What is the equation for the energy needed to operate a bicycle?
Q. I saw in a magazine an equation that detailed the forces for the operation of a bicycle. I can't determine what the equation is because I know nothing about kinetic mathematics or advanced algebra. This is for a tattoo I want to do. There is an equation I know it, I saw it in Tattoo Magazine or somewhere
Asked by scatterbrainedgenius - Wed Aug 19 12:57:37 2009 - - 4 Answers - 0 Comments
A. I tried deriving this: A bike is in static equilibrium, that is it will not fall left or right because it is in forward motion. So start with motion:length over time times mass.l/s*m Umm I forget the rest, I'll see if I can find something found something! lots of lowercase greek letters, you're in trouble
Answered by mike c - Wed Aug 19 13:07:40 2009
Q. I saw in a magazine an equation that detailed the forces for the operation of a bicycle. I can't determine what the equation is because I know nothing about kinetic mathematics or advanced algebra. This is for a tattoo I want to do. There is an equation I know it, I saw it in Tattoo Magazine or somewhere
Asked by scatterbrainedgenius - Wed Aug 19 12:57:37 2009 - - 4 Answers - 0 Comments
A. I tried deriving this: A bike is in static equilibrium, that is it will not fall left or right because it is in forward motion. So start with motion:length over time times mass.l/s*m Umm I forget the rest, I'll see if I can find something found something! lots of lowercase greek letters, you're in trouble
Answered by mike c - Wed Aug 19 13:07:40 2009
How to find equation of tangent to a circle given equation of circle and external point?
Q. I have the equation of the circle, and the external point. I know the radius of the circle and its centre. So how can i find the equation of the tangent? Thanks
Asked by Ed - Sun Nov 15 11:39:53 2009 - - 1 Answers - 0 Comments
Q. I have the equation of the circle, and the external point. I know the radius of the circle and its centre. So how can i find the equation of the tangent? Thanks
Asked by Ed - Sun Nov 15 11:39:53 2009 - - 1 Answers - 0 Comments
What is the equation connecting systolic pressure to pulse rate?
Q. In a recent Question a User asked how pulse rate affected blood pressure, and one answer supplied the equation showing the how pulse rate and diastolic pressure are connected. I asked what the equation for diastolic pressure was, but the Question was closed before I could get an answer. Can someone tell me how pulse rate affects systolic pressure?
Asked by Henry - Sat Aug 8 06:11:00 2009 - - 4 Answers - 1 Comments
A. The equation you seek relating pulse rate (N) to Systolic pressure (Ps) is: Ps = (N x Pp x C x R) + (CO/N.C) where C is aortal wall compliance, Pp the pulse pressure, R being the total resistance presented to the flow of blood (by the vasculature extending from the aortic valve to the venous pool), and CO of course, is cardiac output. Orthodox textbooks will (quite wrongly) contend that the 1st term on the RH side of the equation should read {N.Pp.C.R - Pp/3} but this is quite incorrect, because the pulse pressure wave does no work, (apart from losses) and therefore plays no part in the driving of blood round the loop.. But it doesn't really matter anyway, because the equation works in any case, -with or without the " -Pp/3". … [cont.]
Answered by Huguette - Sat Aug 8 11:15:50 2009
Q. In a recent Question a User asked how pulse rate affected blood pressure, and one answer supplied the equation showing the how pulse rate and diastolic pressure are connected. I asked what the equation for diastolic pressure was, but the Question was closed before I could get an answer. Can someone tell me how pulse rate affects systolic pressure?
Asked by Henry - Sat Aug 8 06:11:00 2009 - - 4 Answers - 1 Comments
A. The equation you seek relating pulse rate (N) to Systolic pressure (Ps) is: Ps = (N x Pp x C x R) + (CO/N.C) where C is aortal wall compliance, Pp the pulse pressure, R being the total resistance presented to the flow of blood (by the vasculature extending from the aortic valve to the venous pool), and CO of course, is cardiac output. Orthodox textbooks will (quite wrongly) contend that the 1st term on the RH side of the equation should read {N.Pp.C.R - Pp/3} but this is quite incorrect, because the pulse pressure wave does no work, (apart from losses) and therefore plays no part in the driving of blood round the loop.. But it doesn't really matter anyway, because the equation works in any case, -with or without the " -Pp/3". … [cont.]
