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As Gokul said I'm not going to tell you the answer. I, or many others, can verify your answer though. I chose 74(because the length of the column was 76) to get you thinking along the lines of the Weight of the column of Hg vs the force of the compressed gas pushing back against the Hg.

This has the distinct taste of a homework question and as such should probably have been posted there. Additionally, you'll get a lot more help if you actually ask a question about the problem rather than simply posting the problem. I know I can solve this type of problem; however, simply...

Your parallel equation is wrong: R_{eq}=(\frac{1}{R_1}+\frac{1}{R_2}+\cdots+\frac{1}{R_n})^{-1} So if you say I=\frac{V}{R_{eq}} you end up with: I=\frac{V}{(\frac{1}{R_1}+\frac{1}{R_2}+\cdots+\frac{1}{R_n})^{-1}} or more simply...

First: Read this again "A 1.20 kΩ resistor and a 6.8 μF capacitor are connected in series" Second: what are the impedance equations? Third: How do you deal with resistances in series circuits (impedance is analogous to resistance). Finally: Reread this "Calculate the impedence of...

How is the tube supported? This is a very important detail when determing bending moments. Cantalevered beams will have different support reactions than simply supported beams... Essentially what you need to do is determine the support reactions on the tube as if it were solid; additionally, you...

Pa has units of N/m^2 thus you need a little more info to convert 20mm of gas to pascals.

What resistance? caps should have infinate resistance between the plates. There is an equation relating charge to the distance between the plates isn't there?

nope, "And finally, I calculated I = .39 for copper and .25 for aluminum using V = IR again" Is wrong. Look into how current is effected in serial components. The rest looks good enough for government work.

You DO have the force acting on the carts. Reread the question you started this thread with.

Well, you have 15N acting toward the center of the circular path and 4.905N acting down. Do you know of any pseudo-forces associated with circular motion? Do you know what force keeps a person stuck to the walls of one of those spinning amusement park rides, or the force you feel when a car goes...

Start with a fbd showing the spring compressed and the spring force (the same force) acting on both carts. Do you know, or can you derive, the potential energy equation for a spring? The energy stored in the compressed spring will equale the energy of motion for the carts. You know the force...

what have you done so far? did you draw a fbd with the ball at the bottom of the circle? Did you sum the forces acting on the ball at that point? I'd say start there and see if you get an answer. Good luck.

Here's another approach(KVL which I'd use this for simple circuits): You have 1.91A going through the series 10 ohm resistor thus that resistor is dropping 19.1V. The parallel legs of the circuit will all have the same voltage drop(by KVL) which will be 25-19.1=5.9V so...

What are the requirements for an electric field to be present? A brand new battery sitting on a countertop--is there an e-field sitting at the top and bottom of the battery?

13.06 is right as is the 1.91A Your current through the R leg is wrong though. How exactly did you find that value?

Here: I'll do more work to show you how to do this rather than simply give you the answer... \delta x=0.1 This is the error in x \delta y=0.05 This is the error in y x=10--value found during an experiment. y=20--value found during an experiment. U=x^3+3x^2y+3y^2x+y^3 the partial of...

It's easy to do. Just follow the steps I supplied in my above post. Pretend the other variables are constants and perform differentiation (just like regular except only one one variable with the others held constant) on the equation. Repeat this until all variables have been differentiated...

Given a function U=x^2+y^2+2xy. To find the partials simply pretend the other variables are simply constants: \frac{\partial U}{\partial x}=\frac{\partial}{\partial x}(x^2+y^2+2xy) So, looking at the above y^2 will act as a constant. The derivative of a constant is zero so...

Do you know how to do regular differentiation?

The answer looks reasonable. Take the partial of x and y then stick those into the original eqn to verify that they do indeed equal something (zero in this case). As far as not understanding what your doing its nothing more than a recipe to follow. If you look at the original eqn then you see...

Did you choose to do it the way you did, or was it prescribed to do it like that? You should be able to solve it via separation of variables.

Algebra is the short answer. 1/(sin x)=a thus (sin x)=1/a Use the value the problem gave directly. This will yield some angle (in radians). find the second angle using the proporties of sin i.e. where in the 4 quadrants is sin + and where is it negative.

First, start off by writing secant as a cosine function: \frac{1}{\cos A}=1.2048 Now, solve the function for A. Remember, \arccos A is only defined from 0 to \pi so you have to find a corresponding angle between \pi and 2\pi good luck.

Thanks! So simple... I used approach number 1 where time is defined t>=0 so there is only one instant where the delta function applies t=0. I have: x(t)=e^{-t}(\cos (\sqrt{3}t)+\frac{\sqrt{3}}{3}\sin ({\sqrt{3}t)) with my interval greater than t=0 as my answer. Thanks for the help.

Anyone else out there in physics land have a hint for me?

Wow, that's a lot of typing. There's a quick way and a slower way. I'll walk you through the latter because it is easier to grasp what is going on. F=ma F=[B]e where B is the field strength and e is the electron charge (use correct sign) Be=ma a=dv/dt thus: (Be)dt=mdv...

What variable do you not have a value for? There's only one. Solve for the unknown plug in your known values. This will yield your initial condition. Apply the condition in the question. Then use: W=\int_{V1}^{V2}PdV

Ideal gas law? You know pressure, temperature and the number of atoms and 1 atm is much less than the critical pressure.

