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 im left with m^2/s^2
Sorry for the confusion.
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 isnt allowed. Sorry for wasting space with this topic.
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 cant seem to get i. Any help would be appreciated.
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 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.
This is the problem that I think is similar to the one labeled 3.71. I just want to know if I'm correct in my assumption the two problems have a similar premise.
I have a couple problems that I can't seem to get.
The first I'm really close, I can tell, because my professor did one that was similar in class. The answer the book gives is 250 lbs, I'm coming up with 225.
The second I think is similar to another problem he did in class but I'm not...
Couple of other questions.
1) This is just to make sure that I answered this question correctly, but in the ferrofluid, there are two ions of Fe 3+ and one ion of Fe 2+ present, correct?
2) Having a bit of difficulty understanding this question:
*You prepared aqueous ferrofluid, but the...
Wow, I feel stupid. Aqueous ferrofluid would result in rust, thereby making the ferrofluid inefficient.
I suppose not. A magnet could be placed in the area of the region outside of the body that the ferrofluid needs to enter, just depends on how strong the attraction would be in order to not...
Are you asking how iron reacts with different ions in general or something different? I'm apologize for not understanding the question, but I'm quite awful when it comes to chemistry.
To localize the delivery, wouldn't you need an actual magnetic piece or magnetic attraction for the...
I recently did a lab report on ferrofluid for chemistry and am having difficulty with some questions. I was wondering if I could get some help.
*Are there advantages to an oil-based ferrofluid compared to an aqueous ferrofluid?
*Researchers have explored ways to attach medications to...
No, this makes sense. I actually came up with this stuff like an hour or two ago, but like you said, it looked so simple, I thought that there was no way that it could be right and that I was missing a crucial step and making dumb mistakes like I usually do. Thank you so much for your help. You...