Professor's note: All is VERY good, to save your time and serve the purpose of this exercise, let's avoid the numbers. Give me the steps to solve a given problem in words - explain as much as you want ! Keep up the good !!
Part 5:
x(t) = Acos(wt) <--- solve for t
0.06m = 0.12m cos(wt) <---- divide each side by 0.06m
0.5m = cos(wt)
cos^-1(0.5m) = wt
pi/3 = wt
(pi/3)/w = t
pi/3w = t
pi/[(3)(9.78rad/s)] = t
t1 = 0.107s
x(t) = Acos(wt) <---- x = 0, solve for t2
0 = A cos (wt) <--- divide each side by A
0 = cos(wt)
cos^-1(0) = wt
(cos^-1(0))/w = t
(cos^-1(0))/(9.78rad/s) = t
t2 = 0.161s
t2-t1 <---- time it takes block to go from x =0 to x=0.06m
0.161s - 0.107s = 0.054s
Problem 2: Given: T = 150N u = 7.2g/m ---> 0.0072kg/m x = 90 cm ---> 0.90m
Part 1:
V= sqrt(F/u)
V = sqrt (150N/(0.0072kg/m)) V = 144.34m/s
Part 2:
Lambda --> (L)
3(L/2) = x ---> find L (wavelength)
3(L/2) = 0.9m
0.9m/3 = (L/2)
2(0.9m/3) = L L = 0.6m
Part3:
V = fL ---> find f (frequency)
V/L = f
[(144.34m/s)/(0.6m)] = f f = 240.57Hz
Period (T) = 2pi sqrt(L/g) --> find L T/2pi = sqrt(L/g) --> square each side (t/2pi)^2 = L/g --> multiply by g g(T^2/4pi^2) = L L =(9.8m/s^2)((2s^2)/(4pi^2) L = 0.993m
Linear Density of String u = m(string)/L u = 0.0006kg/0.993m u = 0.000604kg/m
Tension of string (F) F = m(ball)g F = (5kg)(9.8m/s^2) F = 49N
Velocity of wave V = sqrt(F/u) V = sqrt((49N)/(0.000604kg/m)) V = 284.77m/s
Week 3
Use the equation: fo = fs[(1- (vo/v))/(1-(vs/v))] --> use (-) for numerator to show that observer is moving away from source; use (-) in denominator since the source is moving forward, but the frequency must increase and therefore the denominator should be decreased.
Plug in 440Hz for the frequency of the source (fs)
Plug in 54m/s for the velocity of the source (vs)
Plug in 80m/s for the velocity of the observer (vo)
Plug in 340m/s for speed of sound (v)
Week 4
1. Given:
f = 420Hz
V = 340m/s
L = ?
a) Find length of the tube
Use f = nV/2L --> (tube open at both ends)
--> frequency is given (420Hz)
--> v is given (340m/s)
--> n =1 because it is the fundamental frequency (1st harmonic)
--> Rearrange to find L
L = nV/2f
b) What is the second harmonic?
(Find frequency)
f = nv/2L
--> v is given (340m/s)
-->use L from part a
--> n = 2 because it is the 2nd harmonic
c) What is the fundamental frequency of this pipe when the speed of sound in air is increased to 367m/s?
Find f
Use f = nV/2L
--> v is given (now 367m/s)
--> n = 1 (fundamental frequency)
--> use length of tube from part a
2. How long are the pipes?
Given (Pipe A):
Open at both ends
fA = 60Hz
Fundamental Frequncy (n = 1)
Given (Pipe B):
Open at one end
fB = fA at 2nd Harmonic
First [[#|Step]]:
Use fA = nV/2L --> Rearrange to Find L
L = nV/2f
Second [[#|Step]]:
Find frequency of Pipe A at 2nd harmonic
fA = nV/2L
--> n = 2 (2nd harmonic)
--> V = 340m/s
--> use L from first step
Third [[#|Step]]:
Frequency of 2nd Harmonic of Pipe A = Frequency of Third Harmonic of Pipe B
fB = nV/4L ----> rearrange to find L
L = nV/4fB
--> n = 3 (third harmonic)
--> V = 340m/s
-->fB = answer from Step 2
Week 6
1. Given: 226Ra88 --> 222Rn86 + 4He2 Ra = 226.02540u Rn = 222.01757u He = 4.002603u mass defect = mass of constituents - mass of nucleus mass defect = (mass of Rn + mass of He) - mass of Ra
*1u = 931.5MeV*
Take mass defect and multiply by 931.5MeV to find the energy in MeV
2. Given: N = 8.14x10^14 No = 4.6x10^15 t = 20 days
Start with this equation --> N = Noe^-Lt N = # of nuclei remaining No = # of nuclei at the start L = lambda (wavelength) t = time
Solve for wavelength
N = Noe^-Lt --> N/No = e^-Lt --> ln(N/No) = ln(e^-Lt) --> ln(N/No) = -Lt --> -(ln(N/No))/t = L (plug in given numbers)
To find Half Life:
t1/2 = (ln(2))/L ---> L = number previously calculated.
