How to calculate a weight of circular piece having dia 150cm, d1 120cm & length 300cm?
613586
For calculating weight you will be needing the specific gravity or bulk density of the material it is made of.\nElse you need a weigh balance.
392733
Divide the circumference of a circle its diameter.
100875
divide your weight in pounds by 2.2  this will be your weight in kilograms
950847
the base angles are equal therfore each is (180-25deg17')/2\nso each is 155deg43'/2   which is 77deg51.5' which as a decimal is  is 77.8583degrees\n\nNOW circular measure is The measure of an angle in radians.\nThere are 2*Pi radians in 360deg therefore\n1deg = PI/180 rads\ntherefore\n77.8583degrees= 77.8583 times Pi/180 rads\n\ntake Pi as 3.142 and a calculator gives 1.359 rads
1031813
use the formula pi*r^2*h\nwhere pi=1.34\n r=radius (in inches)\n  h=height of cylinder (inches)
621321
If the density is 'd' kg/m^3 then the weight is '9.81d' Newton. [remember weight is force]
150570
Weight   3,120+ lbs \n Height   5 feet \n Circumference    15.1 feet  \n Duration of project   5 years 2 months \n Cost of materials   approx. $25,000. \n Number of rubber bands   850,000
183956
Half the diameter, or circumference divided by 2.pi
768468
Diameter and height will be equal in this case.  As V = 55\npi r^2. 2r/3 = 55 giving r^3 = 165/4\n\nr = 3.455\n\nh = 2r = 6.91
1079085
use this formula\n55x54x53x52x51/120
481398
sum the forces in the direction that they are applied and it will be equal for a static object. \n\nRule of thumb on this one is that you have 100N down ward, then you need 100 N's up ward.\n\nGivens, legs are equal lengths, same material, and no angles to calculate.\n\n4 legs - 4 forces - each at 25 Newtons.  \n\nNow if you want to get fancy, break out the strain gauges and let us see what we really find.  But it helps if you have preformed a few material tests and experimenters first.
703001
To find the force P, its equal in size, but opposite in direction to the weight of the crate,resolved in the direction of the slope:\n\nP=Wcos60 or P=Wsin30\nP=(80*9.81)cos60=392.4N\n\nR is equal in size, but opposite in direction to the weight of the crate, resolved perpindicular to the slope\n\nR=Wcos30 or R=Wsin60\nR=(80*9.81)cos30=679.66N\n\nPlease note you need to use weight, not mass ( W=m*g)\n\nGo do the math!
576519
tensile strength = (2*P)/(pi*t*d)\nwhere; P =maximum load you get from the machine(kPa); pi = 3.14; t = thickness/height of specimen (m) and d = diameter of specimen(m)
859844
The previous answerer has the right idea but the wrong details.\n\nThe four parts should weigh 1, 3, 9 and 27 pounds.  The logic is that the desired weight is written in "balanced ternary" notation, where the values of the digit positions are 1, 3, 9, 27, 81, . . . (that's "ternary"), but the allowed digit values instead of being 0, 1, 2 (that's regular ternary) are -1, 0, 1 (that's balanced ternary).  Then the digit values for any desired total weight written down in balanced ternary tell you whether the weight for that column is to go in one pan, go in the other pan, or stay on the floor.
759031
I agree with the first answerer...where is the figure? But let's see if we can solve it just the same. We have to guess a little regarding the location of the supports.\n\nGiven: length of board=5.0m\n           mass of board=54kg\n    Hence,  weight of board =54g N.\n    Place this at the middle of the board because it is a uniform board as given in the problem.\n\nFirst support is 3.4m from the end, and the second is 4.6m form the end. \n\nDiver stands at end of the board. He weighs 65g N.\n\nFind: F1 at first support and F2 at second support.\n\nTake torque about F1:\n\n54g*(2.5-0.4)+65g*4.6=\nF2*(1.6-.4)\n\n54g*2.1+65g*4.6=1.2F2      Call this equation 1.\n113.4g+299g=1.2F2\nF2=(1111.3+2930.2)/1.2\n    =3368N upward.\n\nTake torque about F2:\n\n54g*(3.4-2.5)+65g*3.4=\nF1*(1.6-.4)\n\n54g*0.9+65g*3.4=1.2F1       Call this equation 2.\n\n48.6g+221g=1.2F1\nF1=2642/1.2\n    =2201.7N downward\n\nCheck your answer:\n\nF1+F2=65*9.8+54*9.7\n-2201.7+3368=637+529.2\n\n3368=637+529.2+2201.7\n       =3367.9\n0.1 difference is acceptable. so our solution is correct. (Assuming my assumptions as to locations of supports are same as in your figure.)
829185
2000N is the weight of the object...\nOr is that too obvious?
1094108
Use a calculator and raise it to the 1/3 power.\nDon't try to do it by hand...