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Homework Statement Car Dynamics f(t)→ \frac{\frac{1}{M}}{s+\frac{D}{M}}→y(t) Applied Force Velocity Homework Equations M=1,000 kg and D=1000 kg/s Where f(t) represents the input force and y(t) is the output velocity. M is the Mass and D...

Okay, for the grand finale: I1=30.71∠-47.49° A I2=30.71∠169.38° A Thanks a ton for all of your help and for not letting me drown out in the open ocean, LOL. I can see that I would have gotten the incorrect answer anyways even if my equations were correct to begin with. I had errors in...

After changing that induced value in the primary I have: (1+j4)I1+j2I2=100 and J2I1+(-1.2+j2.4)I2=0 Thanks again for all of the work that you have put into helping me.

Is the correct answer: I1=37.21ang-82.875 deg A I2=27.735ang-70.56 deg A Look correct?

I initially have 100-1I1-j4I1+j2I2=0 and then I consolidate and move everything on one side except for the 100 and solve simultaneously...should I rearrange my equation?

(1+j4)I1-j2I2=100 In the Primary j2I1+(-1.2+j2.4)I2=0 In the Secondary

Okay, I got I1=24.5∠-75.96° I2=0.5∠-90° I double checked what I had with dot notation. I would have +M for the primary because my current goes into j2 in the secondary at the dot, and goes from + to - for a voltage drop across the inductor j4 for itself voltage drop. Then for the...

Ok, thanks. I'm going to grab some dinner and then I'm coming back to work on this some, as well as cover everything else for this course' final. Thanks a lot, I played around with the transformer equations a little, letting everything in the secondary equal ZL, minus the inductor equations...

Thanks for your help. Okay, so combining the loops I2 and I3 I have the 2Ω resistor and -j2Ω in series and then in parallel with the 2Ω resistor which totals (1.2-j0.4)Ω in place of the original 2Ω resistor and now that give me the following mesh equations to solve: For I1: I1(1+j4)-j2I2=100...

When I apply the equations in phasor domain V1=jω(L1)(I1)±jωM(I2) and V2=jω(L2)(I2)±jωM(I1) and then mesh analysis should I consider the other mesh on the far right side with the capacitor? In other words, this is the first problem that I've seen or had such as this and since I2 is a...

Thanks Pilgrimman. You are correct; this is how the professor demonstrated the solution to the problem on the board. He used KCL, showing the 6A and 4A in was 10A out. Ix was equal to 10A and then this 10A multiplied by the 1.5 gave the 15volts. Then he skipped straight to 20(6)=120watts...

According to the professor, he says that the correct answer is 120watt; does anyone else get 120watt? Where did I go wrong in my solving?

Thanks Pligrimman, So with Kirchoff's current law (4+6) gives me Ix=10. Then 1.5Ix=15V. The potential difference in voltage is 15V-5V=10V across B. Then, since the current across B is 6A and the voltage is 10V, I would have 60W absorbed by B? Does this sound correct? I got a little...

Find total Power absorbed or delivered to Element "B". I'm supposed to find how much power was absorbed or delivered by element "B". What is the thought process/solving technique for this type of problem? This was a quiz problem, it has already came and gone; if I see a problem like this...

Please, can you guys offer me any guidance? After I find the derivative and plug it into the original DE and I solve the solution for y and plug it into the original DE I'm not getting the solution to be correct.

I apologize...I found my mistake. The common denominator is X2-3x+2 which of course factors to (2-x)(1-x). I then multiply the numerator and denominator by the common denominator and put everything together. I then multiply both sides by (2-x)(1-x) to solve for dX/dt which leaves me with...

Thanks for the reply. I did that and my common denominator is (2-x). With that, I solve for dX/dt and I get dX/dt=1/(2-x). I plug the dX/dt into the original equation which looks like 1/(2-x)=(2-x)(1-x) the (2-x) cancels from both sides and I am left with 1=1-x and then x=0. I...

Homework Statement Verify Differential Equation Solution dX/dt=(2-x)(1-x) Homework Equations The solution is ln ((2-x)/(1-x))=t The Attempt at a Solution The derivative of the solution is -1/(2-x) dX/dt - (-1)/(1-x) dX/dt=1 But I plug this derivative in and I get stuck. How am...

Okay, I figured the derivative of y2=C1(X+(1/4)C1) to be dy/dx=C1/2y But, now I can't verify the solution by plugging the derivative back into the equation, I can't get both sides equal to each other, which is what the problem is asking for. A differential equation with a solution was...

In my latex from my previous post it is actually supposed to be y prime where y is raised to the first power. But yes, this is how it was in the textbook... Thanks for your reply...

Sorry it's not the best Latex, I hope that you can still help me grasp this. y=2xy1+y(y1)2; y2=C1(x+1/4C1) So, the solution says to implicitly differentiate and gives y1=C1/2y So, how did they get the derivative to be this? This is the first chapter in my DE class and I'm rusty with my...

Thanks for your help jacobrhcp! The book's answer is correct, but mine is not. I need to reverse the order of my three matrices. After rearranging, I get the correct answer. My answer should be 0010 1000 1000 0100 0010 0100 1000 0100 0001 0001 0001 0010 And then when multiplied...

The given matrix is 0100 0001 1000 0010 then R3<->R1 1000 0001 0100 0010 then R2<->R3 1000 0100...

Homework Statement Write the given permutation matrix as a product of elementary (row interchange) matrices. \begin{array}{ccc} 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \end{array} The Attempt at a Solution I found the row echelon form to be the identity...