Answered by Huguette - Sat Aug 8 11:15:50 2009
What equation do I need to find the kinematic viscosity of water given the temp and pressure?
Q. I am given the temp and pressure (10 deg C and 35564 kPa) and need to find the kinematic viscosity of water. I know that the kinematic viscosity is dynamic viscosity/density but how do I take into account the temperature and pressure? I don't want an answer, just an equation or explanation. Thanks.
Asked by blueorion3 - Wed Oct 1 12:36:07 2008 - - 1 Answers - 0 Comments
A. I really doubt that pressure will be an issue with either viscosity or density. You may need to consult the Temp./ viscosity tables and the Temp./ density tables and just plug those values into the equation you alteady have. The viscosity will vary with temperature, the density, not much. To be practical, if you can find one , a (rho) at around 20C , just use that one.
Answered by Bomba - Wed Oct 1 15:12:53 2008
Q. I am given the temp and pressure (10 deg C and 35564 kPa) and need to find the kinematic viscosity of water. I know that the kinematic viscosity is dynamic viscosity/density but how do I take into account the temperature and pressure? I don't want an answer, just an equation or explanation. Thanks.
Asked by blueorion3 - Wed Oct 1 12:36:07 2008 - - 1 Answers - 0 Comments
A. I really doubt that pressure will be an issue with either viscosity or density. You may need to consult the Temp./ viscosity tables and the Temp./ density tables and just plug those values into the equation you alteady have. The viscosity will vary with temperature, the density, not much. To be practical, if you can find one , a (rho) at around 20C , just use that one.
Answered by Bomba - Wed Oct 1 15:12:53 2008
What is the differential equation for steady state molecular diffusion of fluid in a laminar flow?
Q. I need the equation in differential form for steady state molecular diffusion of a fluid in a laminar flow. I am looking for an equation that will work for Reynold's Number 35. Also is it possible to derive it from Ficks law of mass transfer?
Asked by Fluorine - Mon Nov 26 19:23:53 2007 - - 2 Answers - 0 Comments
A. Of course you can use Fick's law! Since it is dependent upon the molecular average velocity anyway, so your variable will be included, anyway; your equation is: JA= -DAB(dCA/dCB); where JA is the diffusion flux for component A, which is proportional to the concentration gradient dCA/dCB, DAB is the diffusivity of component A in its mixture with B. Now how we come up with that and in regard to velocities; the molal flux of a mixture is NM=rMuo, where rM is the average MOLAR denisty of the mixture and uo is the average velocity, then, by substituting rM with more measurable variable which is the molar concentration, we obtain -for a molecule of A crossing a stationary plane with a linear velocity the following: NA=cAuA; so JA=cAuA-cAuo… [cont.]
Answered by Dean B - Mon Nov 26 20:12:20 2007
Q. I need the equation in differential form for steady state molecular diffusion of a fluid in a laminar flow. I am looking for an equation that will work for Reynold's Number 35. Also is it possible to derive it from Ficks law of mass transfer?
Asked by Fluorine - Mon Nov 26 19:23:53 2007 - - 2 Answers - 0 Comments
A. Of course you can use Fick's law! Since it is dependent upon the molecular average velocity anyway, so your variable will be included, anyway; your equation is: JA= -DAB(dCA/dCB); where JA is the diffusion flux for component A, which is proportional to the concentration gradient dCA/dCB, DAB is the diffusivity of component A in its mixture with B. Now how we come up with that and in regard to velocities; the molal flux of a mixture is NM=rMuo, where rM is the average MOLAR denisty of the mixture and uo is the average velocity, then, by substituting rM with more measurable variable which is the molar concentration, we obtain -for a molecule of A crossing a stationary plane with a linear velocity the following: NA=cAuA; so JA=cAuA-cAuo… [cont.]
Answered by Dean B - Mon Nov 26 20:12:20 2007
How do I change the point slope equation to a slope intercept of a form equation??
Q. This is the point slope equation: y - 2 = 2 ( x + 4), and I need to change it into a slope intercept equation? Can I do this without subtracting or adding numbers from each side? Is there another way? Thank you for everyone who actually answered my question.
Asked by Katia - Thu Jan 17 13:11:41 2008 - - 6 Answers - 0 Comments
A. y - 2 = 2(x + 4) y - 2 = 2x + 8 y = 2x + 8 + 2 y = 2x + 10
Answered by psbhowmick - Thu Jan 17 13:15:33 2008
Q. This is the point slope equation: y - 2 = 2 ( x + 4), and I need to change it into a slope intercept equation? Can I do this without subtracting or adding numbers from each side? Is there another way? Thank you for everyone who actually answered my question.
