Recent content by Civil_Disobedient

The integral here makes sense, but shouldn't the boundaries of each integral correspond to the integrals of the sides of the rectangle? i.e bounds of x-axis for part a should be between a and b and the integral for the y-axis be between c and d? The integral form applies to both part a and b I...

Yes, and the z isn't in the numerator

I don't, haven't covered double integrals. Will I be able to do this without double integrals? How would I find electric field in this case? I know area of the rectangle is just length times width, within the x and y bounds. Would those be the bounds of the integral? But there are two sets of...

Homework Statement A vector field is pointed along the z-axis, v → = a/(x^2 + y^2)z . (a) Find the flux of the vector field through a rectangle in the xy-plane between a < x < b and c < y < d . (b) Do the same through a rectangle in the yz-plane between a < z < b and c < y < d . (Leave your...

I see, that makes sense now. Thank you so much!

Ah, I forgot to write in exponents for acceleration. a = 3.83*10^13. I also missed out how charge of a proton is positive, so acceleration would be positive. So from there I recalculated y=1/2at^2 y = 1/2 * (3.83*10^13) * (8*10^-9) = 0.00123m I'm a bit confused on finding the height...

That's just a basic kinematic equation, I may have incorrectly remembered it. I ended up using y=1/2at^2 to calculate height.

d = vt t = d/v = (12*10^-2) / (1.5*10^7) = 8*10^-9s Using kinematic y = yo + vo + 1/2at^2, I removed the yo and vo to get y = vo + 1/2at^2. I first solved for the proton without the electric field y = 1/2at^2 y = 1/2(9.8)(8*10^-9)^2 = 3.136*10^-16 Next, I solved for the proton with the field y...

Homework Statement A proton enters the uniform electric field produced by the two charged plates shown below. The magnitude of the electric field is 4.0 × 105 N/C, and the speed of the proton when it enters is 1.5 × 107 m/s. What distance d has the proton been deflected downward when it leaves...

Okay, so that force would be the horizontal force of tension. This creates a right triangle with angle 5 degrees and a hypotenuse of 0.049N, which is the tension. Using trig, 0.049N * sin(5) = 0.0043N (horizontal tension). From there, I ended up with 6.01 * 10^-8C for charge. I used the...

I figured out some of the incorrect unit conversions I made. 50 cm = 0.5 m, so r = 0.087. 5g = 0.005 kg --> 0.049 N But I don't quite get the part about torque. I've tried torque = rFsinθ. For q = sqrt(F*r^2 / k), shouldn't I use force of tension in the diagonal direction instead of torque?

Homework Statement Two small balls, each of mass 5.0 g, are attached to silk threads 50 cm long, which are in turn tied to the same point on the ceiling, as shown below. When the balls are given the same charge Q, the threads hang at 5.0° to the vertical, as shown below. What is the magnitude...

1. Here's the problem on trig integrating that I'm struggling with (Calculus 2 btw) 2. Wanted to see if I did everything right so far and what to do after all this. The part where I'm stuck is how to integrate (integral)cos^(2)udu and (integral)cos^(2)usin^(2)udu. I'm sure these are easy...

Homework Statement A person runs at a constant speed of 4.0 m/s to catch a stationary bus. When she is 6.0m behind it (t=0s), the bus leaves, accelerating with a (constant) acceleration of 1.2 m/s^2. How long does it take her to catch up to the bus? Homework Equations X = Xo + Vot + 0.5at^2...

Homework Statement Because of your technical background, you have been given a job as a student assistant in a university research laboratory that has been investigating possible accident-avoidance systems for oil tankers. Your group is concerned about oil spills in the North Atlantic caused by...