C) V=lambda*f=>f=V/lambda=(144.337m/s)/(0.6m)=240.56Hz Professor's note: All is 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 ! Week 3 Solutions
1. First, the length of the string must be found using the equation T=(2*pi)*sqrt(L/g). Solving for L, we get L=(g*T^2)/((2*pi)^2). We get that L=0.9929 m.
Then, the force in the string must be found using F=m*g, where m is the mass of the ball. We get that F=49 N.
Then, we find the linear density of the string using u=m/L, where m is the mass of the string (which was converted from 0.6g to 0.0006kg). We get that u=0.000604 kg.
Finally, we use all of this information to find V of the transverse wave in the string using V=sqrt(F/u). We get that V=284.76 m/s.
2. This problem is the case of a sound source and observer both moving. Since the police car is chasing the Porsche, and the Porsche is going much faster than the police car, we know that the distance between the two cars is increasing, so the specific case is that of a source and observer moving away from each other. We can use the equation fo=fs*((1-(vo/v))/(1+(vs/v)) to get that fo=29.03 Hz. Week 4 Solutions 1. A) Since we're given the frequency, speed of sound in air, and the harmonic number (n=1), we can use the equation f=(nV)/(2L), solve for L, plug in all the given values, and find that L=0.404 m.
B) Now that we've found L, we can use that in the equation f=(nV)/(2L), where n=2 because the questions asks for the frequency at the second harmonic. Solving for f, we get that f=840 Hz.
C) Now, n=1. V can be changed to 367 m/s. Using the same equation and solving for f gets us f=453.35 Hz.
2. We can find L of the first organ pipe (organ pipe A) by using f=(nV)/(2L), where f=60 Hz, n=1, and V=340 m/s. Solving for L gets us L=2.83 m (the length of organ pipe A).
Since the fundamental frequency of organ pipe B is the same as the frequency of the second harmonic of organ pipe A, we must first find the frequency of the second harmonic of organ pipe A, using n=2, V=340 m/s, and L=2.83 m. Solving for f, we get that f=120 Hz (the second harmonic of organ pipe A).
We can then use the equation f=(nV)/(4L), where n=1, V=340 m/s, and f=120 Hz. Plugging in these numbers and solving for L gets us that L=2.125 m (the length of organ pipe B). Week 5 Solutions
1. First we must find the mass defect in u, which is the mas of Ra minus the total mass of the Rn plus the He, which comes out to 0.0052266u. Then, we must convert this value to kg by multiplying it by (1.661*10/6-27kg)/(1u), which is 8.6813826*10^-30 kg. Then we must find the binding energy using the equation E=m*C^2, where C is 3*10^8, and m is the mass defect found earlier. This gets us 2.60441478 J of energy, which must be converted to MeV by multiplying it by (931.5MeV)/(1kg), to get 2.426*10^-18 MeV.
2. First, the decay constant, lambda, must be found by the equation N=N0*e^(/lambda*t), where N is the new amount of the substance (8.14*10^14), N0 is the initial amount (4.6*10^15), and t is the time passed (20 days). Solving for lambda gets us a decay constant of 0.086592560284. Then, we can use the equation ln(2)=lambda*halflife, where lambda is the value just found. Solving algebraically for halflife gets us 9.00469 days.
3. The 4ohms and 6ohms are in series, so there is a total of 10ohms there. The 9ohms and 8ohms are in parallel, which comes out to 4.235ohms. That is in series with the 3ohms and the 6ohms, which gets us 13.235ohms. The 13.235ohms, 10ohms, and 20ohms, are all in parallel. Using the parallel equation Rt=(R1^-1+R2^-1+R3^-1)^-1 gets us a total of 4.433ohms. To find the current through the circuit, we can just use Ohm's Law, V=IR, where R is the equivalent resistance, and V is the given 6 V battery. Solving for I gets us 1.354A.
Professor's Note : Good Job Szabolcs, keep up the good work !! Week 6 Solutions 1. Let the point charge at the bottom left corner of the square be the negatively charged particle.
FIrst, we must find the force between the negatively charged particle and the other three positively charge particles, which will all be the same because both the distances between the particels and the charges of the particles are the same.
So, using f=k*((q1*q2)/d^2) and plugging in 8.98755*10^9 for k, 2uC for q1, and 2uC for q2, we get that F is 3.595N for.
Next, we must use the summation rule for both the forces in the X direction and the forces in the Y direction.
In the X direction, the summation is SumFx=F12cos(90)+F13cos(45)+F14cos(0). F12cos(90) cancels out because cos(90) is 0, so we are left with F13cos(45)+F14cos(0), which, after plugging in 3.595 for F13 and F14, gives us 7.19N in the X direction.