Asked by Katia - Thu Jan 17 13:11:41 2008 - - 6 Answers - 0 Comments
A. y - 2 = 2(x + 4) y - 2 = 2x + 8 y = 2x + 8 + 2 y = 2x + 10
Answered by psbhowmick - Thu Jan 17 13:15:33 2008
What is the balanced equation for the breakdown of hydrogen peroxide?
Q. I need to know the balanced equation for the breakdown of hydrogen peroxide. Full equation please.
Asked by joeblake15 - Tue Jan 8 16:12:41 2008 - - 3 Answers - 0 Comments
A. 2 H2O2 --> 2 H2O + O2
Answered by Dr. W. - Tue Jan 8 16:17:03 2008
Q. I need to know the balanced equation for the breakdown of hydrogen peroxide. Full equation please.
Asked by joeblake15 - Tue Jan 8 16:12:41 2008 - - 3 Answers - 0 Comments
A. 2 H2O2 --> 2 H2O + O2
Answered by Dr. W. - Tue Jan 8 16:17:03 2008
How can you create an equation that relates the following things about a stoplight?
Q. The basic idea is relating the math involved in stoplights. So far, I have made it into a linear equation for a single stoplight relating time to number of cars, and have been working on creating an equation that takes in other factors as well. I need help coming up with this equation, which must consist of how you can calculate the time of a stoplight through the # of cars, length of an inersection, average speed of cars, length of cars, and any other factor you may think neccesary. Please help me in either coming up with this equation, telling me other factors, or by telling me what type of equation would best suit this. This is for a math research paper, and this is the topic I have to explore.
Asked by snakehawk37 - Thu Jun 8 18:31:23 2006 - - 5 Answers - 3 Comments
A. My thinking is...Cars flow through lights in highs and lows. Let's assume during high there are X Cars (each 25ft long) that pass at Y miles/hr. Then the duration for the Green would be: T = ( [ (25+BD) / 5280)*X] / Y ) * 3600 T=Duration of Green light in secs BD=Break Distance in ft X= Number of Cars Y=Speed of the Cars There is something missing here...just work on it a little more... Bye
Answered by Nito P - Thu Jun 8 18:54:18 2006
Q. The basic idea is relating the math involved in stoplights. So far, I have made it into a linear equation for a single stoplight relating time to number of cars, and have been working on creating an equation that takes in other factors as well. I need help coming up with this equation, which must consist of how you can calculate the time of a stoplight through the # of cars, length of an inersection, average speed of cars, length of cars, and any other factor you may think neccesary. Please help me in either coming up with this equation, telling me other factors, or by telling me what type of equation would best suit this. This is for a math research paper, and this is the topic I have to explore.
Asked by snakehawk37 - Thu Jun 8 18:31:23 2006 - - 5 Answers - 3 Comments
A. My thinking is...Cars flow through lights in highs and lows. Let's assume during high there are X Cars (each 25ft long) that pass at Y miles/hr. Then the duration for the Green would be: T = ( [ (25+BD) / 5280)*X] / Y ) * 3600 T=Duration of Green light in secs BD=Break Distance in ft X= Number of Cars Y=Speed of the Cars There is something missing here...just work on it a little more... Bye
Answered by Nito P - Thu Jun 8 18:54:18 2006
What is a really complicated equation i can ask somebody?
Q. My friend thinks hes really good at maths and can solve any equations. What is a really hard and complicated equation i can ask him?
Asked by usa_fox1234 - Wed Dec 20 14:05:14 2006 - - 8 Answers - 0 Comments
A. well i asked this question a while ago--- a+b+c=4 a^2+b^2+c^2=10 a^3+b^3+c^3=22 find the values of a, b ,and c It is REALLY hard clue: cubic polynominals...??!!
Answered by rahbonzy - Wed Dec 20 14:25:32 2006
Q. My friend thinks hes really good at maths and can solve any equations. What is a really hard and complicated equation i can ask him?
Asked by usa_fox1234 - Wed Dec 20 14:05:14 2006 - - 8 Answers - 0 Comments
A. well i asked this question a while ago--- a+b+c=4 a^2+b^2+c^2=10 a^3+b^3+c^3=22 find the values of a, b ,and c It is REALLY hard clue: cubic polynominals...??!!
