Recent content by J-dizzal

im having trouble solving this. this is my first step from the starting equations; q1+(3.5605E-10/q1)=2.3E-4. and then i factor out q1 from the left side of the equation and solve from there.

I'll try again to solve, this are my 2 equations: q1+q2=2.3E-4 and q1q2=3.5605E-10.

I tried to solve for q1 and q2, and i got q1=0 and q2=23E-5. my answers are true but not the correct solution to this problem.

Homework Statement Two small, positively charged spheres have a combined charge of 23.0 × 10-5 C. If each sphere is repelled from the other by an electrostatic force of 0.800 N when the spheres are 2.00 m apart, what is the charge on the sphere with the smaller charge? Homework Equations...

##\bar x=\frac{1}{A}\int xdA## = ##1/2 \int kx^3dx## does this look ok?

Wouldnt it be easier to integrate with respect to y, because of the shaded region is above the curve.

Ok thanks Dr. Courtney and Nathanael. I now have 38/k. In my final answer the k's cancel out and I am left with the same 7.5/38.7.

(2/3k)(58.1) =(.667k)(58.1) =38.7k where k is a constant. And i treated the other integral similarly.

Homework Statement Find the coordinates of the centroid of the uniform area. Homework Equations equations for centroid coordinates at the top of my paper. The Attempt at a Solution

T2 is in the direction of impending motion, so i assumed it was 190N. Re reading the problem statement the ship is pulling on the rope with a much greater force, so then T2 would be T, the unknown. Thank you Nathanael.

Homework Statement A dockworker adjusts a spring line (rope) which keeps a ship from drifting alongside a wharf. If he exerts a pull of 190 N on the rope, which has 1.25 turns around the mooring bit, what force T can he support? The coefficient of friction between the rope and the cast-steel...

So far, ΣM2=Wcosθ(d2) - Wsinθ(h) - N1(d1+d2). This solves for N1, the normal force at the front tire. Fsfriction=Fgravity on bike parallel to ramp. μs=Fsfriction / N1. This gives the coefficient of static friction for part (a) of the problem statement. part (b) says to double μs, then I...

sum of forces in the x direction; ΣFx=P-(force of friction at A)=0. unknowns;P,NB,NA,and friction force at A. But only 3 equations, moment and two linear force equations.

there is only one case here, this case is when wheel A freezes and there is a moment about A. None of the top wheels would contact their frame because the door is not tipping. It is not tipping because the location of the normal force and friction force are not at the bottom right corner...

This is what i don't understand, if B is free to rotate then how can a moment equation be made? because it only resists movement in the y axis. This is what i learned form previous equilibrium materials, why does it change now? sum of forces in the x direction, does not involve the normal...