Okay, I understand now. I appreciate you pointing that out to me. I did get a positive 1.846A, but I entered a negative into my table and I typed from the table. Thanks for catching my mistake. Matt

Thanks, Alphysicist. You were saying that the middle branch didn't look correct...so, I make I_4ohm flow towards the 3V source making the negative of the resistor closest to the 3V battery. Then, I_5ohm+I_5ohm=I_4ohm?

I've gotten the correct answers, I do believe. left loop equation: -12-5I_1+4I_2+3=0 right loop equation: -3-4I_2+5I_3+18=0 node equation: I_1=-I_2-I_3 I_5ohm= -0.323A I_2ohm= -1.846A I_3ohm= -1.52 I_2ohm= -1.52 V_5ohm= -1.62V V_2ohm= -3.05V V_3ohm= -4.57V V_4ohm=...

Thanks. So far I've figured out that my current at I_1 was headed the wrong direction and you showed me that I had my polarity of the resistor incorrect since the voltage decreases as the current flows through the resistor. I have found I_1 to be -0.323A or -323mA and V_5ohm to be 1.62V. So...

I'll look over my notes and in my book about how I might do this. I think that I'll be doing, for example, I_1=I_2+I_3, something like this. I just checked the image and it is now available, but I had to log in...I guess that it is just available to me at the moment. Thanks for the...

Homework Statement Find the potential differences across the 5 ohm, 4 ohm, 3 ohm, and 2 ohm resistors. Homework Equations 5I_1-4I_2+3-12=0 Left Loop 3-18+5I_3-4I_2=0 Right Loop 12-18+5I_3-5I_1=0 Whole loop The Attempt at a Solution I simplified the first diagram...

Thanks so much! You guys have all been a great help. This is been some great information. I'd like to find a geometry class to go to now, just to get good. It looks like all of these theorems would be very useful in trig, not to mention real situations where this information could come in...

Thanks to everyone for all of the help! Here is what I have gotten, hoping that I've gotten them correct! Angle 1= 12 deg Angle 2= 78 deg Angle 3= 12 deg Angle 4= 90 deg Angle 5= 30 deg Angle 6= 30 deg Angle 7= 30 deg Angle 8= 30 deg Angle 9= 40 deg Angle 10= 60 deg...

So Snipez90, Angle seven= 30 deg, because the arc it intercepts, ED, is 60 deg? Which with the right angle, the three angles of BDE, are 30 deg, 60 deg, and 90 deg?

Or would angle one and angle three be = 4 deg? Since I only need half of the x=1/2(CD-AB) for angle one and angle three?

So, would that make angle one = 8 deg? I took angle one=.5(156-140)=8 deg. Then angle 3 is equal to angle 1, right?

Just wandering, I'm about to begin Cal III, I don't ever remember seeing a problem like this. I found the problem in a tool and die mathematics question book with no explanation of finding the answer. I'm wandering if anyone might be able to help me see how the problem should work. In high...

Thanks HallsofIvy, once again, for the great help. These forums are great to get the wheels turning!

Homework Statement I've uploaded a scan of the questions. Questions 4, 5, and 6 are given in the 3 files uploaded. They all come from the given information from the first scan of the problem. Homework Equations The Attempt at a Solution I've worked everything I could on paper...

Thanks! That was a dandy, for me anyways! I'm sure this practice will help me recognize these sequences a little better in the future. Thanks for sticking in there with me.

does this look correct for a_{n}=4(\frac{n(n-1)}{2})+4

I believe that the first one is 6-4=2 and 7-4=3, right?

I got 12 and 24?

Gentlemen, I'm totally stuck. I sat here for the last 1.5 hrs and tried to figure this one out. I think that I'm making it too difficult. I've looked at the sequence all the way up through a[sub]6 and I'm just not seeing the formula. I do know that to find the sum from 1 to 1000, would be...

[SOLVED] Series/Sequence Problem? I'm trying to figure out a formula for this sequence problem, rather than doing this over and over 1,000 times. Does anyone have a clue for the formula to find this? I've pasted the info and also the answer. Thanks for your time!

I found my error on problem #2. It should be equal to 0 after rewriting the form of the limit because it was indeterminant. I got it finalized and came up with 0 for the answer to number 2. Thanks HallsofIvy.

Am I going about the solutions correctly now?

#1. I should have gotten 0/2 for #1, because cos(pi)=1 and 0/(1-(-1))=0/2 I thought that 0/(any nymber greater than or less than 0) is equal to 0, but if the fraction was 2/0, there would be an infinite, positive for the limit approaching from the right and negative infinity for numbers...

[SOLVED] One-Sided Limits Homework Statement I've been having problems solving for these one-sided limits. 1.\stackrel{lim}{x\rightarrow\pi^-}\frac{sinx}{1-cosx} 2.\stackrel{lim}{x\rightarrow\O^+}xln(x) 3.\stackrel{lim}{x\rightarrow\frac{\pi}{2}^-}(sec (x)-tan(x)) Homework...

I believe my 2 should have canceled out with the \frac{1}{2} I'm now getting the answer =sin^-^1x+\sqrt{1-x^2}+c After following your help, I did get the integral that I started with after I split it into two fractions. Thanks for your help and for catching my mistake!

Thanks to all, does this look correct? This is what I got: =sin^-^1x+2\sqrt{1-x^2}+c Thanks to everyone!