Answered by rahbonzy - Wed Dec 20 14:25:32 2006
How do I neutralize this equation in chemistry and what does it mean?
Q. Write an equation for the neutralization of H2SO4 by KOH. I don't understanding what neutralizing is and how to write this equation.
Asked by hannahfan84 - Fri Nov 14 10:16:34 2008 - - 2 Answers - 0 Comments
A. You'll have an acid reacting with a base. Neutralization will occur when all reactants have gone to the products (stoichiometrically equal). H2SO4 + 2KOH = K2SO4 + 2H2O Just balance both sides of the equation. Sulfate will require two potassium ions to form potassium sulfate. This means two KOH molecules will be required on the left side. As sulfate is balanced (SO4), you just have to add two waters to balance the equation.
Answered by brewer_engineer - Fri Nov 14 10:27:31 2008
Q. Write an equation for the neutralization of H2SO4 by KOH. I don't understanding what neutralizing is and how to write this equation.
Asked by hannahfan84 - Fri Nov 14 10:16:34 2008 - - 2 Answers - 0 Comments
A. You'll have an acid reacting with a base. Neutralization will occur when all reactants have gone to the products (stoichiometrically equal). H2SO4 + 2KOH = K2SO4 + 2H2O Just balance both sides of the equation. Sulfate will require two potassium ions to form potassium sulfate. This means two KOH molecules will be required on the left side. As sulfate is balanced (SO4), you just have to add two waters to balance the equation.
Answered by brewer_engineer - Fri Nov 14 10:27:31 2008
What is it about linear equations of the format y=mx+b that cause the graph of the equation to b a staright li?
Q. What is it about linear equations of the format y=mx+b that causes the graph of the equation to be a staright line? How do you writea linear equation given two points? How do you write a linear equation given the slope and a single point?
Asked by fuzzy - Sat Aug 29 14:57:58 2009 - - 1 Answers - 0 Comments
A. hahahahahaaa im doing that too!!! idk
Answered by devibabyx416 - Sun Aug 30 14:49:56 2009
Q. What is it about linear equations of the format y=mx+b that causes the graph of the equation to be a staright line? How do you writea linear equation given two points? How do you write a linear equation given the slope and a single point?
Asked by fuzzy - Sat Aug 29 14:57:58 2009 - - 1 Answers - 0 Comments
A. hahahahahaaa im doing that too!!! idk
Answered by devibabyx416 - Sun Aug 30 14:49:56 2009
How do you change an equation from polar coordinate to cartesian?
Q. I am in Calc 2 and we are working on polar coodinate graphs and I can't seem to handle them. The question that I need help with is how to change the equation, r=4sec(theta)(1+ tan(theta)), into the cartesian equation. Thank you so much for your help!
Asked by kjwilson - Tue Apr 10 14:42:46 2007 - - 4 Answers - 2 Comments
A. x = r cos (theta) y = r sin (theta) x = 4 sec (theta)(1+tan(theta)) cos (theta) y = 4 sec(theta)(1+tan(theta)) sin(theta) I'm going to use T instead fo "theta" x = 4 sec(T)(1+tan(T)) cos (T) x = 4 (1+tan(T)) since sec = 1/cos, and cos/cos = 1 y = 4 sec(T)(1+tan(T)) sin(T) y = 4 (1+tan(T)) tan(T) since sin/cos = tan y = 4 (tan(T) + tan^2(T)) you might be able to simplify it a bit more since tan^2(T) + 1 = sec^2(T), but I think that just makes it messier. I hope this helps (but double check its accuracy!).
Answered by Stuey - Tue Apr 10 14:47:57 2007
Q. I am in Calc 2 and we are working on polar coodinate graphs and I can't seem to handle them. The question that I need help with is how to change the equation, r=4sec(theta)(1+ tan(theta)), into the cartesian equation. Thank you so much for your help!
