Thank you, jbunniii, for your response. The expansion of e^x certainly shows the terms I was looking for, the terms do not come from the expression, f(x)≈f(0) + f'(0)x, I thought maybe I missed something in the lecture.
Thanks also for giving me more depth on GPS operations, there is quite a...
My questions are from lecture 9, MIT OCW SV Calculus, Jerison, 2009;
At 27:50 he is deriving the linear approximation for the function
e^(-3x)(1+x)^(-1/2)≈(1-3x)(1-1/2x)≈1-3x-1/2x+3/2x^2≈1-7/2x, for x near 0.
In the last step he drops the x squared term since it is negligible(no questions so...
I will provide a summary of what's been done so far;
The platform:
Horizontal forces.
F-F_{fric}=ma_{plat}
Vertical forces.
N=mg+N_{1}
as I stated in post 4, this N_{1}is really just Mg. One way to think of it is that the normal force of the floor must support the weight of both the...
You are getting ahead of yourself a little bit. Parts (a) and (b) will not involve rotational issues, just Newton's Second Law.
Your equations of motion for the platform look good with a subtle tweak. Specifically, N_{1}is the force of the platform on the cylinder, the force of the cylinder on...
The velocities of the two masses are not necessarily the same(most of the time they are not). Look at the equation 15tungalbert posted, the before collision momentum(left side) is equal to the after collision momentum(right side).
You are given the velocity of the 1kg mass post collision. Your...
That's ok.
Remember that since momentum is conserved, the momentum before the collision is the same as that after the collision. You have correctly calculated the momentum before the collision, -2kgm/s, so that's what the total after collision momentum must be as well. Your unknown speed is the...
The question is not a bother, it's the purpose of the forum.
Your analysis of the before collision momentum looks very good. The after part is almost correct. The terms to the left of the equals sign are good and on the right the first term is as well. There is a "+ 2kg", is that a typo?
I would recommend drawing a picture. Identify the forces acting on each object(i.e. the platform and cylinder), and then apply Newton's second law.
F_{net}=ma=F_{1}+F_{2}+...
My technique is to consider all the forces acting on an object that (a)don't "touch" it, like gravity or electrical...
The mode of vibration is the "harmonic". The second allowed harmonic is not necessarily the second harmonic. In this case, the second allowed harmonic is the "third harmonic" corresponding to n= 3. If n=2 were allowed, the second harmonic, there would be a node at the open end of the pipe, which...
I have never heard of "gravitational units", but based on your second phrase, "… she said that its the absolute unit multiplied by acceleration due to gravity…", I would guess that she is referring to what is commonly called "G's"; as in , " The astronauts experienced 4 G's during their...
I think that I’m doing this problem correctly, but the answer seems a bit unreasonable. Can someone else check my work?
A thermometer has a quartz body within which is sealed a total volume of 0.400cm^{3} of mercury. The stem contains a cylindrical hole with a bore diameter of 0.10mm. How far...
I'll take the easier situation first, an endothermic reaction,where we add energy. This is as simple as heating up the container that the reactants are in, increasing the average speed of the molecules. Some of the molecules are moving slower than the average and some faster. If the molecules...
I think you need to consider why the bond is broken, that is, the bond will not spontaneously break. Say, for example, you have a chlorine molecule, it will not suddenly fly apart into its individual atoms, each gaining energy. That would violate the second law of thermodynamics. Imagine that...
While discussing ω^{'}, the angular frequency of a damped harmonic oscillator, given by:
ω^{'}=\sqrt{\frac{k}{m}-\frac{b^{2}}{4m^{2}}}
where k is the "springiness", m is the mass, and b is the damping constant,
my book, Halliday, Resnick and Walker, says if b is small but not...
I'm not quite with you here.
You don't have to divide the initial velocity by 2. You know Vi,Vf and t, but don't know a(I am still referring to just the last second of travel, once we know a, we can move to the whole 10 second period). Is there an equation that relates these four variables?
That looks right to me. Do you see how you can use this Vi to find acceleration?
I'm not following flyingpig(although I think we must be related), the equation he gave seems to me to be unusable at this point, only two of the four variables are known.
No, that would give the average velocity during the last second of travel. Do you see an equation that has initial and final velocity, time and distance that you could use for the last one second of travel?
Yes, there are too many unknowns to solve in one step. The information given for the last second of travel will allow you to find the acceleration, and since it is constant you will then be able to apply it to the whole 10 sec roll. If you could find the velocity at the beginning of the last...
No, that would be uniform(or constant) velocity, not acceleration. Constant acceleration means the the velocity is changing at a constant rate.
Have you been introduced to the equations of motion with constant acceleration? You will need to use one or more of them. They will look like these...
A simple pendulum of length L and mass m is suspended in a car that is traveling with constant speed v around a circle of radius R. If the pendulum undergoes small oscillations in a radial direction about its equilibrium position, what will its frequency of oscillation be?
I don’t know how to...
Regarding section (c)(i),
I will try to help by offering an analogy; imagine someone standing in the rain with an umbrella, the rain falls straight down and the umbrella is held conventionally with the protective surface horizontal. If you were to ask how much rain is hitting the umbrella it...
I haven't gone through the whole problem but I see at least one part that is not correct.
"...23.2km/hr = 23,200/60 = 386.7 m/s..."
Look at your conversion factor going from hours to seconds.
You are right down to where you found T total. In your last equation where you are calculating the value for x, you use an acceleration of -9.8. What is the acceleration along the x axis?(hint, you answered this when you state your value for ax.)
Regarding part (a), your basic equation U=mgh is correct. But one must use integration for the reason gneill gave, the water varies from height H to H + D. Do you know calculus?