Quiz #3

```Name___________________________________
Physics 120 Quiz #3, February 3, 2012
Please show all work, thoughts and/or reasoning in order to receive partial credit. The
quiz is worth 10 points total.
1. Consider a gold wire that is suspended from the ceiling with dimensions 1cm by 1cm
by 1m. A mass of 10kg is suspended from the wire stretching it by 12.6µm.
a. What is the interatomic spacing (dIAB) between the gold atoms in the wire?
(Hints: Assume that you have a 1cm3 volume to do the calculation, the density of
gold is 19300 kg/m3 and the molar mass of gold is 0.197 kg/mol.
€
Assuming a 1m3 volume the total mass of gold (ρAl = 19300 kg/m3) is 19300kg.
The number of atoms in this mass is given by
1mole
# atoms
atoms
, where the
× 6.02 ×10 23 = 5.9 ×10 28 volume
volume =M T ×m × N A = 19300kg ×
0.197kg
molar mass of gold is m = 197g/mol. Then the number of atoms on a side of
length L = 1m, is the cubed root of the number of atoms in the volume. Thus the
3
5.9 ×10 28 = 3.89 ×10 9 atoms
number of atoms on a side is N atoms
side =
side . Lastly, in the
side of length L = 1m, there are N(atoms/side) each spaced by dIAB. Thus the
length of an interatomic bond is
L
1m
L = N × dIAB → dIAB€= =
= 2.57 ×10−10 m.
9
N 3.89 ×10
b. What is the mass of a single atom of gold?
€
€
matom =
0.197kg
1mol
×
= 3.27 ×10−25 kg
23
1mol
6.02 ×10 atoms
c. What is Young’s modulus for gold?
Stress = Y × Strain → Y =
Y=
Stress
FL
mgL
=
=
Strain AwireΔL AwireΔL
10kg × 9.8 sm2 ×1m
(0.01m)
2
−6
×12.6 ×10 m
= 7.8 ×1010 mN2
d. What is the speed of sound in gold?
€
vs =
€
3
k IAB
YdIAB
dIAB =
=
matom
matom
7.8 ×1010 mN2 × (2.6 ×10−10 m)
−25
3.27 ×10
m
3
= 2048 ms
Useful formulas:


p = γmv
k eff , parallel = n parallel kindividual
1
γ=
keff ,series =
2
v
1− 2
c
 
vi + v f

v avg =
2


Fg = mg
kindividual
n series
stress = Ystrain →
vs =

GM1 M 2
Fgravity =
rˆ12
r122

 
Fspring = −ks; s = ( L − Lo ) sˆ
 
W = ∫ F ⋅ dr =ΔKE = −ΔU
F
ΔL
=Y
A
L
kIAB
d
matom
k IAB = Yd
GM1 M 2
r
U g = mgy
Ug = −
1
U s = ks2
2
1
KE = mv 2
2
KE = (γ −1) mc 2
Useful Constants
€
Momentum Principle:
Position-update:
€

 
p f = pi + Fnet Δt; Δt = large

 
Δt
p f = pi + Fnet dt; dt = = small
n
  

rf = ri + v avg Δt = ri +

p
p2
m 1+ 2 2
mc
  
Δt
rf = ri + v f dt; dt =
= small
n
g = 9.8 sm2
2
G = 6.67 "10#11 Nm
kg 2
Δt; Δt = large
%o = 8.85 "10#12 NmC 2
µo = 4\$ "10#7 TmA
c = 3"10 8 ms
h = 6.63"10#34 Js
€€
Energy principle:
Geometry /Algebra
Circles
Triangles
A = πr 2
A = 12 bh
0.511MeV
c2
937.1MeV
m p =1.67 "10#27 kg =
c2
948.3MeV
mn =1.69 "10#27 kg =
c2
931.5MeV
1amu =1.66 "10#27 kg =
c2
23
N A = 6.02 "10
me = 9.11"10#31 kg =
Spheres
A = 4 πr 2
V = 43 πr 3
Quadratic equation : ax 2 + bx + c = 0,
whose solutions are given by : x =
−b ± b 2 − 4ac
2a
Vectors
€
€
2
1eV =1.6 "10#19 J
ΔE = W = ΔU g + ΔU s + ΔKE
C = 2πr
1e =1.6 "10#19 C
1
C2
k=
= 9 "10 9 Nm
2
4\$%o

magnitude of a vector : a = ax2 + ay2 + az2


writing a vector : a = ax ,ay ,az = a aˆ = ax iˆ + ay ˆj + az kˆ
Ax 2 + Bx + C = 0 & x =
!
#B ± B 2 # 4AC
2A
```
Symmetry

23 Cards

Tensors

20 Cards

Yamanote Line

29 Cards