Asked by kjwilson - Tue Apr 10 14:42:46 2007 - - 4 Answers - 2 Comments
A. x = r cos (theta) y = r sin (theta) x = 4 sec (theta)(1+tan(theta)) cos (theta) y = 4 sec(theta)(1+tan(theta)) sin(theta) I'm going to use T instead fo "theta" x = 4 sec(T)(1+tan(T)) cos (T) x = 4 (1+tan(T)) since sec = 1/cos, and cos/cos = 1 y = 4 sec(T)(1+tan(T)) sin(T) y = 4 (1+tan(T)) tan(T) since sin/cos = tan y = 4 (tan(T) + tan^2(T)) you might be able to simplify it a bit more since tan^2(T) + 1 = sec^2(T), but I think that just makes it messier. I hope this helps (but double check its accuracy!).
Answered by Stuey - Tue Apr 10 14:47:57 2007
What is the equation for a Sodium hydrogen carbonate reaction?
Q. And how can we tell from this equation that it removes acid?
Asked by Siren Productions - Tue Feb 5 16:52:01 2008 - - 1 Answers - 0 Comments
A. NaHCO3 + HCl --> NaCl + H2O + CO2 where did the acid go?
Answered by skipper - Tue Feb 5 16:56:07 2008
Q. And how can we tell from this equation that it removes acid?
Asked by Siren Productions - Tue Feb 5 16:52:01 2008 - - 1 Answers - 0 Comments
A. NaHCO3 + HCl --> NaCl + H2O + CO2 where did the acid go?
Answered by skipper - Tue Feb 5 16:56:07 2008
How do i calculate the coefficients of a quadratic equation?
Q. Using the variables x1,x2,x3,y1,y2,y3 (from any three points on the quadratic) what are the equations that will give me the values of a,b and c in the quadratic equation ax^2+bx+c? I asked this question earlier but its hard to word correctally. I want to be able to plug in any 3 point and get the values nessisary to create the quadratic equation on which these points lie. Thanks. Yes i forgot y = ax^2+bx+c. I know i can solve the simultanious equations to isolate a, b, and c. But if you look at it, its a lot of work when im positive there is a simple equation already out there. In the thousands of years we have known these equations, someone has put a,b and c in terms of 3 coordinates. i want to know the equation, not how to get it.
Asked by Game Theorist - Tue Oct 17 21:02:13 2006 - - 3 Answers - 0 Comments
A. I guess you wanted a group of formulas like: a = function of (x1, x2, x3, y1, y2, y3), b = other func (x1, x2, x3, y1, y2, y3), c = third func (x1, x2, x3, y1, y2, y3), but honestly it's not a good idea to do it that way usually. In the frequent case when the middle point is quite close to the straight line between the two outside points, you can end up with way-out coefficients and a formula that's correct but very hard to evaluate accurately, like it gives you 170893.771 minus 170893.756. When I did this in a computer program, I used to first get the straight line between the two outside points. That's X = x1 + Y * (x3 - x1) / (y3 - y1). Then I would imagine the final quadratic as the sum of my straight line plus a quadratic… [cont.]
Answered by bh8153 - Wed Oct 18 17:23:35 2006
Q. Using the variables x1,x2,x3,y1,y2,y3 (from any three points on the quadratic) what are the equations that will give me the values of a,b and c in the quadratic equation ax^2+bx+c? I asked this question earlier but its hard to word correctally. I want to be able to plug in any 3 point and get the values nessisary to create the quadratic equation on which these points lie. Thanks. Yes i forgot y = ax^2+bx+c. I know i can solve the simultanious equations to isolate a, b, and c. But if you look at it, its a lot of work when im positive there is a simple equation already out there. In the thousands of years we have known these equations, someone has put a,b and c in terms of 3 coordinates. i want to know the equation, not how to get it.
Asked by Game Theorist - Tue Oct 17 21:02:13 2006 - - 3 Answers - 0 Comments
A. I guess you wanted a group of formulas like: a = function of (x1, x2, x3, y1, y2, y3), b = other func (x1, x2, x3, y1, y2, y3), c = third func (x1, x2, x3, y1, y2, y3), but honestly it's not a good idea to do it that way usually. In the frequent case when the middle point is quite close to the straight line between the two outside points, you can end up with way-out coefficients and a formula that's correct but very hard to evaluate accurately, like it gives you 170893.771 minus 170893.756. When I did this in a computer program, I used to first get the straight line between the two outside points. That's X = x1 + Y * (x3 - x1) / (y3 - y1). Then I would imagine the final quadratic as the sum of my straight line plus a quadratic… [cont.]
Answered by bh8153 - Wed Oct 18 17:23:35 2006
What is the equation for the rule of 72 in investing?
