what load will come on mast of 8 meter long by wind velocity of 80km/hr?
449818
You would need the cross-sectional area of the mast (i.e. the resisting face into the wind) in order to fully analyze and solve this problem.
459122
the side length is 2\nso the altitude should be 3^(1/2)\ngood luck
487670
None. It takes no power to maintain that speed (on level ground, no friction). The object has kinetic energy (0.5 * m * v^2). Power was required to put it in motion and power can be recovered by decelerating it. If you do the math, the energy is 3551.86 ft-lb, equivalent to 1 horsepower for 6.458 seconds. Answer 1 neglected the fact that weight must be converted to mass by dividing by g, and energy is proportional to v^2.
1023836
GOOGLE EARTH or NASA WORLD WIND, depending on what exactly you need, both are free of charge, downloadable and partially renewable.\ncheers
658354
http://www.lpl.arizona.edu/~rhill/conversion/windspd.htm\n\nanytime :]
401461
permitter would be 83 cm\narea would be 415 cmXsq
66785
Latitude, altitude, prevailing winds.
1161499
A Lot of factors to consider here.\nSize of the Windmill?\nWeight of the car?\nWeght of the cargo?\nWind speed?\nFor discussion's sake, let's figure the the windmill is 6' high.\nMounted to the top of the vehicle going 'with the wind'. and you're using a generator capable of producing 5 hp..\nA 15 mph wind should do it...\nThat's, of course, taking advantage of the push from the wind as well.
584010
E = 1/2 M x v^2 x F x H \n \nWhere v is the velocity of the load hitting the surface and H is the time period in which the frequency continues.
37097
Technically that question is impossible to answer.  It depends on any wind that was present, the atmospheric pressure, any air resistance, etc.
564626
Which depends on the stres acted on the load and material used
4105
Mars Tension-leg Platform -Gulf of Mexico  1432m(4700ft)\n\nBurj Dubai     --UAE --- 800 m (est.) (2,624 ft)\n\nKVLY-TV mast  ---USA ---629 m (2,063 ft)
581432
atmospheric pressure, winds, light
950679
(8.5m/s)/.008s = 1062.5 m/s^2 = a\nm = 12\nF=ma=12750N. <----- b)\nImpulse is Fdt, so 12750*.008 = 102 N/sec
917981
The pressure P(in pounds per square foot) from wind blowing at S miles per hour is given by P=.00256e^2.\n\nDon't you think the formula ought to have S in it?
948754
we juss need to take vector sum of the two forces. assuming the wind is directly in horizontal direction....... sqrt(9.25sq+1.05sq)=9.3N
1075431
Strength of material, velocity, distance; any of these ring a bell!