i have a question where i have to compute the distances under the specified conditions. Here is the information given:\n\nTemperature: 26 degrees celcious\nTape Length: 30.009 m\nrequired horizontal distance: 114.691m\n\nim not sure what it means...i am looking for the uncorrected distance, hence they have me the required horzontal distance ( which i assume to be the corrected distance)\n\nother information needed\n\ntape standardization length: 30.000m\nstandardiztion temperature: 20 degrees celcious\nmodulous of elasticity for the tape: 11.6 times 10 to the negative6\n\ncan n e body help me...
131586
The coefficient of thermal expansion of steel (assuming you are using a steel tape) is about 0.0000065/ degree F. A difference of 15 degrees in temperature will cause a change in length of nearly 0.01 feet in a 100 foot tape. Therefore a measurement made with a steel tape on an extremely hot day in the summer cannot be expected to agree with one mage during zero weather.\n\nThere is a formula for this (in English system) for the amount for correction for temperature:\n\nC=Lα(T-t)\nWhere C is the correction applied\nL is the length in feet\nT is the temperature at which the tape was used\nT is the temperature at which the tape was standardized\nΑ is the coefficient of thermal expansion.\n\nThere should be a formula just like this but with a different coefficient of thermal expansion in metric.
486034
I assumed the fingers are the pivot in this problem, so all distances in the torque equation are measured relative to them.\nThe results are the same, regardless of where you take the pivot.  \n\nLet clockwise torque be negative, and counterclockwise be positive.\n\na)\nBalance of torques:\nτ(positive) = τ(negative)\nτ(thumb) = τ(tray) + τ(plate) + τ(cup)\nF(th)*x(th) = F(tray)*x(tray) + F(plate)*x(plate) + F(cup)*x(cup)\n(and remember, all distances, x, are relative to the fingers)\n(.04)*F(th) = [(.193)(.1) + (1.04)(.14) + (.229)(.28)]*9.8\n──► F(th) = 56.11 N\n\nb)\nBalance of forces:\nF(upward) = F(downward)\nF(fingers) = F(thumb) + F(plate) + F(tray) + F(cup)\nF(f) = F(th) + (0.193 + 1.04 + 0.229)*9.8\n= F(th) + 14.33\n= 70.44 N\n\nNote: Try to understand what is going on.  Balance of torques is used to ensure the tray is not rotating on an axis.  Balance of forces is used to ensure the tray is not moving staight up or straight down.
14632
here you assume there is almost no point of intersection...or at least that it is exactly a point therefor no measureable space so you must determine the two circumfrences...use the equation\nc=pi * diameter\nc1 = pi *6\nc2 = pi * 18\n\ntotal distance is 24pi\n\nhope that helps!\n\nactually, that is a good point, when i first read it i assumed that we would want to put the rope around the circle...meaning that we would have the rope touch the circle at all times...the problem with stating that there is a line from one circle to the other will be slightly inaccurate, the line from the two circles will be different from the horizontal by several degrees, that would be the need for teh trig function.  I would say that for this point in the school year for a high school geometry class they probably havnt gotten to trig yet...\n\nhope this helps!!
491970
Someone posted the exact same set of questions a few days back, either here or on the physics page - you ought to be able to find the methods in the responses. Have a look.\n\n\nBest of Luck - Mike
527006
I won't show in details since it's hard to type out the work, but you'll get the rest, here it goes:\n\na)\nWork = mgd  *d=change in height\nW = (10kg) (9.8m/ss) (sin15) (-7.5m) = -190J\n\nb)\nW=(friction coef)(natural force)(distance traveled)\nnatural force = (cos15)(10kg)(9.8m/ss)\nW= (0.4) (cos15) (10kg) (9.8m/ss) (-7.5m) = -280J\n\nc) \nW=Force x distance\nW=(100N)(7.5m)=750J\n\nd)\n(initial KE)+ (puller work)=(gravity work) + (frictional work) + (final KE)\n(final KE) - (initial KE) = 750J - 280J - 190J = 280J\n\ne)\n(final KE) - (initial KE) = 280J\n*solve for final v\nv = squareroot (2 x [280j+ (1/2)(10kg)(1.5squared)] / 10kg) =7.63m/s
548445
maybe what your teacher is trying to say here is the body temperature. now search for that answer. it's your homework.
