Recent content by jackarms

You're right about the force being away from the origin from points B and D, but it's also away from the origin for the right half of the graph as well. Think about it -- the slope is opposite, yes, but it's also on the other side of the origin. Try working out the sign of the electric field and...

No problem! Glad you got it all sorted out. :)

You're close with setting the two forces equal, but they aren't quite the correct forces to use. You want q2 to be in equilibrium, which means the forces you're going to be dealing with are those that act on q2. So that will be the force between Q and q2 and between q1 and q2. If you set up the...

I almost didn't see the partial derivatives. Those certainly make this more complicated. I've actually asked this on math.stackexchange to see what the people over there think. One answer I've found is that you can use the Biot-Savart law to solve for ##\vec{B}##: $$\bf{B}(r) = \int...

You might be confusing the angle a resultant makes with the angle the individual vectors make. In this case, ##\vec{a}## does make a 90 degree angle with ##\vec{b}##, but this is separate from the angle a resultant makes. I attached a picture below (really should have made this earlier! It's so...

Well, it looks like you'll have three unknowns -- ##B_{x}, B_{y}, B_{z}## -- so you'll need three equations to solve for all of them. If you find the equation for the second component you should be able to solve for ##\vec{B}## in full.

Well, what work have you done so far with the limit? It's not enough here on PF to simply give the problem statement and nothing else.

It means the angle that gets traced if you start at the x-axis and then go counterclockwise to the vector. Imagine a rotating bar that is hinged at the origin and only rotates counterclockwise, and starts out overlaping the positive x axis. How much of a full rotation would it take to get the...

Yes, you can use that method to add the vectors together and then find the angle of the resultant vector. You could also find the resultant with components, but since you're given all the vectors, it's probably easiest to just draw the additions off to the side. So you want to draw the vectors...

Hey there! Your approach with components is on the right track, but slightly off. And the distance left won't quite be 4.58 km -- that would be if they traveled in a straight line toward the destination, but since they were at an angle, it will actually be a greater distance for them to travel...

For the general formula (in terms of n), you're right that the sequence going both ways complicates things. The answer is that you have to arbitrarily designate a term to be ##a_{1}##, or perhaps ##a_{0}##. You would write the whole formula based around that starting term, in a form like this...

Yes, it's just vector addition as usual. The key part of the problem is that it gives you vectors in terms of forces. The thrust of 675N is one vector, with magnitude of 675N at 0° above the forward direction, and the other is a vector with magnitude 450N at 20.4° above the forward direction...

Indeed -- it can be difficult to confuse the change in velocity with the acceleration. Mathematically: ##\Delta v = at## Acceleration is in fact a scaled copy of the change in velocity, where the scaling factor is time. This is how the different units come into play, even though...

Well, the first part involves finding a resultant force, so you have to find the components of the two vectors given using: ##v_{x} = |v| \times cos(\theta)## and ##v_{y} = |v| \times sin(\theta)## And then add them together, and then find the magnitude using: ##|v| = \sqrt{v_{x}^{2} +...

Well, like you said, if the situation is UCM like the example you gave, then speed won't change, and so there's no calculation to be done. If the situation is non-UCM, and you want find the final velocity at some point, what you first want to do is resolve the acceleration into tangential and...