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1. ### Electric field of a washer (hollow disk)

I was attempting to get the E field of a disc with radius b--aka integrating from radius of 0 to b, but I don't think that's right. I'm confused by what you mean by "the general expression as a function of a and b." I don't understand how I can get a and b in the same equation... From the E = k...
2. ### Electric field of a washer (hollow disk)

Ok so if I use E = k ∫(σ/b²) dA (from kq/r2 and Guass' Law using r=b) where dA=dθdb and θ is the angle between the point z and a point along radius a, I can differentiate it from a →b and from θ=0→360°. Is that what you mean?
3. ### Electric field of a washer (hollow disk)

Homework Statement A washer made of nonconducting material lies in the x − y plane, with the center at the coordinate origin. The washer has an inner radius a and an outer radius b (so it looks like a disk of radius b with a concentric circular cut-out of radius a). The surface of the washer is...
4. ### Solving for Min Speed on Loop-the-Loop Rides

The normal force will equal the force of gravity, m*g?
5. ### Solving for Min Speed on Loop-the-Loop Rides

Homework Statement In a loop-the-loop ride a car goes around a vertical, circular loop at a constant speed. The car has a mass m = 260 kg and moves with speed v = 15 m/s. The loop-the-loop has a radius of R = 10 m. What is the minimum speed of the car so that it stays in contact with the track...
6. ### Two accelerating crates

Great, thank you so much for all your help (and your patience...)!
7. ### Two accelerating crates

So would I just use F=m1a to find the acceleration? Aka: Ffr=m1a --> 123.48 N = (20 kg)a --> a=6.17 m/s2
8. ### Two accelerating crates

The forces acting on the top crate are the force of friction from the bottom crate, the force of gravity downward, and the normal force upward, correct? So the sum of the forces in the horizontal direction would be... only the force of friction?
9. ### Two accelerating crates

Alright so I used Ffr=μm1g using μk=0.63 and got Ffr=(0.63)(20 kg)(9.8 m/s2) = 123.48 N This would be the force of friction acting on the top crate, so would I then use Ffr=Ft-Fnet to find the Fnet, the sum of the forces acting on the top crate? I don't see another way to incorporate the new tension
10. ### Two accelerating crates

Hm... I suppose it would be the force of friction? In that case, would I use Ffr=μmg using μk?
11. ### Two accelerating crates

Homework Statement Two wooden crates rest on top of one another. The smaller top crate has a mass of m1 = 20 kg and the larger bottom crate has a mass of m2 = 80 kg. There is NO friction between the crate and the floor, but the coefficient of static friction between the two crates is μs = 0.79...