298840
Remaining length of string\n= 7m - 2.56m\n= 4.44m\n= 444 cm\n\nHow did I do it, below is the working.\n7m is also same as 7.00m.\n\n  7.00\n- 2.56\n------\n  4.44\n------\nCancel the last 0 to 10.\nCancel the middle zero to 9.\nCancel the 7 to 6.\n\n1m = 100cm\n4.44 m = 100 cm * 4.44 = 444cm
43656
Your equation was partially cut off. This is what it looks like in my browser:\nInside Diameter=Outside Diameter + [2(Height-Tolerance)(1-Compres...\n\nIf this is the final part of the equation:\n...(1-Compression)]\nthen the equation, solved for Height, is:\n\nLet:\nI = Inside Diameter\n0 = Outside Diameter\nH = Height\nT = Tolerance\nC = Compression\n\nI=O+[2(H-T)(1-C)]\nI-O=2(H-T)(1-C)\nI-O=2(1-C)(H-T)\nI-O=(2-2C)(H-T)\n(I-O)/(2-2C)=H-T\n[(I-O)/(2-2C)]+T=H\n\nAnswer:\nH=T+[(I-O)/(2-2C)]
103896
I believe the bounds are where the numbers are deviated but still close. Say, in the problem, 11 cm is measured to the nearest centimeter. So, it may not be 11cm. It could be 10.6 cm, 11.3 cm, or even 10.51294803 cm. But they'll all be 11cm in the measurement. \n\nSo for the question, if it's asking for the minimum possible area:\nLet l be length and w be breadth. \n10.5 <= l < 11.5\n8.5 <= w < 9.5\nThe lower bound would be composed of the shortest length and breadth. \nA = 10.5(8.5) = 89.25 cm^2
42839
The length is 10 miles and the width is 8 miles.\n\nOkay... this is what you do:\n\nLet w = width\nLet w + 2 = length, because the length is 2 miles larger than the width.\n\nArea = (Length)(Width)\n80 = (w+2)(w)\n80 = w^2 + 2w\n0 = w^2 + 2w - 80\n0 = (w+10)(w-8)\nw = - 10 miles or 8 miles. Since there is no such thing as a negative distance, it's 8 miles.\nw + 2 = length, so...\n8 + 2 = length\n10 = length
528935
first why do we need his moving 30 meters and what does it mean anyway? i don't understand the question here.. first he let the kite and it was flying and the string made an angle of 47 degrees.. and then he walked somewhere 30 m? so now it is an angle of 39 degrees? i could say some little things to help only:\n\n-we watch a man as a spot that doesn't move\n-a string has first brace at force T1, and it is equal the strength of wind that is flowing against kites dimension S.. so it is T1=S(paralel) * P(paralel).. \n-this task is too complicated, i'll write it down, see your mail tomorrow and send you, it is 2 am here..\n-good night
538577
Range = High - low:\n\n119° - (-54°) = 119° + 54° = 173°F\n\nThe range is 173 degrees from -54°F to 119°F\n\nThe only hard part is in subtracting the negative temperature, which is the same as adding the positive value...
510480
Convert all this to MKS units and find the cross-section of that wire.\n\nE (stainless-steel) is given to be 18 x 10^10 N/m^2.\nCross-section is from pi(r^2).  This must be in m^2, not mm^2.\n3.14159(0.125*10^-3 m)^2 = 4.909 x 10^-8 m^2 cross-section.\n\nYoung's modulus (E) is given as,\n\nE = (delta F / cross-section) / (delta length / original length)\n\nDelta F is the tension in the wire, thats the first part of solving this problem.  So rearrange the equation for delta F, renaming delta F as "tension" and substitute the value given in place of E:\n\nTension = (18 x 10^10 N/m^2) * (8 x 10^-5 m / 0.028 m) * (4.909 x 10^-8 m^2)\n\nTension in the wire is therefore 25 N.\n\nThe magnitude of force on the tooth is 25 sin 30 deg or 12.5 N\n\nWarning!:  While the reasoning is sound, double-check all of my math.  Also, note that I rounded the intermediate answer 25 N.
444932
1.\nConsider horizontal distance\nv(horiz)=25 m/s\nDistance=50 m\nTime=50/25=2 sec\nNow consider vertical distances\nsince acceleration due to gravity is g=9.8 m/s/s,\ndistance, S=ut-gt^2/2  where u=initial vertical velocity=0\nS=0x2 - 9.8x2^2=39.2 m.\n\n2. \nConsider vertical distance\nS=ut-gt^2/2=4\nu=0 (horizontal)\nt^2=4x2/9.8=0.8163\nt=sqrt(0.8163)=0.9035\nNow consider horizontal distance\nThe ball has travelled 25 m in 0.9035 sec,\ntherefore the horizontal speed is 25/.9035=27.67 m/s
107785
You need to find half of the accuracy (nearest) which is 50 cm, and add it to then subtract it from the measurement.\n\nSo upper bound = 1450 cm; lower bound = 1350 cm.\n\nAgain, the reason it's half is because that's where rounding changes the answer.
607518
i am no genius but i did try. here's how i went about it.\nspeed = distance/time.\n(1/30)s = 525 lines, therefore time for 1 line = ((1/30)/525)\n=(1/15750)s\n20in = 1 line.\nresweeping the line, therefore distance = 40in.\n40in = 3.33ft (1ft = 12in)\nagain, speed = distance/time, so\nspeed = (3.33ft)/(1/15750s) = 52500 (ft/sec)\nhopefully, this is the right answer. please let me know. thanks.
477468
initial temperature is when t=0\ne=400+121=521\ntemperature at the end of 5.8 min\ne=400+121/e^(0.3)(5.8)\nwhen you give a formula like this it is better to say what the various symbols stand for and their units\nsince i do not know in what units t has to be taken i have taken it in min.\nmake suitable correction if necessary