i figured it out thanks to oxygen's logic...I checked all 32 combinations for all the letters related to the problem...and all of the outputs came out correct.
final condensed form: c(de + da + ab) + b(ae + de)
heres my diagram:
original form: ##ab \bar ce## + ##b \bar cde## + ##cde## + ##abc## + ##acd \bar e##
condensed form: ##b \bar c(ae + de)## + ##c(de + ab + ad \bar e)##
partial circuit diagram using NANDS ##c(de + ab + ad \bar e)##:
I made that portion with 7 NANDS...I still have to finish the ##b \bar...
thank you. I have one more expression for practice problems:
ab~ce + b~cde + cde + abc + acd~e (The ~ symbol represent "not")
I'm supposed to do this expression in 12 gates.
I currently have the last 3 condensed to: c(de + ab + ad~e)
The first two: b~ce(a + d)
so the final...
Here is what I got....this is a practice problem to prepare for a midterm, but I would like to see how you used 6 NAND gates.
http://img827.imageshack.us/img827/7766/schematic.jpg [Broken]
I'm trying to convert this expression: ga + za + sgz
using just 2-input nand gates...more specifically the 7400 ic chip.
I'm trying to use as little NAND gates as possible. I've got (ga + za) down to 5 NAND gates currently...I can only use 8 total NANDS for this.
Homework Statement
I'm trying to convert the 3-input AND gate shown below using only NAND gates...but am having troubles. Is it possible to use only 2 NANDS for the conversion?
http://www.doctronics.co.uk/images/4081_03.gif [Broken]
I'm intending on creating a solenoid that will project a bb for a class project. I'm using a flash capacitor from a disposable camera as my current source. Here is my question...what is the best "setup" for the solenoid to increase the magnetic force on the bb without increasing its current...
yes, the force applied depends upon its position since we are dealing with electromagnetic force. It seems without an equation given...I would have to calculate the variables separately per second, which won't be as accurate.
Homework Statement
I'm not sure how to set up the differential equation. I've got 2 point charges...both positive. One charge is fixed in position (0, -250m)...the other is travelling at an initial speed in the -x direction (10,000m, 0) with 0 acceleration. At time t=0, I calculated the...
Homework Statement
Lets say I have a .5kg mass traveling at 1200 m/s in the -x direction with 0 acceleration. The force applied to this mass is (2.3i + .093j)N. I have to find position, velocity, and acceleration at say 1 second after the force is applied. No gravitational force involved...
Homework Statement
A water balloon takes .22 s to cross the 130 cm high window, from what height above the top of the window was it dropped?
The Attempt at a Solution
I'm using:
v^2 = vo^2 + 2a(x - xo)
v = distance/time
v = 1.30 m / .22 s = 5.91 m/s
(5.91 m/s)^2 = 0 +...
ok so...if you integrate a(t) = 2t...just becomes v(t) = t^2...
v = vo + at
would this be the correct approach?
21 m/s = 3 m/s + 2t(t)
18 m/s = 2t^2
9 m/s = t^2
3 seconds = t
i understand it goes
x(t) = position
x'(t) = velocity
x''(t) = acceleration
to get the velocity from the acceleration just integrate...i understand that...but setting up the problem is the thing i'm having difficulty with even though it is something simple...the thing that makes me...
ok...where does the initial velocity get plugged into that equation though.
When you integrate dont you just get 1/2*at^2
I got caught up on the cubed part of the acceleration...because im used to seeing it squared...like 9.8 m/s ^2.
Homework Statement
An object is traveling in a straight line with an initial velocity of 3 m/s and an acceleration a(t) = st, where s = 2 m/s ^3 and t is measured in seconds. Find a time T such that v(T) = 21 m/s.
I wanted to use v = vo + at ... to find t...but the function of a(t) is...
thanks guys...i knew it was something simple.
The original problem was:
\int(csc^4 3\theta)(tan^4 3\theta)
simplified this to:
\int\sec^4 3\theta
and now i know how easy the rest was.
Homework Statement
Using this polar equation:
r = \theta + sin(2\theta)
Find the angle \theta that relates to the point on the curve when x = -2
I'm not sure where to start...but my guess is to convert the equation to another form...any help is appreciated.
yep...i just concluded guys that I was inputting the equation wrong into my calculator...4.38 is the right answer and i was doing it right by hand all along...what a relief.
I have one more question though...
how do I find the angle at which the graph is at x = -2 ?
I'm attempting to solve this area problem
1/2\int_{0}^{\pi }(\theta + sin2\theta)^2 d\theta}
The area found by my calculator comes out to be 4.93...but by hand I get 4.38
The original polar equation: r = \theta + sin(2\theta) from 0 to pi.
I think it may by the use of my input...
\theta^2 + 2\theta\sin2\theta + sin^2(2\theta)
after integrating....
\frac{1}{3}\theta^3-\theta\cos2\theta+ 1/2sin2\theta + \frac{1}{2}\theta - \frac{1}{8}sin4\theta
i forgot to integrate the last part...but this still doesn't seem correct
Homework Statement
trying to integrate this...the second term is the difficult one here.
\theta^2 + 2\theta\sin2\theta + sin^2(2\theta)
The Attempt at a Solution
I attempted the problem and ended up with this but it doesnt seem right...