Recent content by jnbfive

W=q(Delta V) = Change in kinetic energy, which is (-1.6*10^-19 C)(3000V) = -4.8*10^-16 N*m -4.8*10^-16 N*m = .5*(9.109*10^-31 kg)*(v^2) Since a Newton is kg*m/s^2, the kg's cancel and I am left with m^2/s^2 Sorry for the confusion.

-4.8*10^-16 = .5*(9.109*10^-31)*(v^2) So v = 3.246 * 10^7 m/s?

I need to find the force. Once I find that everything else isn't difficult. Part I) Find the speed of the electron: v=F/(eB) W=q(Delta V) = F*d, where W is -4.8*10^-16 N*m Part II) Find the distance between the been entering and exiting the field. r=mv/(eB) where r is the radius so...

Homework Statement An electron gun (applied voltage of 3000 volts) is emitting electrons. THe electrons enter a region of constant magnetic field (B=.025 Tesla.) The magnetic field is perpendicular to the velocity of the beam. Homework Equations W=Q(Delta V) = F*d V=F/eB The...

Homework Statement I have a problem that I need assistance with: There is a well insulated rigid tank with 3 kg of saturated liquid-vapor mixture of water at 200 kPa. Initially, half of the mass is in the liquid phase. An electric resistor placed in the tank is connected to a 120-V source...

I scanned a sheet and it was saved as a jpeg file. I went to attach it, and the screen that pops up said it was loading and the suddenly went blank. Thats it. For some reason my scanner produced an image size that isn't allowed. Sorry for wasting space with this topic.

For some reason, I am unable to attach anything in a post. I was wondering why this is.

I was given a question about a certain material. The idea was that when it was heated from zero degrees C to 200 degrees C, it would be 10.06 mm from its original length of 10. Doing it in reverse order, length of 10.06 mm from 200 C to 0 C gave me an answer of 9.99964, which is less than the...

This is my work for solving part a) when a=2 Sum of the Moment about A = 20(2)-16(1.5)+[5.5(sin(30))+2(cos(30)]E -8=[5.5(sin(30))+2(cos(30)]E E= -8/[5.5(sin(30))+2(cos(30)] or 3.57 lbs at a 60 degree angle, which is what the book says. Now how come when I plug in 7.5 in a, I can't...

Just figured I'd throw this up here. I was able to solve for the reaction at E when a=2 but when I tried for a=7.5, I wound up with something completely different. I was wondering what I'm not incorporating into the steps to solving this. Also, it says determine the reactions at A and E. The...

In response to my above comment, I really realize that I overlook things too much. 3.94 I have the j and k of the force couple system (use the force that I obtained when breaking down the 220 N in the cross product. Duh.) Somehow I can't seem to get i. Any help would be appreciated.

I'm amazed at my ability to overlook the simple things in problems. Many thanks for your help with that.

If I could actually ask for help in another question, any help would be appreciated. Excuse my chicken scratch work. I solved for the angle at which the force is acting at, which is 30 degrees in the xz plane. I solve for the actually distance from the x-axis that said force was, which was...

*I took the inverse tangent of 11.4/15.2 *The angle of 53.1 is correct; that's the answer the book got. *Why only 15.2? *Thanks

I need help on part c of 3.71. I have the angle from b, which I believe I need to use. I originally thought that the way to set it up would be 86.2 = x(22.36)*sin(53.1) ^22.36 coming from the sqrt of 17.6^2+ 13.8^2 Any help on what I'm doing wrong would be appreciated.