Recent content by uazaga

does anyone have any idea on how to solve this electric force problem that involves a spring?

Homework Statement A block having mass m and charge + Q is connected to a spring having constant k. The block lies on a frictionless horizontal track, and the system is immersed in a uniform electric field of magnitude E, directed as shown in Figure P25.11. If the block is released from rest...

1005 = 1000 (1500-5/1500-v) I need to isolate for v but I have no clue how to do this. How would i go about doing so?

and no air resistance is not considered. Oh and I must find this using ONLY kinematics and no forces.

alright guys sorry, let me a bit more clear. I need to find the average acceleration during thrust of my rocket. I have these values. Thrust Time = 1.05 s Thrust Angle = 28 degrees Max Time = 3.6 s Max Angle = 49 degrees This is what i have so far. We stood 50 m away horizontally from...

so how would I calculate that into when I'm trying to find the acceleration during thrust?

I don't know the weight of the rocket its not given. and I can't use forces and all that. I have to use ONLY kinematics

k my main problem is i need to find the maximum height of the rocket i launched using kinematic equations. I'm assuming i use d=vt+0.5at^2 for the distance during thrust and then again after thrust to its maximum height. Then I would just add these two distance and get the max height. (btw i...

I just have a quick question. If an object is moving upwards at constant acceleration, does it take into account the acceleration of gravity pulling downwards? For example if I have to find the acceleration of a rocket during its thrust. Would its acceleration include the force of gravity...

This is the diagram. http://tinypic.com/r/2i7r2pg/3

A weight of mass 12.5 Kg hangs from very light pulleys as shown in fig. What force must be applied to the rope so that the mass stays at rest. M=12.5 Kg W=mg a=g=9/8 m/s^2 M=12.5 Kg W = mg = 12.5 x 9.8 = 122.5 N I've done this much but I have no clue what to do now. Can someone please...