Q. I have just figured out the rate, doubling time and product of a variety of percents. 72 is the product of many and I do not know how to write an equation explaining it. Any help?
Asked by muskgirlrun - Thu Feb 26 16:54:37 2009 - - 3 Answers - 0 Comments
A. t = 72/r where t = time to doubling and r = interest rate per annum
Answered by Jaminio - Thu Feb 26 17:24:08 2009
Q. I have just figured out the rate, doubling time and product of a variety of percents. 72 is the product of many and I do not know how to write an equation explaining it. Any help?
Asked by muskgirlrun - Thu Feb 26 16:54:37 2009 - - 3 Answers - 0 Comments
A. t = 72/r where t = time to doubling and r = interest rate per annum
Answered by Jaminio - Thu Feb 26 17:24:08 2009
How would you use the quadratic equation to solve this inequality involving time?
Q. A bicyclist is riding at a speed of 20 mi/h when shestarts down a long hill. I've been given the equation d(t) = 5t^2 + 20t, where t is the time in seconds. I know that the hill is 585 feet long. How would I set up the equation? How would I set it up if the hill were half as long? Thanks.
Asked by lapointetj - Sun Feb 15 21:07:44 2009 - - 4 Answers - 0 Comments
A. is the hill 585 long or 585 high? I'm assuming your equation reads 585 high d(t) = 5t^2 +20t 585 = 5t^2 +20t 5t^2 + 20t -585 = 0 t^2 + 4t - 117 = 0 (t + 13)(t - 9) = 0 possible solutions are {-13, 9} toss negative time so {9 sec} seconds to get to the bottom of the hill second question set problem: 292.5 = 5t^2 +20t 5t^2 + 20t - 292.5 = 0 t^2 + 4t - 58.5 = 0 [-4 +/- sqrt(16 - 4(1)(-58.5)] / 2 [-4 +/- sqrt(250)]/2 [-4 +/- 5*sqrt(10)]/2 x = {5.91, -9.91} toss negative time Answer = 5.91 seconds to get to the bottom of the hill
Answered by Mark - Sun Feb 15 21:12:41 2009
Q. A bicyclist is riding at a speed of 20 mi/h when shestarts down a long hill. I've been given the equation d(t) = 5t^2 + 20t, where t is the time in seconds. I know that the hill is 585 feet long. How would I set up the equation? How would I set it up if the hill were half as long? Thanks.
Asked by lapointetj - Sun Feb 15 21:07:44 2009 - - 4 Answers - 0 Comments
A. is the hill 585 long or 585 high? I'm assuming your equation reads 585 high d(t) = 5t^2 +20t 585 = 5t^2 +20t 5t^2 + 20t -585 = 0 t^2 + 4t - 117 = 0 (t + 13)(t - 9) = 0 possible solutions are {-13, 9} toss negative time so {9 sec} seconds to get to the bottom of the hill second question set problem: 292.5 = 5t^2 +20t 5t^2 + 20t - 292.5 = 0 t^2 + 4t - 58.5 = 0 [-4 +/- sqrt(16 - 4(1)(-58.5)] / 2 [-4 +/- sqrt(250)]/2 [-4 +/- 5*sqrt(10)]/2 x = {5.91, -9.91} toss negative time Answer = 5.91 seconds to get to the bottom of the hill
Answered by Mark - Sun Feb 15 21:12:41 2009
What is the equation to find out your carbon footprint?
Q. I know that there are calculators out there to find it out but I need the equation(s). Especially if it has to do with only the primary footprint, but whatever you can give me would be nice.
Asked by Kristal F - Mon Nov 12 21:48:44 2007 - - 1 Answers - 0 Comments
A. The site has a calculator and gives an explanation of how each component works. Perhaps it will help you.
Answered by Bad Brain Punk - Thu Nov 15 15:14:35 2007
Q. I know that there are calculators out there to find it out but I need the equation(s). Especially if it has to do with only the primary footprint, but whatever you can give me would be nice.
Asked by Kristal F - Mon Nov 12 21:48:44 2007 - - 1 Answers - 0 Comments
A. The site has a calculator and gives an explanation of how each component works. Perhaps it will help you.
Answered by Bad Brain Punk - Thu Nov 15 15:14:35 2007
What is the equation for position given a particular acceleration?