Professor's Note : Shelby, you are doing good, but missing the circuit problem. If you need help on that, please get back to me
3. In series: 4ohms, 6ohms --> 4ohm + 6ohm = 10ohm In parallel 9ohms, 8ohms --> 1/9 +1/8 = 1/0.236 = 4.235ohm
Now 10ohm and 4.235ohm are in parallel --> 1/4.235 + 1/10 = 1/0.3361 = 2.975ohm
In series: 3ohm, 2.975ohm, 6ohm --> 3 + 2.975 + 6 = 11.975ohm
Now 11.975ohm is in parallel with 20ohm --> 1/11.975 + 1/20 = 1/0.1335 = 7.49ohm
Net Resistance = 7.49ohm
To find Current:
V = IR ---> Given: V=6V; R = Net resistance from part 1
I = V/R
I = (6V)/(7.49ohm)
I = 0.801A
__
Week 7
1.
Four point charges in a square (10cm a side)
--> Q1 at bottom left corner (-2uC)
--> Q2 at top left corner (+2uc)
--> Q3 at top right corner (+2uc)
--> Q4 at bottom right corner (+2uc)
*Q1 and Q3 are not directly touching within the square... they act on each other through the center of the square; cutting the square diagonally in half (in two triangles)
-----> to find hypotenuse: a^2 + b^2 = C^2 (a is side 1 of triangle, b is side 2, c is hypotenuse)
----> 10^2 +10^2 = c^2 => c = 14.14cm --> this is the distance to find F13
*1uC = 10^-9C
Use eqnt: F = (kQaQb)/r^2 for each force acting on Q1:
Now Find the X and Y Components:
SumFx = F14cos(0) + F13cos(45) + F12cos(90) ---> Q14 is directly on x axis, thus why cos(0); Q13 is between 0 and 90, thus cos(45); Q12 is 90 degrees from x axis thus cos(90)
SumFx = 4.85e-6N
2.
Three point charges in corners of an equilateral triangle (4cm each side)
-->Q1 in bottom right corner (+3uC)
--> Q2 top middle (+3uC)
--> Q3 bottom left corner (+3uC)
Use eqn: F = (kQaQb)/r^2 for each force acting on Q1
Professor's note : All is VERY good, to save your time and serve the purpose of this exercise, let's avoid the numbers. Give me the steps to solve a given problem in words - explain as much as you want ! Keep up the good !!
Week 1:
Problem 1:m = 1.5kg
k = 120N/m
Part a:
Fn (normal force) = mg
Fn = (1.5kg)(9.81m/s^2)
Fn = 14.715N
Fn = kx <---- have Fn and k, need to find x
x = Fn/k
x = (14.715N)/(120N/m)
x = 0.1226m
Part b:
xb = xa + 0.25m
xb = 14.715m + 0.25m
xb = 0.3726m
1/2mv^2 = 1/2kx^2 <---- have m, k, and x; need to find v
v^2 = (1/2kx^2)/(1/2m) <---- 1/2's cancel
v^2 = (kx^2)/(m)
v = sqrt[(kx^2)/(m)]
v = sqrt[((120N/m)(0.3726m^2))/(1.5kg)]
v = 3.3328m/s
Problem 2:
Length (L) = 0.78m
T = 1/4[2pi sqrt(L/g)] <---- the ball only travels 1/4 of the period
T = 1/4[2pi sqrt((0.78m)/(9.81m/s^2))]
T = 0.4429s
Week 2 (Sept. 17, 2012):
Problem 1:Given:
m = 680g ---> 0.680kg
k = 65N/m
x = 12cm ---> 0.12m
t = 0s
Part 1:
w = sqrt(k/m)
w = sqrt[(65N/m)/(0.680kg)]
w = 9.78rad/s
T = 2pi sqrt(m/k)
T = 2pi sqrt[(0.680kg)/(65N/m)]
