finding temperature at distant space at a position (3d)
128620
temperature itself is a property of matter. so, in space, where there is a near perfect vacuum, there is no temperature.\n\ntemperature of stars / dark matter in space can be measured by studying the radiation that is emitted / absorbed by it.
87941
Average Power = Work/Time\nWork=Force*Distance\nDistance=2*pi*radius=2*pi*(0.31 m)=(1.95 m)\nAverage Power = ((25 N)*(1.95 m))/(1.6 s)=30.4 W
254510
Well as far as a formula i can't help you there. i just happen to own a program that allows me to draw perfect shapes of any size and diminsion them accordingly (auto CADD). and at the top of your 300 ft house, or where ever it is that your measuring from the horizon at sea level is a tad over 21 miles away (21.20389 to be analy exact). first you must know the the earths circumfrance (i know i can't spell) which is 20,890,637 ft, next you have to know the distance you are away from the earth (which you do, 300 ft). now you have to know the formula for finding a line that intersects your hieght and crosses tangentally across the earths surface. know you have your triangle and apply the Pythagorean theorem. you have an answer.
74741
W=F*d,  d=W/F,  d=600J/400N = 1.5m
548151
P1V1/T1 = P2V2/T2\nP1 = 1 atmosphere = 101325 N / (m^2)\nV1 = (4*10^3 cm^3)(1 m^3/10^6 cm^3) = 4*10^-3 m^3\nP2 = P1 + h*2.00103/0.01\nV2 = V1 + 0.01h\nT1 = 20°C = 293.15°K\nT2 = 260°C + 273.15 = 533.15°K\nA = 0.01 m^2\nP1V1/T1 = (P1 + hk)(V1 + Ah)/T2\nP1V1/T1 = P1V1/T2 + ((kV1 + AP1)h + kAh^2)/T2\nP1V1(T2 - T1)/T1 = (kV1 + AP1)h + kAh^2 \nkAh^2 + (kV1 + AP1)h - P1V1(T2 - T1)/T1\nh^2 + ((kV1 + AP1)/kA)h - P1V1(T2 - T1)/(kAT1)\n((kV1 + AP1)/kA) = ((2.00103*4*10^-3 + 0.01*101325))/(2.00103*0.01)\n((kV1 + AP1)/kA) = 50636.82\nP1V1(T2 - T1)/(kAT1) = (101325)(4*10^-3)(533.15 - 293.15)/(2.00103*0.01*293.15)\nP1V1(T2 - T1)/(kAT1) = 16582.28\nh = (- 50636.82 ± √(50636.82^2 + 4*16582.28))/2\nh = - 25318.41 ± √(25318.41^2 + 16582.28)\nh = - 25318.41 ± 25318.74\nh = 0.34 m = 340 cm\n\nb)\nP2 = 101,325 + 0.34*2.00103/0.01\nP2 = 101,393 N/m^2
207313
the side length is 2\nso the altitude should be 3^(1/2)\ngood luck
410473
Use the Gay-Lussac gas law: V/T is constant. Note that T here is the temperature in Kelvin, not in Celsius. Luckily 27 C = 300 K
213835
1. 1080 degrees\n\n2. 1800 degrees
389673
You asked this question twice, and I posted a detailed solution in answer to your other question.  I know your teacher's answer is 49 degrees, but if you go to the nearest tenth of a degree, 48.9 is a better answer.
676043
I can think of two ways to measure the absorption and reflection coefficient of a material.  The first would only work with a polished surface.  First measure the intensity of a laser beam with a photodetector, then shoot the laser beam at the surface and aim the reflection at a photodetector.  Coefficient of reflection = transmitted intensity/initial intensity  (coefficient of absorption =1-reflection).  The other, harder way, which would work with any material, is to measure the heat gain in the target.\n\nFor this you're gonna want as small a piece of the target material you can get so that the laser still completely hits it.  Now attach a thermistor to the back of the target (again the smaller and thinner the target the better, ideal would be a foil just thick enough to be completely opaque).  Now shoot the laser at the target and measure the temperature as a function of time  It should be linear up until it starts radiating nearly as fast as it absorbs energy, so take points until it starts to level off and drop points after that.  Now find the specific heat of the thermistor-target apparatus by dipping it in water with a thermometer in it, using the initial and final temperature of the water, the volume of the water, and the initial temp of the apparatus, determine the heat it absorbed per degree it's temperature changed (this will actually give you specific heat*mass which is fine).  Now convert your temperature vs. time data to heat vs. time using the specific heat you determined (temp*(specific heat*mass)).  The slope of the line of best fit for this data will give you the rate it absorbed energy in watts.  Divide this by the lasers output in watts and you get the absorption coefficient.
343722
sorry i am no good at math, but this may help:\nhttp://www.fourmilab.ch/earthview/pacalc.html
501185
http://www.webreference.com/3d/ Try that, good luck with the assignment.
413406
looks like 240 degrees or  300 degrees. cause sine is +ve in 1st and second quadrant and -ve in the third and fourth quandrant and sin 60 = sqrt3/2. so i think above two answers are some possibilities( note there are infinite angles at which this is possible these are just some possibilities)
454831
If you are asking how to measure temperatures this cold (practically), you would use a T thermocouple.\n\nhttp://instserv.com/rmocoupl.htm
403397
°oO****Oo°°oO****Oo°°oO****Oo°\n\n\n 4 ∫[g(x)] g'(x) dx = 4 x 1/2g(x)² +c = 2g(x)²+c\n\nits the first suggestion that u made but i think you forgot the number 4...\n\n\n°oO~<<_P~'*'_ I~'*'_M~'*'_O~'*'~L _~'*'_A>>~Oo°
548445
maybe what your teacher is trying to say here is the body temperature. now search for that answer. it's your homework.
217226
http://www.nasa.com\nsee this