Here's how it works in detail: Each path that the girls takes can be represented by a vector. When drawn out in an x-y coordinate system you can express each vector in terms of its x and y components. The resultant vector is found by adding up all of the x components and then adding up all the y...
That is the correct approach. The resulting displacement is just the sum of those individual vectors. If you drew them to scale, you can connect the starting point to the ending point and measure that vector to get your answer.
Here are a couple of hints: (1) The time frame for the velocity and the acceleration is the same: 5 sec. (2) Find the distance that the tip of the second hand travels in 5 sec. You can do this by knowing that it travels 360 degrees, or 2 Pi radians in 60 seconds. From that tells you can figure...
Yes, you calculate each potential separately and then add them all up. For the curved pieces construct an integral for a potential dV from a bit of charge dq. then integrate. Your limits of integration depend on how you choose to draw your diagram, but the total angle is the same no matter how...
Besides the approach that I mentioned in my previous post, you can also use the fact that Work = qV where q is the charge that's being moved, and V is the change in electric potential.
What you calculated in your work below, was the change in electric potential, not the work. All you have to...
In brief, Work = Force times distance. So your equation for work should be
W_{\vec{E}} = -\int_{r_{1}}^{r_{2}} q \vec{E}\cdot d\vec{r}
Where the force is qE, q being the charge that's moved, and E is the field due to the charge Q.
Your idea is correct, but your first equation is wrong. Usually gravity is ignored because the gravitational force is small compared to the electric force.
I would make the approximation that the expansion of the glass can be ignored. Then you don't have to worry about the expansion of the cross section and the linear formula will give you an acceptable answer.
Okay, draw a picture of the girl on the plank. Label the girl's center of mass. Label the plank's center of mass. Look up how to calculate the center of mass of the girl + plank. That point doesn't move. However, the girls and the plank DO move. They have to move so that in their final position...
Here's a couple of things you might want to consider. (1) What is the temperature of the mixture at the triple point? (2) What is the pressure at the triple point? (3) What is the pressure at the normal freezing point of water? What does that imply about the difference between the triple point...
The essence of this problem is that since all forces involved are internal to the system, the center of mass of the system (plank plus girl) does not move even though the girl and the plank do move. The velocity of the girl or the plank is immaterial, {though as a practical matter, she should...
Here's some hints: The frequency is the number of revolutions per second (or fractions of a revolution per second). That depends on the velocity of the mass because the faster it goes the more complete revolutions it can make in a given time period. But you have this nice little equation that...
The components of the momentum in the x and y directions are separately conserved. Which component of the system's momentum is zero before the collision? Whenyou figure this out, you will be able to answer your question.
Well, here are a couple more hints. The t variable is the same in both equations. The time taken to get to the top of the parabolic path is one-half the total time of flight. You should write two equations:
X = X_0 + V_{0x} t + \frac{1}{2} a_x t^2
and
Y = Y_0 + V_{0y} t + \frac{1}{2} g...
That is what I was referring to. Your values for the initial velocity components are correct. I would place the origin of my coordinate system either at the point on the roof where the ball leaves the roof, or directly under it on the ground. Now write two equations:
X = X_0 + V_{0x} t...
Here's some hints: the trajectory of the dart is a parabola. The maximum height of the dart is at the vertex of the parabola. Your initial velocity will be inclined upwards from the horizontal and you will need to use two kinematic equations.
Well, what you need to do is to use two of your kinematic equations. One for the horizontal direction and one for the vertical direction. You assume that the acceleration in the horizontal direction is zero (neglect air resistance). and take the vertical acceleration to be the acceleration of...
Start out by finding the velocity of the ball as it hits the water. Assume that the ball starts from rest when dropped from the diving board and use an equation that involves the acceleration due to gravity, the velocity of the ball, and the distance from the diving board to the water.
Once...
A couple of things are not clear in your problem statement. What is it you are asked to find? Since you didn't provide a picture, you need to describe the arrangement of the charges a little better. Are q1 and q2 on the x axis while q3 is above them on the y axis? Is L the length of the sides of...
It strikes me that you could do this most simply by conservation of energy. The potential energy of the falling cylinder goes into kinetic energy. Don't forget that the KE of cylinder (1) is both KE of translation and KE of rotation and that the linear velocity of the center of mass of cylinders...
Have you ever heard of the "Method of Undetermined Coefficients"? It is covered in introductory differential equations textbooks.
You might try setting
f = A e^x sin(x) + B e^x cos(x)
and substituting it into your DE. Then collect like terms and see what values A and B have to be...
Okay, you've confused the force between two point charges with the force on a charge in an electric field. The force on a charge in an electric field is
F = qE
You can combine this with F = ma
That gives you qE = ma .
This last equation allows you to solve for the acceleration a...
I would think so. The contribution of the electrons to the binding energy it probably small compared to the nuclear binding energy (I haven't calculated it myself in a long long time, so I don't recall the exact comparison.)