T = 0.643s
f = 1/T
f = 1/0.643s
f = 1.56Hz
Part 2:
x(t) = Acos(wt) <--- t=0, x=0.12m
0.12m = Acos[(9.78rad/s)(0s)]
0.12m = Acos(0)
(0.12m)cos^-1 = A
A = 0.12m
Part 3:
Vmax occurs at x = 0m
Vmax = Aw
Vmax = (0.12m)(9.78rad/s)
Vmax = 1.174m/s
Part 4:
Amax = Aw^2
Amax = (0.12m)(9.78rad/s)^2
Amax = 11.478m/s^2
Part 5:
x(t) = Acos(wt) <--- solve for t
0.06m = 0.12m cos(wt) <---- divide each side by 0.06m
0.5m = cos(wt)
cos^-1(0.5m) = wt
pi/3 = wt
(pi/3)/w = t
pi/3w = t
pi/[(3)(9.78rad/s)] = t
t1 = 0.107s
x(t) = Acos(wt) <---- x = 0, solve for t2
0 = A cos (wt) <--- divide each side by A
0 = cos(wt)
cos^-1(0) = wt
(cos^-1(0))/w = t
(cos^-1(0))/(9.78rad/s) = t
t2 = 0.161s
t2-t1 <---- time it takes block to go from x =0 to x=0.06m
0.161s - 0.107s = 0.054s
Problem 2:
Given:
T = 150N
u = 7.2g/m ---> 0.0072kg/m
x = 90 cm ---> 0.90m
Part 1:
V= sqrt(F/u)
V = sqrt (150N/(0.0072kg/m))
V = 144.34m/s
Part 2:
Lambda --> (L)
3(L/2) = x ---> find L (wavelength)
3(L/2) = 0.9m
0.9m/3 = (L/2)
2(0.9m/3) = L
L = 0.6m
Part 3:
V = fL ---> find f (frequency)
V/L = f
[(144.34m/s)/(0.6m)] = f
f = 240.57Hz
Week 3 (September 24, 2012)
Given:m(string) = 0.6g --> 0.0006kg
L(string) = ?
m(ball) = 5kg
g = 9.8m/s^2
T(period) = 2s
V(transverse wave) = ?
Period (T) = 2pi sqrt(L/g) --> find L
T/2pi = sqrt(L/g) --> square each side
(t/2pi)^2 = L/g --> multiply by g
g(T^2/4pi^2) = L
L =(9.8m/s^2)((2s^2)/(4pi^2)
L = 0.993m
Linear Density of String
u = m(string)/L
u = 0.0006kg/0.993m
u = 0.000604kg/m
Tension of string (F)
F = m(ball)g
F = (5kg)(9.8m/s^2)
F = 49N
Velocity of wave
V = sqrt(F/u)
V = sqrt((49N)/(0.000604kg/m))
V = 284.77m/s
Week 3
Use the equation: fo = fs[(1- (vo/v))/(1-(vs/v))] --> use (-) for numerator to show that observer is moving away from source; use (-) in denominator since the source is moving forward, but the frequency must increase and therefore the denominator should be decreased.
Plug in 440Hz for the frequency of the source (fs)
Plug in 54m/s for the velocity of the source (vs)
Plug in 80m/s for the velocity of the observer (vo)
Plug in 340m/s for speed of sound (v)
Week 4
1. Given:
f = 420Hz
V = 340m/s
L = ?
a) Find length of the tube
Use f = nV/2L --> (tube open at both ends)
--> frequency is given (420Hz)
--> v is given (340m/s)
--> n =1 because it is the fundamental frequency (1st harmonic)
--> Rearrange to find L
L = nV/2f
b) What is the second harmonic?
(Find frequency)
f = nv/2L
--> v is given (340m/s)
-->use L from part a
--> n = 2 because it is the 2nd harmonic
c) What is the fundamental frequency of this pipe when the speed of sound in air is increased to 367m/s?
Find f
Use f = nV/2L
--> v is given (now 367m/s)
--> n = 1 (fundamental frequency)
