PHY231 Review - Final
Skip problem 84
Problem 62: was done, but don’t worry about it.
Problem 61, 63-67,70,71-73: see lon-capa
74,75 see 49,50
A few questions concerning midterm 2 have been added
ΔL=αLoΔT
The right strip has a higher coefficient of expansion, so will increase its
length more than the left strip when heating. Hence the strip must bend
left.
Use pV=nRT (ideal gas law)
n=pV/(RT)
p=3.30 atm = 3.30 x 1.015x105 Pa = 334950 Pa
V=8.6 L = 8.6 x 1x10-3 m3=8.6x10-3 m3
T=23.0 oC = 23.0+273.15=296.15 K
R = 8.31 J/molK
So n= 334950 x 8.6x10-3 / (8.31 x 296.15) = 1.17 mol
Use P=kA(Th-Tc)/Δx=kAΔT/Δx
For left side:
k: kaluminum
A: 2a with “a” area of single bar
Th-Tc=600 K
Δx: Length (L) of single bar
For right side:
k: kaluminum
A: a with “a” area of single bar
Th-Tc=600 K
Δx: 2 x Length (L) of single bar
Pleft= k x 2a x 600 / L = 1200 ka/L
Problem 50:
Pright= k x a x 600 / (2L) = 300 ka/L
Similar.
So Pright is 300/1200 = 1/4 times that of Pleft
Make sure to read carefully; don’t turn question around!!!
Vrms=√ (3kBT/m) = √ (3RT/M)
kB: Boltzmann constant =1.38x10-23 J/K
T: temperature (Kelvin)
m: mass (in kg) of 1 molecule
M: molar mass
R: 8.31 J/molK
Here: T=46+273.15 = 319.15
m=7.3x10-26 kg
So v=425 m/s
cwater=1.0 kcal/kgoC
(a) Q=2.2 x 0.5 x 11 = 12.1 kcal
(b) Q=2.2 x 79.7 = 174.9 kcal
(c) Q=2.2 x 1 x 71.5 = 157.3
Total
344.3 kcal
P=σAeT4 : Stefan’s law (J/s)
σ=5.6696x10-8 W/m2K4
A: surface area
e: object dependent constant emissivity (0-1)
T: temperature (K)
P: energy radiated per second.
Px = σAe(6300)4
Py = σAe(12300)4 Note that σ,A,e are the same for both.
So: Py/Px= (12300)4 / (6300)4 = 14.53
54. Change in momentum: Δp=mΔv=FΔt (mass constant)
Δv=FΔt /m=1.47x106 x 18.6/(7.76 x 106) =3.54 m/s (slowing down)
Initial velocity: 81.1 km/hr = 81.1 x 1000/3600=22.7 m/s
Final velocity: 22.7-3.54 = 19.1 m/s
55. Acceleration a= Δv/Δt = -3.54/18.6 = -0.19 m/s2
Distance traveled: Δx=v0t+0.5at2=22.7x18.6+0.5x(-0.19)x 18.62=389 m
Parabolic motion
horizontal
x(t)=x0+v0xt
vx(t)=v0x
vertical
y(t)=y0+v0yt-0.5gt2
Vy(t)=v0y-gt g=9.81 m/s2
V0x=vocos(initial angle)
V0y=v0sin(initial angle)
56. x(t)=x0+v0xt = 0 + 338cos(38.1) x 5.57 = 1481.5 m
57. y(t)=y0+v0yt-0.5gt2=0+338sin(38.1)x5.57 –0.5x9.8x5.572=
= 1162 – 152 = 1010 m
Total Distance traveled in horizontal direction?
Y(t)=0 so 338sin(38.1)t –0.5x9.8t2 =0 209t-4.9t2=0 t(209-4.9t)=0
So, t=0 (start) or t=42.65 so x(42.65)=338cos(38.1)x42.65=11344 m
Height at highest point?
Vy(t)=0 at highest point, so 0=338sin(38.1)-9.8t=0 t=21.3 s
Y(21.3)=338sin(38.1)x21.3 – 0.5x9.8x(21.3)2=4442-2223=2219 m
Perfect inelastic collision: objects stick together after collision and
only momentum is conserved. m1v1+m2v2=(m1+m2)vf
Here:
m1=3.28 kg m2=???
v1=????? v2=0 vf=v1/3
So: 3.28v1 = (3.28+m2)v1/3 v1 drops out
m2 = 3(3.28-3.28/3) =6.56 kg
W=FΔx: work performed against the drag force when swimming Δx.
P=W/Δt: Power needed (work per second carried out)
W=58x137=7964
P=7964/(2.15 x 60)=61.6 J/s (W)
To have the same weight as on earth, the centripetal acceleration
must match the earth’s gravitational acceleration (9.81 m/s2)
For objects rotating at constant angular velocity (linear/tangential
speed):
F=mac F: force can be tension, friction, normal force, gravity…..
ac=centripetal acceleration=v2/r = ω2r (used ω=v/r)
Here: F is provided by the normal force of the hull of the space
station, but must match the gravitational force on earth: F=n=mg
mg = mω2r ω=√(g/r)= √(9.8/25.5)=0.62 rad/s
(don’t forget conversion from diameter to radius)
A 0.400-kg object is swung in a circular path
and in a vertical plane on a 0.500-m-length
string. If the angular speed at the bottom is 8.00
rad/s, what is the tension in the string when the
object is at the bottom of the circle?
