Recent content by Mushroom79

Ah, I may be a little too fast sometimes. Thank you both for great help :smile:

I see, so: F = -7,290*10^-14 F/Fx = 8,6/0,7 Fx = (-7,290*10^-14)/(8,6/0,7) = -5,934*10^-15 N F/Fy = 8,6/0,5 Fy = (-7,290*10^-14)/(8,6/0,5) = -4,238*10^-15 N F = (5,934*10^-15 N, 4,238*10^-15 N )

Okay, let's say I want to calculate the force on Q2. To get the magnitude (length) isn't it the same as F = squarerot(5^2+7^2) = 8,602 mm ? After that I use the uniformity F/Fx = F/0,7 F/Fy = F/0,5 Not sure how I know what axis (x or y) should be negative or positive?

Updated picture with direction of the force vector I need (acting on Q2): http://imgur.com/c7hu8 I'm trying to calculate the vectorforce F Q1, Q2 = squarerot(5^2+7^2) = 8,602 mm would be the distance between the 2 charges

Homework Statement http://imgur.com/opD71 Q1 = -2 C Q2 = 3 C Decide the forcevector F Homework Equations F = (k*Q1*Q2)/r^2 k = 8,99*10^9 Nm^2/C^2 The Attempt at a Solution Q1, Q2 = squarerot(5+7) = 9,24 mm = 0,924 cm = 0,00924 m F = (8,99*10^9*-2*3)/0,00924 = -5,838 * 10^12 N...

Okay, thank you for the help :smile:

Oh, I forgot about that Probably not the right solution considering the answer: F = m*g = 1,672*10^-27 * 9,82 = 1,642 * 10^-26 E = F/Q = (1,672*10^-26)/(1,602*10^-19) = 1,044*10^-7 E = U/d --> U = E*d = (1,044*10^-7)*2 = 2,088*10^-7

Homework Statement A proton just floating in a field in a between two boards (g=9,82 m/s^2) What is the voltage U? distance d = 2 m charge for protone Q = 1,602*10^-19Homework Equations E = F/Q E = U/dThe Attempt at a Solution I've tried two different ways of which I don't know is correct...

Ah! That gives me almost 16 sec, which makes more sense Thank you!

Homework Statement Friction=0.8 r=50 m T=? Homework Equations T=distance/speed Perimeter=2*pi*r Fc=(mv^2)/r Ffriction=m*g*friction The Attempt at a Solution Perimeter 2*3.14*50=314 Fc=Ffriction (mv^2)/r=m*g*friction v=r*g*friction = 50*9.82*0.8=391.8 T=314/391.8 = 0.8 s Is this...

Ah, but that is the part I get stuck on, √25^2+(20-10t)^2

Homework Statement An object moves so it's coordinates at the time t is given by the relationships x = 25t y = 20t-5t^2 What is the object's speed and direction at 3 sec? t = 3 sec Homework Equations v = √(dy/dt)^2 / (dx/dt)^2 Pythagoras theorem The Attempt at a...

So I take the value of t I got from finding out the flying distance and divide it by 2? If that is correct I got my answer! Thank you.

It's above and between the two points.

A quadratic curve?