Q. What is the equation for position given a particular acceleration?
Asked by Jack H - Mon Oct 20 10:20:25 2008 - - 2 Answers - 0 Comments
A. D=|1/2|at^2
Answered by nozar nazari - Mon Oct 20 10:32:38 2008
Q. What is the equation for position given a particular acceleration?
Asked by Jack H - Mon Oct 20 10:20:25 2008 - - 2 Answers - 0 Comments
A. D=|1/2|at^2
Answered by nozar nazari - Mon Oct 20 10:32:38 2008
What is the equation for atmospheric pressure?
Q. I am wondering what the equation for atmospheric pressure because my physics teacher gave me a take-home test with random questions on it. I need help with this one. What will be the pressure in atmospheres felt by a bathysphere at 35,813 ft in the Marianas Trench?
Asked by demarr_10 - Sun Feb 22 09:10:44 2009 - - 1 Answers - 0 Comments
A. for every 33 feet down the pressure increases one atm so 33 ft would be 2atm 66 ft 3 atm 99 ft 4 atm etc
Answered by Lee S - Sun Feb 22 09:47:07 2009
Q. I am wondering what the equation for atmospheric pressure because my physics teacher gave me a take-home test with random questions on it. I need help with this one. What will be the pressure in atmospheres felt by a bathysphere at 35,813 ft in the Marianas Trench?
Asked by demarr_10 - Sun Feb 22 09:10:44 2009 - - 1 Answers - 0 Comments
A. for every 33 feet down the pressure increases one atm so 33 ft would be 2atm 66 ft 3 atm 99 ft 4 atm etc
Answered by Lee S - Sun Feb 22 09:47:07 2009
What is the equation to find exterior angle measurements in any polygon?
Q. Hey, I was wondering if anyone knew on how to find the equation of any polygon's exterior angles? Thx Ty, but that's how to find the sum of the interior angles =P. I need exterior angles. TY both for your time to help me. =D
Asked by no name - Mon Dec 15 19:52:03 2008 - - 2 Answers - 0 Comments
A. The sum of the *exterior* angles of a convex polygon is always 360 . If you have a *regular* polygon (all angles the same), then you can find the measure of any exterior angle by dividing by the number of angles (which is the same as the number of sides). The exterior angles of an n-sided regular polygon are all: 360 /n (If you don't have a regular polygon, you can have different measurements for each of the exterior angles. All you can definitively say is they add up to 360 )
Answered by Puzzling - Mon Dec 15 19:58:22 2008
Q. Hey, I was wondering if anyone knew on how to find the equation of any polygon's exterior angles? Thx Ty, but that's how to find the sum of the interior angles =P. I need exterior angles. TY both for your time to help me. =D
Asked by no name - Mon Dec 15 19:52:03 2008 - - 2 Answers - 0 Comments
A. The sum of the *exterior* angles of a convex polygon is always 360 . If you have a *regular* polygon (all angles the same), then you can find the measure of any exterior angle by dividing by the number of angles (which is the same as the number of sides). The exterior angles of an n-sided regular polygon are all: 360 /n (If you don't have a regular polygon, you can have different measurements for each of the exterior angles. All you can definitively say is they add up to 360 )
Answered by Puzzling - Mon Dec 15 19:58:22 2008
From Yahoo Answer Search: 'equation'
Fri Nov 20 15:12:07 2009 [ refresh local cache ]
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Solving the India Equation
Global Supply Chain Council
Guido Nassimbeni and Marco Sartor (GN & MS): The world has changed significantly in past few years. Two of the most important factors in that change have ...
Regional situation a look back Pakistan Observer
all 85 news articles »
Global Supply Chain Council
Guido Nassimbeni and Marco Sartor (GN & MS): The world has changed significantly in past few years. Two of the most important factors in that change have ...
Regional situation a look back Pakistan Observer
all 85 news articles »
CitE* Watch: The Notre Dame Equation
catholicsensibility
Sat, 14 Nov 2009 18:24:50 GM
Rock has news on the two latest bishop-appointments stateside. However, no new prelates will be making last-minute hotel reservations. Both new bishops come to Milwaukee and northern Indiana via that grand tradition of modernism: from ...
catholicsensibility
Sat, 14 Nov 2009 18:24:50 GM
Rock has news on the two latest bishop-appointments stateside. However, no new prelates will be making last-minute hotel reservations. Both new bishops come to Milwaukee and northern Indiana via that grand tradition of modernism: from ...
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