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1. Simple Coulombs Law Problem

alright got it now, thanks a lot for your help. i was about 80% through typing out my reworking to see if you could check it when i accidentally clicked the back button on my mouse and lost it all, and i really can't put myself through typing it again lol.
2. Simple Coulombs Law Problem

1. Homework Statement Three point charges lie along the x axis. A positive charge q1=16.0mC is at x=2.00m. Another positive charge q2=9.00mC is at the origin. Where should we put a third charge so the resultant force acting on it is zero? What should be the sign of the charge? 2...
3. Sliding wire problem

really confused sorry :(
4. Sliding wire problem

9.8Msin180=ILBsin90??
5. Sliding wire problem

and the current would go to the left, determined by the right hand rule?
6. Sliding wire problem

ok so force pulling the wire down: F=masinθ = -9.8Msinθ so for the rest: F=ILBsinθ 9.8Msinθ=ILBsinθ 9.8M=ILB I=9.8M/LB ??
7. Sliding wire problem

1. Homework Statement A straight piece of conducting wire with mass M and length L is placed on a frictionless incline tilted at an angle theta from the horizontal. There is a uniform, vertical magnetic field vecB at all points (produced by an arrangement of magnets not shown in the...
8. Current carrying wire

ΔB = μo/4π(IΔLsinθ)/R^2 so I = 27 θ = 45 ΔL = 2mm??? and R can be found using trig. so what is μ and o?? soz just really lost. any help would be great.
9. Current carrying wire

1. Homework Statement A wire carrying a 27.0A current bends through a right angle. Consider two 2.00mm segments of wire, each 3.00cm from the bend. Find the magnitude of the magnetic field these two segments produce at point P, which is midway between them. 2. Homework Equations...
10. Photon Magnetic field question

that was correct :) thanks heaps for your help
11. Photon Magnetic field question

r = mv/qb 0.008276 = (1.67x10^-27 x 1200) / (1.602×10^−19 x b) (1.602×10^−19 x b) = (1.67x10^-27 x 1200) / 0.008276) (1.602×10^−19 x b) = 2.42146x10^-22 b = 1.5115x10^-3
12. Photon Magnetic field question

would i be correct in saying that using the arc length formula: pi/2 radians is equivalent to 1.3cm θ = s / r r = s / θ = 1.3 / pi/2 r = 0.8276 cm t = 0.013m / 1200m/s = 1.083x10^-5 s then w = θ / t = (pi/2) / 1.083x10^-5 = 145041.2121 w = w0 + at 145041.2121 = 0...
13. Photon Magnetic field question

yeh i was about to ask how you knew the radius. ok i'll work on those and get back to you in a few mintues. **digs deep into the memory bank for equations**
14. Photon Magnetic field question

i got 1.129x10^-13 and failed my first attempt.
15. Photon Magnetic field question

ok so: F = qv x B B = F/qv = F / (1.602 × 10^−19) x 1.2 (charge of a proton x velocity of proton) F = ma? = 1.67x10^-27 x a (mass of a proton x acceleration) however, as: m(v/t) / q(v) = m(1/t) / q = (1.67x10^-27 x (1/t)) / 1.602 × 10^−19 (where t = distance / velocity??)
16. Photon Magnetic field question

1. Homework Statement A beam of protons traveling at 1.20 km/s enters a uniform magnetic field, traveling perpendicular to the field. The beam exits the magnetic field, leaving the field in a direction perpendicular to its original direction. The beam travels a distance of 1.30 cm while in...
17. Special relativity assignment topic help

hey all, i've recently been given an assignment of special relativity, the question is as follows:- Topic B: SPECIAL RELATIVITY Write a short report on any experiment, published in the literature (ie journals), which proves or depends on a relativistic effect. Include essential experimental...
18. Spring-loaded toy gun problem

similar question... A toy gun uses a spring to project a 5.3 g soft rubber sphere. The spring constant k = 8.0 N m-1. The barrel of the gun is 15 cm long and there is a constant frictional force of 0.032 N between the barrel and the projectile. If the spring is compressed 5 cm what is the speed...
19. Toy gyroscope problem

1. Homework Statement The rotor (flywheel) of a toy gyroscope has mass 0.140 kilograms. Its moment of inertia about its axis is 1.20x10^{-4} kilogram meters squared. The mass of the frame is 0.0250 kilograms. The gyroscope is supported on a single pivot with its center of mass a horizontal...
20. Spring-loaded toy gun problem

ok great thanks heaps. got it correct, and finished off the next 3 questions which required that answer also.
21. Spring-loaded toy gun problem

hmm ok. Initially Ei = Ki + Ui = 0 + (Ugravity + Uspring) = 0 + (0 + 1/2kx^2) = (1/2)667 x .25^2 = 20.84375J Final Ef = Kf + Uf = (1/2)mv^2 + (Ugravity + Uspring) = (1/2)mv^2 + (mgh + 0)...
22. Spring-loaded toy gun problem

So potential energy = kinetic energy? like: kx = .5mv^2 667x0.25=.5(1.5)v^2 166.75=.75v^2 v^2=222.33 v=14.91m/s
23. Deceleration problem!

a) try the formula u^2 = v^2 + 2as where u=initial velocity, v=final velocity, a=acceleration (will be a negative), and s=the distance traveled. b) try using v = u + at using the final velocity that you found in section a. t will give you time.
24. Spring-loaded toy gun problem

1. Homework Statement A spring-loaded toy gun is used to shoot a ball of mass m= 1.50kg straight up in the air, as shown in the figure. The spring has spring constant k = 667N/m. If the spring is compressed a distance of 25.0 centimeters from its equilibrium position y=0 and then released, the...
25. Tension of a cord problem

just subbed those in, and for Tb i got w/1.115? same as in example a. and then for Ta = w/0.724
26. Tension of a cord problem

horizontal component: -Tacos60+Tbcos45=0 Tacos60=Tbcos45 Ta=Tb(cos45/cos60) =Tb(2^.5) vertical component: -Tasin60+Tbsin45-Tcsin90=0 -Tb(2^.5)(sin60)+Tbsin45=w 0.707Tb-((2^.5)(sin60))Tb=w Tb=-w/0.5176 which i put in and was wrong. so i misc decided to make the a component positive and put...
27. Tension of a cord problem

sweet thanks. got them both right. now need to try for part b :p do i apply the same rules or does it change up since one cord is coming from the side now?
28. Tension of a cord problem

ok so say i take the first of the 2 options: x component: Ta+Tb+Tc=0 Ta+Tb=0 (as there is no x direction of Tc) -Tacos30+Tbcos45=0 Tacos30=Tbcos45 Ta=Tbcos45/cos30 y component: Tasin30+Tbsin45-Tcsin90=0 (Tc is in the negative direction?) Tb(cos45/cos30)(sin30)+Tbsin45=w (subbing Tc for w...
29. Tension of a cord problem

ha i know. nothing wrong with starting now and seeking a little help to get on top of it though right?
30. Tension of a cord problem

ok so i'm trying to follow an example from my textbook, but it's only dealing with 2 strings so that is probably where i am getting stuck. Ta+Tb+Tc+w=ma as net force is 0, ma = 0. then into x and y components: x: Tax+Tbx+Tcx+wx=0 -Tacos30+Tbcos45+Tccos90+0=0 y: Tay+Tby+Tcy+wy=0...