In the Y direction, the summation is SumFy=F12sin(90)+F13sin(45)+F14sin(0). F13sin(0) cancels out because sin(0) is 0, so we are left with F12sin(90)+F13sin(45), which, after plugging in 3.595 for F12 and F13, gives us 7.19N as well.
Finally, to get the net force on the negatively charged particle, we must us the Pythagorean theorem in the following form: Fnet=sqrt((Fx^2)+(Fy^2)), where Fx and Fy are both 7.19N. Solving for Fnet gives us a net force of 10.168N.
2. As in the last problem, the forces between all of the particles are equal. So, solving the equation F=(8.98755*10^9)*((3uC*3uC)/0.4m^2) gives us a force of 0.5055496875N.
Since this is an equilateral triangle, each angle in the triangle is 60 degrees. So, if we take the bottom left particle as the reference, the particle that is not directly to the right of it (i.e. at 0 degrees) must be at 60 degrees.
So, SumFx=F12cos(60)+F13cos(0). Plugging in the force found earlier for both F12 and F13, we get that Fx is 0.75832453125.
SumFy=F12sin(60)+F13sin(0), where F13sin(0) cancels out because sin(0) is 0. Plugging in the force found earlier for F12, we get that Fy is 0.619169386999.
Again, using the Pythagorean Theorem, we find that Fnet=sqrt((0.75832453125^2)+(0.619169386999^2)), which comes out to approximately 0.9789 N.
1. A) F=kx=>mg=kx
x=(mg)/k=(1.5kg*9.8m/s^2)/(120N/m)=0.1225m
B) (1/2)mv^2=(1/2)kx^2=>(1/2)(1.5kg)(v^2)=(1/2)(102N/m)(0.3725^2m)
v=sqrt((120N/m*0.3725^2m)/(1.5kg))=2.9292m/s
2. V=Vmax;x=0;t=(1/4)T
T=(2pi)sqrt(L/g)=(2pi)sqrt(0.78m/9.8m/s^2)=1.77261382385
t=(1/4)T=(1/4)*1.77261382385=0.443s
Week 2 Solutions :
1. m=680 g=0.680 kg
k=65 N/m
x1=A=12 cm=0.12 m
A) (1/2)*m*Vmax^2=(1/2)*k*A^2
Vmax=sqrt((k*A^2)/m)=sqrt((65 N/m*0.12^2 m)/0.680 kg)=1.173 m/s
Vmax=A*w=>w=Vmax/A=1.173 m/s/0.12 m=9.776 rad/s
w=(2pi/T)=>T=(2pi/w)=(2pi)/(9.776 rad/s)=0.642s
f=1/T=1/0.642s=1.556Hz
B) A=0.12m
C) Vmax=1.173m/s
x=0m
D) amax=A*w^2=0.12m*9,776^2rad/s=11.47m/s^2
E) x(t)=A*cos(wt)
0.06m=0.12m*cos(9.776rad/s*t')=>t'=(cos^-1(0.5))/(9.776rad/s)=0.1071s
0m=0.12m*cos(9.776rad/s*t'')=>t''=(cos^-1(0))/(9.776rad/s)=0.1606s
t=t''-t'=0.1606s=0.1071s=0.05355s
2. u=7.2g/m=0.0072kg/m
F=150N
L=0.90m
A) V=sqrt(F/u)=sqrt(150N/0.0072kg/m)=144.337m/s
B) 0.9m=3lambda/2=>lambda=(2*0.9m)/(3)=0.6m
C) V=lambda*f=>f=V/lambda=(144.337m/s)/(0.6m)=240.56Hz
Professor's note : All is 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 !
Week 3 Solutions
1. First, the length of the string must be found using the equation T=(2*pi)*sqrt(L/g). Solving for L, we get L=(g*T^2)/((2*pi)^2). We get that L=0.9929 m.
Then, the force in the string must be found using F=m*g, where m is the mass of the ball. We get that F=49 N.
Then, we find the linear density of the string using u=m/L, where m is the mass of the string (which was converted from 0.6g to 0.0006kg). We get that u=0.000604 kg.
Finally, we use all of this information to find V of the transverse wave in the string using V=sqrt(F/u). We get that V=284.76 m/s.
2. This problem is the case of a sound source and observer both moving. Since the police car is chasing the Porsche, and the Porsche is going much faster than the police car, we know that the distance between the two cars is increasing, so the specific case is that of a source and observer moving away from each other. We can use the equation fo=fs*((1-(vo/v))/(1+(vs/v)) to get that fo=29.03 Hz.
Week 4 Solutions
1. A) Since we're given the frequency, speed of sound in air, and the harmonic number (n=1), we can use the equation f=(nV)/(2L), solve for L, plug in all the given values, and find that L=0.404 m.
