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Can you see the link, if not here it is http://m.imgur.com/a/TQ8MS sorry ]]>

I don't even have a clue what to do here. In our one example, we weren't seeking velocity, and our instructions were clear.

1. Find Rex

2. fx from Rex

3. Stx from fx and Pr

4. avgSt from Stx

5. avg Nu from avg St

6. avg h from avg Nu

7. q_y=h*A(delta T)

I guess you could say my attempt is to work it backwards. I have a chart provided that has info at a given T: rho, C_p, k, alpha, visco, kinematic visco - v, Pr, g*b/v^2 (from grashof). This is all i have to work it. Not sure where to go. As I said, I was working it backwards from steps above, 7-1, but hit a road block between 6 and 5. not even sure how well 6 worked out. Please help with anything you can. I'm not sure if I'm going in the right direction, and if so, where/how to go next. I messed up my schedule, and I thought this was due thursday, and my intention was to meet with my professor tomorrow. I found out late yesterday, that it was due today, and he was nice enough to let me email it to him before midnight, 3 hours and counting, but now I can't get the help I need. Thanks in advance. ]]>

A 60kg woman walking at a speed of 1.0 m/s steps onto a long plank that has a mass of 60kg. Upon stepping on the plank, both she and the plank begin to slide with speed v and spin with angular rate ω on a frictionless surface.

1. How far from the woman is the center of mass of the woman + plank system?

a. 0.0m, b. 2.5m, c. 5.0m, d. 7.5m, e. 10m

Relevant equations

regarding center of mass m1x1= m2x2, x is center of mass from side of plank where woman is standing

The attempt at a solution

for woman+plank: [60 kg + (x/10)*60kg] * x

& for plank on other side: [60 kg + ((10-x)/10)*60kg] * (10 - x)

These 2 equal each other by definition of center of mass

We get 60x + 6x^2 = 1200 - 120x - 60x + 6x^2

60x = 1200 - 180x

240x = 1200

x = 5m

This doesn't conceptually seem right to me, but I'm not sure how else to go about this. 5m is the center of the 10m plank, that would be the center of the mass without the woman standing on one end, no? So I'm thinking it wouldn't be with her there? But what is wrong with my calculations if this is true?

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high value of temperature cofficient of linear expansion or low value of that. Or is there

something else the truth ]]>

a) Determine what spacing should be made between the glass in order to have heat loss rate less than 100W/(m^2k)

b)If we wanted to replace the composite window with a single plane of glass, what thickness would be required.

I'm struggling to get set up here. I have a solid interacting with a fluid, so I know I need to track down that equation, but part of the problem is I don't have a text book. Our professor has hand-written notes that are used as the class text (not joking). His writing isn't the best. I don't know what I need to get going. Please help. ]]>

I am modelling an inclined rectangular cavity and got confused with Grashof and Rayleigh number. I've reviewed many research papers - some researchers use Rayleigh, some Grashof number to determine the type of flow.

This is probably very simple but I can't find where my mistake lies...

I get 10^6 for Rayleigh which indicates Laminar flow and I also get 10^12 for Grashof which indicates Turbulent flow. This got me confused.

My problem involves upper wall exposed to a heat flux, bottom wall and side walls adiabatic. Heat flux warms up the upper wall and the fluid inside. The conditions become:

Tupper = 306 K (average)

Tbottom = 303 K (average)

Tfluid = 300 K (average)

Dimensions: L=2m, W=1m, H=0.007m (space between plates)

From literature and texbooks, I use the following formulas:

Ra=( g*B* (Tu-Tf)*L^3 ) / (v*a)

Gr = (g*B (Tu-Tf)*H^3) / (v^2)

where Tu - upper plate temperature

Tf - fluid temperature

L - Length of the cavity

g - 9.81m/s

B - thermal expansion coeff for WATER @ 20C (0.0033 [1/K])

v - kinematic viscosity 1x10e-7 m/s2

a - thermal diffusivity 1x10e-7 mē/s

Thank you for you help. ]]>

I'm completely stuck :(

I've been trying to find the temperature of air in an otto cycle before the fuel is ignited. I know

exhaust air is 733^C,

Cp is 1005,

air flow mass rate is 0.00460

fuel mass flow rate is 0.000393.

The cal val for fuel is 46.5 Mj/kg,

and im using the equation Q=m x cv (t3-t2)

but...i just get a ridiculous answer every time.

46.58106 x 0.000393=18 kj/kg

1.225 * 0.00460=0.005635

cv=718

Q=m x cv (t3-t2)=>(Q/(m x Cv))-t2=-t1

(18 x 103 / (718 x 0.005635))-1006=3442k for the lower temperature,

I'm not sure whats going wrong.

If anyone could shed some light that would be great

Many thanks ]]>

So I'm in IB physics, and to be completely honest I'm not really a science guy so I had some trouble coming up with an idea for my physics IA. I ended up doing something with radio waves because I had access to somebody who operated an amateur radio...I figured I would put together an antenna, connect it to light bulb, and try to see the effect of using different frequencies to try to power it.

Needless to say I quickly realized that I would need a massive transmitter to get it to light up so I connected an oscilloscope to the whole shebang to try to salvage some data from the experiment.

My results were that 420 and 450 Mhz waves generated the most voltage (~1mV), 440 Mhz generated the 2nd greatest (~0.6mV), and 460 and 480 Mhz generated voltages of 0.2 and 0.1 mV respectively. All of this was when the transmitter was held around 10cm from the antenna.

This data is all over the place and I don't know exactly what conclusions I could possibly draw from it. I know that I probably should have done my research before jumping in with the idea that differences in frequency will somehow affect the resulting voltage in an antenna at all, but this is all I have at this point so I would really appreciate if somebody could give me some pointers on what I could do with this data..

Thanks ]]>

a) the mass flowrate of the steam.

b) the work output per kilogram , assume that kinetic-energy and potential-energy changes are negligible. ]]>

I feel like there is a shortcut to this answer that I'm missing, because I've done several pages of algebra and simplifications and I can't seem to get a form that either I or Wolfram can simplify into something reasonable.

I'm calculating the uncertainty by using the equation with <A^2> - <A>^2 for some operator A.

I'm using the form of Sx below and corresponding forms of Sy and Sz

And I'm using the definition of expectation value below

However, despite double and triple checking all my algebra, this is the equation I get for the product of the two uncertainties, which is not really workable.

https://www.wolframalpha.com/input/?...cosx*sinx)%5E2))

I feel like I'm certainly missing some major shortcut.

Thanks in advance. ]]>

I'm trying to complete a static force analysis on the bar linkage shown below. The goal is to find the payload mass in terms of the cylinder forces and everything else in a static environment.

To simplify things I've made the assumption that the mass of the payload in the bucket acts at the centre of mass of the bucket. Also, the bucket cylinder is attached to the boom, and the boom is attached to an anchor point.

From what I understand, to do the static force analysis, you freeze everything in place, and then calculate the force and torque that each link must exert on each other for it to hold that position.

My attempt at the problem has been to break the linkage apart and try to work my way through each component and write down the forces and moments acting on them. For example, the bucket has a force from the boom, a force from the bucket link, and a force downwards from payload mass.

Then, I've tried to do the next component like the bucket link, and repeat the process. My attempt for the bucket and bucket link are below in the photo.

However, I'm terrible at this, so would anyone be able to help me? To reiterate, my end goal is to get an expression for the payload mass in the bucket, in terms of cylinder forces and everything else. I'm thinking the best way would be to take a sum of the torques around the boom pivot point, but I'm not really sure how to get there.

Many thanks in advance. ]]>