Recent content by xanthym

To supplement O-Dan's circuit info, the following link is specific to Bipolar Junction Transistors (BJT), and in particular, covers their special switching applications (which is probably what you're looking for). Check the links below. (First link is main article, in which you'll need to...

SOLUTION HINTS: This problem is designed to distinguish between RMS and Peak values of Voltage, Current, and Power for a purely resistive load. We are given: {RMS Voltage} = (240 V) {RMS Power Rating for Resistive Load} = (1000 W) For standard AC systems, we know that: {RMS Power} = {RMS...

Good observation. "Sunshine" should make that correction for the final result (see Msg #2). ~~

SOLUTION HINTS: You did very well so far! When indexing a vector result, remember to indicate the LEFT side component with the FREE index ("i" in this case). Thus, your result should have been (note the "i" subscript added on the LEFT side): (Note: To save time, the constant fraction...

For Problem #1, solve for "c" which makes the estimator unbiased, which (in this case) involves setting the Eq #2 integral equal to (1/θ). See Problem #1 statement for other info. An estimator \hat{\omega} of distribution parameter \omega is Consistent if 2 conditions are satisified: 4...

SOLUTION HINTS: For both cases, an Unbiased Estimator \hat{\omega} of distribution parameter \omega satisfies: 1: \ \ \ \ \ \ \mathbf{E}(\hat{\omega}) \, \ = \, \ \int \hat{\omega} \, f(y; \, \omega) \, dy \, \ = \, \ \omega \ \ \ \ \ \ \mbox{(Unbiased Estimator)} where f(y; ω) is...

Now solve the equation for spherical balloon radius "R": {Weight of Helium} + {Weight of Balloon + Load} = {Weight of Displaced Air} ::: <--- Buoyant Force ::: ⇒ {ρhelium*Vballoon*g} + {5905 N} = {ρair*Vballoon*g} ::: ⇒ {ρhelium*(4*π*R3/3)*g} + {5905 N} = {ρair*(4*π*R3/3)*g} ~~

HINT: Since fluid pressure ⊥ Hatch Surface: 1: \ \ \ \ \textsf{Torque} \ \, = \, \ \int_{Hatch} r \, P \ dA where "r" is the distance of Area Element "dA" from the Reference Point or Axis, and "P" is the fluid pressure on Area Element "dA". Both "r"and "P" will be functions of Area...

{Volume of Container} = V = (24.8 liter) {Moles of Water in Container} = n = (5.7 grams)/(18 grams/mole) = (0.3167 moles) {Temperature} = T = (115.8 degC) = (388.9 degK) {Ideal Gas Constant} = R = (0.08206 Liter*atm/(mol*degK)) Use the Ideal Gas Law to calculate Pressure "P" in units of...

When the valve is opened, the situation becomes {P2 = P1}. Recalculate your results using this fact together with the other given values. ~~

The graphs should be different. Unfortunately, without seeing what you've done or the results therefrom, it's difficult to help. Can you provide more details concerning your transformations and provide images of the graphs?? To provide images of your graphs, upload to the site shown below...

O-Dan: You're certainly correct that there might be some confusion regarding notation in this thread. Nevertheless, the objective of Msg #9 (and subsequently of Msg #11) was to present this concept: 5: \color{red}\ \ \ \ \ \ \left | \begin{array}{ccc} a_{1} & a_{2} & a_{3} \\ b_{1} &...

SOLUTION HINTS: Factor 720 to help find the solution: 720 = (24)*(32)*(5) = n*(n - 1)*(n - 2) = (?)*(?)*(?) (Hint: Try 23) ~~

. For Mathman23: *** POP QUIZ *** Given two (2) vectors {V1, V2 ∈ \mathbb{R}^{3}}, state a necessary and sufficient condition that these vectors are Linearly Independent. CLICK BETWEEN DASHED LINES BELOW To Reveal Answer In Pop-Up Window --------------------- \color{white}...

Three (3) vectors in \displaystyle \mathbb{R}^2 are NEVER Linearly Independent. ~~