B) Now that we've found L, we can use that in the equation f=(nV)/(2L), where n=2 because the questions asks for the frequency at the second harmonic. Solving for f, we get that f=840 Hz.
C) Now, n=1. V can be changed to 367 m/s. Using the same equation and solving for f gets us f=453.35 Hz.
2. We can find L of the first organ pipe (organ pipe A) by using f=(nV)/(2L), where f=60 Hz, n=1, and V=340 m/s. Solving for L gets us L=2.83 m (the length of organ pipe A).
Since the fundamental frequency of organ pipe B is the same as the frequency of the second harmonic of organ pipe A, we must first find the frequency of the second harmonic of organ pipe A, using n=2, V=340 m/s, and L=2.83 m. Solving for f, we get that f=120 Hz (the second harmonic of organ pipe A).
We can then use the equation f=(nV)/(4L), where n=1, V=340 m/s, and f=120 Hz. Plugging in these numbers and solving for L gets us that L=2.125 m (the length of organ pipe B).
Week 5 Solutions
1. First we must find the mass defect in u, which is the mas of Ra minus the total mass of the Rn plus the He, which comes out to 0.0052266u. Then, we must convert this value to kg by multiplying it by (1.661*10/6-27kg)/(1u), which is 8.6813826*10^-30 kg. Then we must find the binding energy using the equation E=m*C^2, where C is 3*10^8, and m is the mass defect found earlier. This gets us 2.60441478 J of energy, which must be converted to MeV by multiplying it by (931.5MeV)/(1kg), to get 2.426*10^-18 MeV.
2. First, the decay constant, lambda, must be found by the equation N=N0*e^(/lambda*t), where N is the new amount of the substance (8.14*10^14), N0 is the initial amount (4.6*10^15), and t is the time passed (20 days). Solving for lambda gets us a decay constant of 0.086592560284. Then, we can use the equation ln(2)=lambda*halflife, where lambda is the value just found. Solving algebraically for halflife gets us 9.00469 days.
3. The 4ohms and 6ohms are in series, so there is a total of 10ohms there. The 9ohms and 8ohms are in parallel, which comes out to 4.235ohms. That is in series with the 3ohms and the 6ohms, which gets us 13.235ohms. The 13.235ohms, 10ohms, and 20ohms, are all in parallel. Using the parallel equation Rt=(R1^-1+R2^-1+R3^-1)^-1 gets us a total of 4.433ohms. To find the current through the circuit, we can just use Ohm's Law, V=IR, where R is the equivalent resistance, and V is the given 6 V battery. Solving for I gets us 1.354A.
Professor's Note : Good Job Szabolcs, keep up the good work !!
Week 6 Solutions
1. Let the point charge at the bottom left corner of the square be the negatively charged particle.
FIrst, we must find the force between the negatively charged particle and the other three positively charge particles, which will all be the same because both the distances between the particels and the charges of the particles are the same.
So, using f=k*((q1*q2)/d^2) and plugging in 8.98755*10^9 for k, 2uC for q1, and 2uC for q2, we get that F is 3.595N for.
Next, we must use the summation rule for both the forces in the X direction and the forces in the Y direction.
In the X direction, the summation is SumFx=F12cos(90)+F13cos(45)+F14cos(0). F12cos(90) cancels out because cos(90) is 0, so we are left with F13cos(45)+F14cos(0), which, after plugging in 3.595 for F13 and F14, gives us 7.19N in the X direction.
In the Y direction, the summation is SumFy=F12sin(90)+F13sin(45)+F14sin(0). F13sin(0) cancels out because sin(0) is 0, so we are left with F12sin(90)+F13sin(45), which, after plugging in 3.595 for F12 and F13, gives us 7.19N as well.
Finally, to get the net force on the negatively charged particle, we must us the Pythagorean theorem in the following form: Fnet=sqrt((Fx^2)+(Fy^2)), where Fx and Fy are both 7.19N. Solving for Fnet gives us a net force of 10.168N.
2. As in the last problem, the forces between all of the particles are equal. So, solving the equation F=(8.98755*10^9)*((3uC*3uC)/0.4m^2) gives us a force of 0.5055496875N.
Since this is an equilateral triangle, each angle in the triangle is 60 degrees. So, if we take the bottom left particle as the reference, the particle that is not directly to the right of it (i.e. at 0 degrees) must be at 60 degrees.
So, SumFx=F12cos(60)+F13cos(0). Plugging in the force found earlier for both F12 and F13, we get that Fx is 0.75832453125.
SumFy=F12sin(60)+F13sin(0), where F13sin(0) cancels out because sin(0) is 0. Plugging in the force found earlier for F12, we get that Fy is 0.619169386999.
Again, using the Pythagorean Theorem, we find that Fnet=sqrt((0.75832453125^2)+(0.619169386999^2)), which comes out to approximately 0.9789 N.