If this is a calculus based class then you should know that acceleration is the slope of a velocity function. given a function y(x) then the slope of said function is dy/dx. Given a function of v(t) then the slope is dv/dt which is acceleration. Think slope and grab a ruler.

Nope. My professor flew through this the other day (a similar problem) and looking at my notes it seems I missed something while he was talking. edit: Should I use the Homo eqn and the characteristic eqn for a delta function and solve using variation of parameters? edit:edit: I can't...

velocity is in m/s is it not? Acceleration is the change of velocity with respect to a change in time or \Delta v/\Delta t. Loot athe the graph and equate changes in velocity with their corresponding change in time.

Ok, I was given: Solve the following using superposition: \ddot{x}+2\dot{x}+4x=\delta(t) bounded by \dot{x}=0, x(0)=0 I solved the Homo eqn and got the following: x(t)=e^{-t}(\cos (\sqrt{3}t)+\frac{\sqrt{3}}{3}\sin ({\sqrt{3}t)) I also know that : \ddot{x}+2\dot{x}+4x=u(t)...

Ok, simplifying the above I get: \ddot\theta+\dot\theta^2\frac{4\sin\theta+4\cos\theta}{1+4\sin\theta+4\cos\theta}+\frac{8g}{R(1+4\sin\theta+4\cos\theta)}=0 Thus the linear function for small angles of theta becomes: \ddot\theta+\frac{8g}{R(5+4\theta)}=0 Is this correct? Thanks

The question: A 2-kg point mass is welded on the interior of a 2-kg thin ring at point P. The ring has a radius R = 160 mm and rolls on the surface without slipping. (a) Draw a free body diagram for the ring and point mass. Develop the equations of motion for the system. (b)...

Thanks. Additional delving into the topic showed us both that other factors such as porosity cooling rate grain structure and and mold flow are all 'typically' improved with thin wall castings versus thicker wall castings. Thanks for the response.

Quick question: Why does the strenght to wieght ratio increase for die cast parts with decreasing wall thickness? My view is that the weight decreases with nominal change in strength; however, a classmate of mine disagrees. Any thoughts?

The answer here depends on how the question is actually phrased. I gave a general response to finding an area using integrals; however, the usage depends on the question. If the absolute area is desired then you must break the function into discrete areas and sum the individual areas. If the...

sin and cos functions intersect every (x)pi/4

Area can be found by \int_{x(0)}^{x(1)} (F(x)-G(x))dx Simply let one of the functions be F(x) and the other be G(x). The absolute value of your integral over your limits will be the area (we don't owe the universe area if it comes out negative). Also, use symetry to your advantage here...

Here's the question: so, I said the mean (X) of delta is 0.0015 and the standard deviation (S) of delta is 0.000092 X_d=0.0015, S_d=0.000092 X_l=2.000, S_l=0.0081 I said Z=d/l\ thus\ X_z=X_d/X_l and the S_z^2=(C_d^2+C_l^2)/X_z^2 So, I did the following...

Ok, I don't understand why you WANT to put the zero on the top? Lets walk through an easy example the way you want to do it and the correct way: 116/4 ___2 4)116 2*4=8 11-8=3 Now 3/4=0 ___20 4)116 bring down the 6 36/4=9 ___209 4)116 Does that look like the...

Use the princlipple of conservation of energy. You have potential energies for both locations and a kinetic energy ao one location. The problem states the force acting on the particle is conservative so POCOE is fairly easy to implement here. Just realize that the sum of the potential and...

You're just trying to get the derviative of x^{1/3} You have the procedure written down correctly in the second line then you went way overboard after that. \frac{d}{dx}x^n=nx^{n-1} where n is a constant Replace n with 1/3--that's all there is to it. Is your question written correctly?

Your first question deals with the conservation of energy. The kinetic energy of a particle is completely converted to the potential energy stored in a spring. Knowing that, you should be able to solve this. use:V_{spring}=\frac{1}{2}k(\Delta S)^2 where delta S is the change in spring state...

P of the box needs to equal the normal force of the little box onto the big box. Draw two FBDs. On of a little box showing the effects of friction, a normal force and weight. Remember, the little box is in static equalibrium when referenced to the big box. The entire two box system may be...

Assume the engine shut down as soon as trouble occured. Furthermore, you'll have to assume that drag is zero (unless this is given to you). Draw a FBD. What forces are acting on the car and in which directions? Once you know the forces and how they are working with relation to the direction of...

This is only true for 3D space. You can't use the cross product in spaces greater that 3D or less than 3D.

OK, I've got this so far: m\ddot x+Kv^2\cos \theta=0 m\ddot y+Kv^2\sin \theta+mg=0 I have to use MATLAB to display a graph of this using the ODE45 function. Unfortunetly, I've never used MATLAB before so this is quite a problem. From whay I understand, I have to set up an m-file...

In know also that \dot{x}(0)=100\frac{m}{s}\cos \theta,\ \dot{y}(0)=100\frac{m}{s}\sin \theta and x(0)=0,\ y(0)=0

i direction: \ddot{x}=\frac{-Kv^2\cos \theta}{m} j direction: \ddot{y}=\frac{-Kv^2\sin \theta}{m}-g Is this correct thus far?