For objects rotating at constant angular
velocity (linear/tangential speed):
F=mac F: force can be tension, friction,
normal force, gravity…..
ac=centripetal acceleration=v2/r = ω2r
T
(used ω=v/r)
Fgrav
At the bottom: F=T-Fgrav=mac
So: T-mg=mac and T-mg=mω2r
And: T=m(g+ ω2r)=0.4(9.8+82x0.5)
T=16.7 N
Note: if mass at the top: T+mg=mω2r
If the mass of Mars is 0.107 times that of Earth, and its radius is
0.530 that of Earth, estimate the gravitational acceleration g at the
surface of Mars. (gearth = 9.80 m/s2)
F=mg and F=Gmmplanet/rplanet2
So g=Gmplanet/rplanet2
gearth=Gmearth/rearth2
gmars=Gmmars/rmars2=G x (0.107 mearth)/(0.530rearth)2
gmars/gearth=0.107/0.5302 = 0.38
gmars=0.38gearth=0.38x9.8=3.733 m/s2
Pascal’s principle: pressure in an
enclosed volume is the same
everywhere
P=F1/A1=F2/A2
68) Force on piston 2: mg=818x9.81 = 8024 N
A1=πr12=π(0.0192/2)2 =2.9x10-4 m2 A2=πr12=π(0.094/2)2=6.94x10-3 m2
So F1=F2A1/A2=335 N
69) To make a torque τ=F1d at a distance d=2” from the rotation point a
torque τ=Fhanddhand is required by the hand (d=12”)
So F1d=Fhanddhand Fhand=F1xd/dhand=55.8 N
A uniform, horizontal beam of length 6.0 m and weight 120 N is attached
at one end to a wall by a pin connection (so that it may rotate). A cable
attached to the wall above the pin supports the opposite end. The cable
makes an angle of 60° with the horizontal. What is the tension in the
cable needed to maintain the beam in equilibrium?
Equilibrium: sum of torques must be zero
Fvert
60o
Fgrav
τbeam+τcable=0
τbeam=Fgravdcenter-of-gravity=120 x 3 =360 N
τcable= -(Fvertd)=-(Tsin(60o) x 6)
6m
Solve for T
360 = Tsin(60o) x 6
T=60/sin(60o) =69.3 N
A solid cylinder (I = MR2/2) has a string wrapped around it many
times. When I release the cylinder, holding on to the string, the cylinder
falls and spins as the string unwinds. What is the downward acceleration
of the cylinder as it falls?
a. 0
T
VERY HARD!!!!!!!!
2
b. 4.9 m/s
c. 6.5 m/s2
d. 9.8 m/s2
Newton’s law: ΣF=ma so T-mg = -ma
(a will be a positive number, the – sign indicates
It goes down).
Torque: τ=Iα=Ia/r=mr2a/(2r)
Fgrav
Also
> combine: T=ma/2
Torque: τ=F x d = T x r
So: T-mg = -ma
And: ma/2-mg = -ma
So mg=1.5ma and a=g/1.5=9.8/1.5=6.5 m/s2
For floating: gravitation force = buoyancy force
So: weight (mg) = weight of displaced water (mwaterg)
ρwoodVwoodg = ρwaterVdisplacedg
(used: ρ=M/V so M= ρV
So
Vdisplaced/Vwood= ρwood/ρwater=770/1040=0.74
This is the fraction of the tree under water.
77. Efficiency of an engine (any) = 1-Qcold/Qhot=1-75.9/93.8=0.191
Note that for a Carnot engine only : efficiency = 1-Tcold/Thot
78. Nothing is known about the volume, pressure or temperature
(internal energy), so here one must use:
Efficiency = (Qhot-Qcold)/Qhot = W/Qhot so 0.191 = W/93.8 W=17.9 kJ
Easier is: W= Qhot-Qcold=93.8-75.9=17.9
79. What is the amplitude
80. What is the period
81. What is the angular frequency ω
82 what is the frequency (f)
79. Amplitude: distance between equilibrium position (x=0) and
maximum displacement. Read from graph: 1.10 m
80. Period: time that it takes to make one full oscillation. Read from
graph T=4.6
81. ω=2π/T = 2π/4.6 = 1.37 rad/s
82. f=1/T = 1/4.6 = 0.217 Hz
ω=√(k/m) = √(17.9/0.664) = 5.19 rad/s
ω=2πf so f=ω/2π and f=0.826 Hz
Sound level β=10log(I/I0) I0=10-12 W/m2
= 10log(1x10-6/1x10-12) = 60 dB
New Sound level = 10log(26x10-6/1x10-12) = 74.15 dB
Or use log(ab)
=log(a)+log(b)
New sound level = 10log(26) + 60 = 14.14+60=74.15
⎛ v + vobserver
f ′ = f ⎜⎜
⎝ v − vsource
⎞
⎟⎟
⎠
vobserver: positive if moving
towards to source
vsource: positive if moving
towards the observer
Here: f=250 Hz f’=275 Hz
Vobserver=0 vsource=??? V=340 m/s
Rewrite doppler equation: vsource=(f/f’)v x ( f’/f –1)
= (250/275) x 340 x (275/250-1)=30.9 m/s (towards observer)