J⃗ and K
Page 9 of 14
Semester One Examinations, 2022
PHYS1001
Vectors – continued
If X⃗, Y⃗, and ZU⃗ are mutually perpendicular unit vectors then
U⃗ =Z
U⃗ ∙X⃗= 0 and X⃗×X⃗=Y⃗×Y⃗=Z
U⃗ ×ZU⃗= 0 and X⃗×Y⃗=Z
U⃗ , Y⃗×Z
U⃗ =X⃗, Z
U⃗ ×X⃗=Y⃗.
X⃗∙X⃗=Y⃗∙Y⃗=ZU⃗∙ZU⃗= 1 and X⃗∙Y⃗=Y⃗∙Z
X⃗
⃗
U⃗
J×K= [J+
K+
Kinematics and Dynamics
vs = ds / dt
as = dvs / dt
w = dq / dt
a = dw / dt
tf
tf
s f = si + ò vs dt
ti
tf
Y⃗
J,
K,
s
r
vt = ω r
θ=
q f = qi + ò w dt
ti
tf
v f = vi + ò as dt
w f = wi + ò a s dt
v fs = vis + as Dt
w f = wi + a Dt
1
s f = si + vis Dt + as (Dt ) 2
2
2
2
v fs = vis + 2as Ds
1
q f = qi + wi Dt + a (Dt ) 2
2
2
2
w f = wi + 2a Dq
ti
ZU⃗
J- [
K-
at = α r
ti
vt2
ar =
r
2Ï€ r 2Ï€
T=
=
v
ω
Impulse, Momentum, Energy, Work


Pf = Pi


F = d p / dt
tf
ΔEsys = ΔK + ΔU + ΔEth = Wext
K f +U f + ΔEth = K i +U i +Wext
J x = ∫ Fx (t)dt
ΔK = Wnet = Wc +Wdiss +Wext
Δpx = J x
W = ∫ Fs ds
K f +U f = K i +U i
ΔEth = f k Δs
 
W = F • Δr
ti
1
K = mv 2
2
U g = mgy
1
U s = k(Δs) 2
2
Fs = −kΔs
sf
si
Fs = −dU / ds
P=
dEsys
dt 
P = F •v
Page 10 of 14

 Fnet
a=
m
FG = mg
f s,max = µ s n
f k = µk n
f r = µr n
D≈
1 2
Av
4
Semester One Examinations, 2022
PHYS1001
Rigid Body Rotation
α=


L = Iω

dL
= Ï„ net
dt
1
xcm =
∫ x dm
M
1
ycm =
∫ y dm
M
I = ∑ mi ri 2 = ∫ r 2 dm
Ï„ net
I
1
1
2
E = K rot + K cm +U g = Iω 2 + Mvcm
+ Mgycm
2
2
! ! !
τ =r×F
τ = rF sin θ = rFt = dF
vcm = Rω
I = I cm + Md 2
i
Moments of Inertia
1
ML2
12
1
I = ML2
3
1
I = Ma 2
12
1
I = Ma 2
3
I=
Thin rod, about centre.
Thin rod, about end.
Plane or slab, about centre.
Plane or slab, about edge.
1
MR 2
2
I = MR 2
2
I = MR 2
5
2
I = MR 2
3
I=
Cylinder or disk, about centre.
Cylindrical hoop, about centre.
Solid sphere, about diameter.
Spherical shell, about diameter.
Oscillations
( Fnet ) s = – ks
w=
k
m
T = 2p
m
k
x(t ) = A cos(wt + f0 )
æ mg ö
( Fnet )t = – ç
÷s
è L ø
w=
g
L
T
= 2p
vx (t ) = -vmax sin(wt + f0 )
L
g
1 2 1 2 1
1
mvx + kx = m(vmax ) 2 = kA2
2
2
2
2
– t /t
E = E0 e
E=
f = 1/ T
w = 2p f = 2p / T
ax = -w 2 x
vmax = w A
x(t ) = Ae – bt / 2 m cos(wt + f0 )
t = m/b
Page 11 of 14
Semester One Examinations, 2022
PHYS1001
Fluids and Elasticity
r = m /V
p=F/A
p = p0 + r gh
v1 A1 = v2 A2
1
1
p1 + r v12 + r gy1 = p2 + r v22 + r gy2
2
2
Thermodynamics
ΔEth = W + Q
pV = nRT
Vf
pV = Nk BT
W =−∫
p2V2 p1V1
=
T2
T1
γ = C P / CV
M
m
N M (in grams)
n=
=
NA
M mol
Q = ±ML
Q = Mc ΔT
Q = nC ΔT
Q / Δt = (kA / L) ΔT
Vi
p dV
C P = CV + R
N=
Number density = N /V
Q / Δt = eσ AT 4
9
TF = TC + 32o
5
TK = TC + 273
pV γ = const
TV γ −1 = const
1N
2N
2
p=
mvrms
=
ε
3V
3 V avg
3
εavg = k BT
2
3
3
Eth = Nk BT = nRT (Monatomic gas)
2
2
5
5
Eth = Nk BT = nRT
(Diatomic gas)
2
2
Eth = 3Nk BT = 3nRT (Elemental solid)
vrms = (v 2 )avg
Tpγ /(γ −1) = const
W = −nRT ln(V f /Vi )
η=
Wout
QH
T
η ≤ 1− C
TH
K=
QC
Win
K≤
TC
TH − TC
Ws = ∫ p dV
Page 12 of 14
F
DL
=Y
A
L
DV
p = -B
V
Semester One Examinations, 2022
PHYS1001
Numerical Values
= 9.80 m s&#
^ = 6.672 × 10&.. m/ kg &. s&# or N m# kg &#
Earth Mass = 6.0 × 10#’ kg
Earth Radius = 6.4 × 100 m
Solar Mass = 2.0 × 10/1 kg
Solar Radius = 7.0 × 102 m
Earth − Sun mean distance = 1.5 × 10.. m
one (metric) tonne = 1.0 × 10/ kg
Z! = 1.3801 × 10&#/ J K &.
p = 5.67 × 10&2 W m&# K &’
c = 4190 J kg &. K &. (for water)
r3 = 3.33 × 104 J kg &. (for water)
r5 = 22.6 × 104 J kg &. (for water)
r5 = 2.00 × 104 J kg &. (for carbon dioxide, dry-ice sublimation)
s = 1.6606 × 10&#6 kg
t7 = 6.022 × 10#/ particles mol&.
v89: (H) = 0.001 kg mol&.
x = 8.314 J mol&. K &.
?” =
/
#
x (monatomic gas)
4
?” = # x (diatomic gas)
y = 1.67 (monatomic gas)
y = 1.40 (diatomic gas)
1 atm = 101.3 kPa = 1.013 × 104 Pa
Uncertainty Analysis
Summary of rules for combining uncertainties for dependent measurements
If / = < + { + | + ⋯ or / = < − { − | − ⋯ then Δ/ = Δ< + Δ{ + Δz + ⋯. If / = < × { × | × … or / =
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