--> use length of tube from part a
2. How long are the pipes?
Given (Pipe A):
Open at both ends
fA = 60Hz
Fundamental Frequncy (n = 1)
Given (Pipe B):
Open at one end
fB = fA at 2nd Harmonic
First [[#|Step]]:
Use fA = nV/2L --> Rearrange to Find L
L = nV/2f
Second [[#|Step]]:
Find frequency of Pipe A at 2nd harmonic
fA = nV/2L
--> n = 2 (2nd harmonic)
--> V = 340m/s
--> use L from first step
Third [[#|Step]]:
Frequency of 2nd Harmonic of Pipe A = Frequency of Third Harmonic of Pipe B
fB = nV/4L ----> rearrange to find L
L = nV/4fB
--> n = 3 (third harmonic)
--> V = 340m/s
-->fB = answer from Step 2
Week 6
1. Given:226Ra88 --> 222Rn86 + 4He2
Ra = 226.02540u
Rn = 222.01757u
He = 4.002603u
mass defect = mass of constituents - mass of nucleus
mass defect = (mass of Rn + mass of He) - mass of Ra
*1u = 931.5MeV*
Take mass defect and multiply by 931.5MeV to find the energy in MeV
2. Given:
N = 8.14x10^14
No = 4.6x10^15
t = 20 days
Start with this equation --> N = Noe^-Lt
N = # of nuclei remaining
No = # of nuclei at the start
L = lambda (wavelength)
t = time
Solve for wavelength
N = Noe^-Lt --> N/No = e^-Lt --> ln(N/No) = ln(e^-Lt) --> ln(N/No) = -Lt --> -(ln(N/No))/t = L (plug in given numbers)
To find Half Life:
t1/2 = (ln(2))/L ---> L = number previously calculated.
Professor's Note : Shelby, you are doing good, but missing the circuit problem. If you need help on that, please get back to me
3. In series: 4ohms, 6ohms --> 4ohm + 6ohm = 10ohm
In parallel 9ohms, 8ohms --> 1/9 +1/8 = 1/0.236 = 4.235ohm
Now 10ohm and 4.235ohm are in parallel --> 1/4.235 + 1/10 = 1/0.3361 = 2.975ohm
In series: 3ohm, 2.975ohm, 6ohm --> 3 + 2.975 + 6 = 11.975ohm
Now 11.975ohm is in parallel with 20ohm --> 1/11.975 + 1/20 = 1/0.1335 = 7.49ohm
Net Resistance = 7.49ohm
To find Current:
V = IR ---> Given: V=6V; R = Net resistance from part 1
I = V/R
I = (6V)/(7.49ohm)
I = 0.801A
__
Week 7
1.
Four point charges in a square (10cm a side)
--> Q1 at bottom left corner (-2uC)
--> Q2 at top left corner (+2uc)
--> Q3 at top right corner (+2uc)
--> Q4 at bottom right corner (+2uc)
*Q1 and Q3 are not directly touching within the square... they act on each other through the center of the square; cutting the square diagonally in half (in two triangles)
-----> to find hypotenuse: a^2 + b^2 = C^2 (a is side 1 of triangle, b is side 2, c is hypotenuse)
----> 10^2 +10^2 = c^2 => c = 14.14cm --> this is the distance to find F13
*1uC = 10^-9C
Use eqnt: F = (kQaQb)/r^2 for each force acting on Q1:
F14 = (kQ1Q4)/r^2 --> F14 = [(8.98755e9)(2e-9C)(2e-9C)]/(0.10m^2) ---> F14 = 3.59e-6N
F12 = (kQ1Q2)/r^2 --> F12 = 3.59e-6N
F13 = (kQ1Q2)/r^2 --> *Be sure to use 14.14cm instead of 10cm* --> F13 = 1.79e-6N
Now Find the X and Y Components:
SumFx = F14cos(0) + F13cos(45) + F12cos(90) ---> Q14 is directly on x axis, thus why cos(0); Q13 is between 0 and 90, thus cos(45); Q12 is 90 degrees from x axis thus cos(90)
SumFx = 4.85e-6N
SumFy = F14sin(0) + F13sin(45) + F12sin(90)
SumFy = 4.85e-6N
To find Fnet:
Fnet = sqrt(Fx^2 + Fy^2)
Fnet = 6.86e-6N
2.
Three point charges in corners of an equilateral triangle (4cm each side)
-->Q1 in bottom right corner (+3uC)
--> Q2 top middle (+3uC)
--> Q3 bottom left corner (+3uC)
Use eqn: F = (kQaQb)/r^2 for each force acting on Q1
F12 = (kQ1Q2)/r^2 --> F12 = 5.06e-5N
F13 = (kQ1Q3)/r^2 --> 5.06e-5N
Now find X and Y Components:
SumFx = F13cos(0) + F12cos(60) ---> equilateral trial has all three angles at 60degrees
SumFx = 7.59e-5N
SumFy = F13sin(0) + F12sin(60)
SumFy = 4.38e-5N
To find Fnet:
Fnet = sqrt(Fx^2 +Fy^2)
Fnet = 8.76e-5N