Recent content by Zashmar

So Rh has to equal Ro? wow thanks man

I am a high school student, how would i simplify that, and in terms of maximum efficiency of a wire what does this mean? ie in terms of matching impedance

Here is the working leading up to this: The circuit is a series circuit, with Ro being the circuit's internal resistance and Rh being the heater's resistance. We're only concerned about the power transfer to Rh. We have an input with a constant voltage V. Ohm's law: V = IR, where R is the...

Well I will show you how i did it because I cannot find an error: i j k 0-3 1 2-1 1 ^that is a matrix So i got: -2i+2j+6k =Sqrt 44 =2 sqrt 11

Homework Statement Find the are of the parallelogram ABCD where A is (1,2,-3), B is (-1,3,-4) and D is (1,5,-2)Homework Equations Area=\left|AxB\right| where A and B are the vectors AD, and AB respectively.The Attempt at a Solution I have calculated AD to be= (0,-3,-1) and AB=(2,-1,1) ∴ to...

No I don't mean it like that, I mean if I wanted to define a set of polynomials that had x intercepts between say 38 and 42 on the x axis. In the my particular question i am trying to find a set of polynomials that the particle would land on a 4m^2 square trampoline on the 2d plane,

Okay, So I have attached a screenshot of my two graphs of a particle shot from a cannon. The blue one has had an air resistance constant of 0.1 applied to it and, as you can see, has 'shrunk'. For the particular question I am investigating a range of answers are plausible ( ie the x-intercepts...

The second option :)

Im subbing it into another equation

Homework Statement please help me solve the following, I do not have a good enough calculator :( 15=35 sinΘ(-35 sin\vartheta \pm( \sqrt{(35sin \vartheta)^2 -294)}/-9.8) -4.9((-35 sin\vartheta \pm( \sqrt{(35sin \vartheta)^2 -294)}/-9.8)^2 It is a massive equation, I tired using wolfram...

all terms except 35cos theta, should be divided by -9.8. I tired to show the parentheses but it wouldn't work

Okay So I have it now, how would i simplify the following (I have subbed it into x=35cos\vartheta*t)? x=35cos\vartheta(-35 sin\vartheta \pm \sqrt{(35sin \vartheta)^2 -294}/-9.8 I tried to put in the parentheses but it would not let me

15=35 sin (Θ)*t -4.9t2 a=35 sinΘ b=-4.9 c=-15

So... x= (4.9 \pm \sqrt{24.01+2100 sin\vartheta}/(70 sin \vartheta)

I know the quadratic formula, but how will I do it if j have theta as well?