Useful equation: F(net)=dp/dt
Given:
Initial VELOCITY (initial MOMENTUM) is in the direction of g^vector.
A net FORCE is applied in the direction of the a^vector.
Ask yourself: did you think about it appropriately at first? Can you truly add those vectors quantities?
I have no idea why you would take ln of anything with this.
You're graphing d=mt^2...m=0.5a.
So you would want to graph d v. t^2 (d is y-axis, t^2 is x-axis) to give a straight line and take its slope.
Certainly you can think of (1,1,1) as the start of your vector, then (3,4,5) as the end of your vector...thus subtraction yields a vector that passes through both points, just with a certain magnitude.
The magnitude of a vector is given as sqrt(x^2+y^2+z^2)...you can find that you won't get a...
Since they have STDEV as a separate row, I suspect they want the absolute uncertainty for the mean:
x(avg) = sum(x)/N...so if you add up all the uncertainties in quadrature, and then scale it by N, you have
uncert(avg)=(1/N)*sqrt(uncert(x1)^2+uncert(x2)^2+uncert(x3)^2...)...
I would find the power series for 1/(x^2+1) then multiply each term by x, then look at it and write the summation notation.
f(x)= f(0)+f '(0)*x +f ''(0)*x^2/2 + f '''(0)*x^3/6 +...
f(x) = (x^2+1)^-1
f '(x) = - 2x*(x^2+1)^-2
f'' (x) = -2x*-2*2x*(x^2+1)^-3 + (x^2+1)^-2*-2
f(0) =...
Without knowing the relation of dynes to newtons off the top of my head, let me write out your DE using variables (I assume this is not a simpler case of horizontal motion, rather, by saying "above" and "directed down" you are dealing with a vertical spring system so the force of gravity...
As far as I can tell, (a) refers to f(x) as a series (thus a sum), but since f(x) is not itself a summation, it is more a sequence, a function for discrete values of n (if you plotted f(x) v. n for a fixed value of x, it'd look like a discontinuous function...a bunch of points).
When x=0...
Just a bit further down the same page, the point is that the velocity selector allows a charged particle with a certain velocity to shoot straight down the finely tuned system between the plates and in the magnetic field unimpeded. Presumably, if it veers off the middle line just a bit, the...
Not knowing anything about wavefunctions, just on the information you gave me, I think you multiplied incorrectly (left out the A):
(A-Ae^(ikx))*(A-Ae^(-ikx)) = 2A^2-A^2*(e^(ikx)-e^(-ikx)) = 1
From there, I'm seeing a chance to divide by 2A^2, so you have something like 1-sinh(u) = (1/(2A^2))...
Right, so if you rewrite that so it looks a little nicer:
You have the limit AS N GOES TO INFINITY of the absolute value of x*(n+2)/(3n+3).
That's the key here: you're taking the limit for n, seeing what the "final" term looks like compared to the one before it (remember to converge, the...
They used the trig identity to replace sin(x)^2 with (1-cos(2x))/2...then pulled the 1/2 out of the integral and integrated each term. Integral of 1 = x. They did a u-sub on cos(2x) with u=2x, du=2dx, thus it gets another 1/2 out front, and integral of sin = cos.
Right. It's much easier to think about F=ma really being:
\sum_{\text{all}}\bold{F}\,=\,m \bold{a}
So it's a sum of all your vector forces (the bold meaning it's a vector...so for your problem, the +/- indicates the direction is all, since there are no components, it's just in one...
Indeed! F=ma is your net force = m* net acceleration. If you apply your maximum tension upward, gravity still acts downward (sum the forces for your net force, don't forget that gravity gets a minus-sign because it acts in the opposite direction of your upward tension). Then use F=ma.
Only Ft*sinX. In the horizontal circle (draw it in, why not?), the only component of your forces is the horizontal component of tension. Assuming it is not falling or moving up, the vertical component of tension balances with gravity.
Consider the edge far from the applied force (i.e. the one against the table that it would intuitively tip around) as an axis of rotation...then you can find the torque applied by the force at that point a distance from the axis at an angle (associated with the geometry of the cube). Also...
Yep. It might help to visualize the exponential graphs:
Charging:
http://www.wolframalpha.com/input/?i=16*%281-e^%28-x%2F0.000636%29%29+from+x%3D0+to+x%3D.00318
Discharging:
http://www.wolframalpha.com/input/?i=16*%28e^%28-x%2F0.000636%29%29+from+x%3D0+to+x%3D.00318
V_0 is always...
Right...if you draw an amperian loop (a circle that is coaxial with the torus) that is smaller than the inner radius, you enclose no current, thus no B-field. If you draw it so the radius of the loop is between the inner and outer radii of the torus, you enclose N*I, and r*d(theta) (or your...
Watch your units...your kinematics equations deal with m/s and m/s/s. You have km/h.
Also, from an energy standpoint, I verified that 1.87kN is correct for the force.
I guess it's partly the weird idea of integration with such a thing.
The actual problem involves a cylinder with circular cross-sectional area, inner radius a, outer radius b. I am working on the inside of the cylinder (a<r<b) and so that's why I'm doing this.
I want to integrate from a...
This is not a homework problem in itself. In my physics homework, I wanted to write the difference of two areas (thus yielding a differential disk) as:
Pi*(r+dr)2-Pi*r2
It reduces to
2*Pi*r*dr+Pi*(dr)2
Now, I seem to recall from a prior class that a quick hand-waving made (dr)2 =...
You need to know the series expansions of sin(theta) and cos(theta). You can tell from Eq3 that they approximated sin(i) ~~ i. What then does cos(i)~~ ?
Look up the series expansions of sin and cos and truncate them to one term. Then plug that in for sin and cos and then a small amount of...
Well then you need to look at the total energy before and after.
K(i)=K(deu,i)+K(pro,i)
But also there was energy in the masses:
E(i)=E(deu, rest) +E(pro, rest)
I assume they brought the particles together from infinity because there was no electrical potential energy mentioned in the...
Just work it simply as if you're "massaging" Eq1 to Eq2. You know they're equal, just do whatever you can mathematically to make it LOOK the same (which means, BE the same, but still).
I:
Distributed Sd=Ssin(i)cos(r)-Scos(i)sin(r)
Substituted your supplied t=Scos(r)...
The total energy remains the same. Since the fusion reaction hasn't taken place, I would say the rest energies aren't important...also because the kinetic energy is defined as the total relativistic energy minus the rest energy.
So what kinds of energy changes could there be? Since they've...
If there are 0.254 meters in 1 inch, then how many square meters are there in a square inch?
And just to keep your work nice, note that on your conversion from lbs to N, you get
20459.77 N/in2, not just 20459.77 N.