What is the deflection fields?
83027
Here's a link that might help you:  http://www.mebs.co.uk/deflection_fields.htm
42505
For a fully supported beam with a uniformly distributed load, the inflection point would be in the center of the beam. Inflection points are by definition points of zero moment or places where deflection goes from positive to negative.
871516
mass of the object\nforce of the magnet in a magnetic field\nthe object's own interference with the magnet
684438
print this out, it's in cross ref form\n\nhttp://www.iop.org/EJ/abstract/0031-9155/48/12/401
799966
the external angle\n\ni hope that this helps
656305
Heat distortion temperature (HDT) is determined by placing the the sample material under a defined pressure load and then exposing it to heat which is uniformly raised. The temperature at which a specified deflection occurs is the HDT.
344162
It depends upon the design of the particular meter and is specified by the manufacturer as the amount of current required for full-scale deflection of the meter. This commonly falls in the microamp or millamp range.\n\nAn external resistor called a shunt may be placed in parallel across the meter to multiply its full-scale range by several magnitudes for measurement of much larger currents using the same meter.\n\nWK
641362
coriolis effect\ndeflection relative to Earth: the observed deflection of something such as a missile in flight relative to the Earth’s surface, caused by the Earth’s rotation beneath the object. \nThe deflection is to the right in the northern hemisphere and to the left in the southern hemisphere.
493310
Refraction: Deflection from a straight path undergone by a light ray or energy wave in passing obliquely from one medium (as air) into another (as glass) in which its velocity is different.
687946
In fatigue analysis where you apply a load range to your FEA model, i.e. if you have a load of +/- 10kN then you would apply a single load of 20kN to represent the whole load amplitude.\n\nWhen you come to look at the stresses you look at maximum principal and minimum principal stresses to determine the location and level of peak stress range.  The rationale behind this is that you could just as easily apply your load in the opposite direction, in which case all the stresses would be reversed.\n\nDon't forget that the magnitude of the minimum principal stress isn't necessarily smaller than the maximum, it can be large and negative.
919976
Apparently we are ignoring any pitching motion due to their getting in the car and due to the bump, since you'd need to know a lot more about mass distribution if it was pitching. So we're only considering vertical movement and the distance between axles is irrelevant. I'm assuming "a total of 8.9 cm" means the sum of the 4 spring deflections, so their average deflection is 2.225 cm. If it actually means the average deflection, then k and m are 1/4 what you'll get with the following calculations. Stiffness k (of the 4 springs in parallel) in n/M = 146*g/.02225. Natural radian frequency w = sqrt(k/m), so m = k/w^2 = k/(2*pi/1.7)^2. (That's the total m; mass of car = m-146.) \nThis was not a well-phrased question, because "a total of 8.9 cm" also could mean 8.9 cm from their no-load length (and thus due to the weight of the car and the students). However this would result in an insoluble problem. \nBTW, why are they geeky? Did they figure all this out before me?
440395
Refracted light rays bend toward the normal when the material being entered has a higher index of refraction then the material you are coming from.  Reflected light behaves according to total internal reflection, and bounces off the surface at the same angle that it approached at, much like skipping a stone off a lake's surface.  Check out snell's law on wikipedia for more.
1137202
Cosidering that the load (2P) has been applied to the midpoint of the bar, the load which will be carried by each cantilever beam will be 'P'.\nIn that case, the maximum deflection at the free edge of the system will be  \n\n(P*L*L*L )/ (3EI)\nwhere L is the span of each beam \nEI is the flexural rigidity
179131
Try this:\nhttp://kr.cs.ait.ac.th/~radok/physics/f3.htm
644077
The voltage at which the attractive forces between the capacitor plates overpowers the fasteners.\n\nTake a look\n\nwww.math.ubc.ca/~ward/papers_ps/memspap.pdf
1018810
thats deflection,,,and mesurement
240384
http://hyperphysics.phy-astr.gsu.edu/hbase/hframe.html