how to calculate a right lenght for the alpin ski poles?
106614
Wearing normal footware or in your socks. Turn the ski pole upside down, grasp it in your fist under the basket.  Place the top of hand grip on the floor.  The correct length pole will be when you forearm is horizontal with the ground.\n\nI have a wrist injury and have mine slightly shorter so that my wrist is angles downwards - but that is my personal preference.
695563
You're dealing with right triangles here. If you stand at the tip of the shadow, the length of the shadow is the adjacent side of the triangle and the height of the tree or pole is the opposite side of the triangle.  In any case, the ratio of the height to the length of the shadows will be constant.  If T = height of the tree, P = height of the pole, St = length of the shadow of the tree, and Sp = length of the shadow of the pole, then:\n\nT / St = P / Sp\nT = (P x St) / Sp\nT = (9 ft x 25 ft) / 15 ft\nHeight of the tree is 15 ft.
837697
1.  hard description...but if the cables are on opposite sides of the "weight" than the answer is...\n\nwell, if the 30 degrees is to the vertical than tension is equal to  29/cos30 degrees\n\nif it is to the horizontal than it is 29/sin 30 degrees.\n\n2.  again depends on the instructions...if thre 30 degree cable is to the vertical its    29 x tan30   \nhorizontal is 29/tan30\n\nsorry my first one was wrong, had to edit it
909305
you might try a figure 8 knot, or a stopper knot.\n\nhttp://www.chockstone.org/TechTips/Stopper.htm
269764
p = 540\np = 2*(l + w)\nw = l - 150\n\np = 540 = 2*(l + w)\n540 = 2*(l + l - 150)\n540 = 2*(2l - 150)\n540 = 4l - 300\n840 = 4l\n\n210 m = length\n\n\nw = l - 150\n\nwidth = 60 m
721808
If you are trying to make a triangle out of three equal lengths, all you do is devide 180 by 3.....and taddaaa you get your angles.\n\nIf does not matter how long each side is.
842774
well i am 5'2 and have 147s. I would try maybe well check out this website!http://www.alpineskicenter.com/skisizes.htm \nhope you have a good winter♥
189082
If this isn't a math question and you don't want to buy a tool and you don't have a protractor, there should be a leveled board on the ground, and your tilted board should be long enough to reach it. Mark the ground board where the tilted board touches it. Mark a point on the tilted board (the higher it is the more accurate your answer will be), drop a plumb line from there to the ground board, and mark where it touches. Measure the length from the bottom of the tilted board to the upper plumb-line mark (call it "A") and the distance between the two marks on the ground board (call that "B"). The cosine of the angle you want is B/A. Many calculators will give you the angle (using an arc-cosine function).
1057845
Well, first you should draw a diagram of the flag pole and the dog.\n\nAssuming the sun hits the dogs and pole at the same angle, you have similar triangles by ~AA (angle-angle) since the angle of the sun to the object is the same, and they are both perpendicular to the ground (creating a right angle ).\n\nThen set up a proportion of height to length for the pole and the dog (let x be the height of the pole):\nx/18 feet = 3 feet/8.1 feet\n\nTherefore x, the height of the pole should be (3/8.1)(18) = 6.7feet
318413
OOO, We did this project in IPC, except our was 20 meters -.-;;. You're ideal angle is 45 degrees. That usually gets you the farthest. It would be pretty smart to use elastic potential energy by using piece of bamboo or just a flexible stick (haha, flexible stick) or just a rubber band. Don't put anything on the egg. Our model was made with a giant rubber band that I found on the floor, and a little seat attached to the crossbow model. If your teacher is strict about a catapult model, your best bet would be to use the bamboo stick and research trebuchets.
777446
If as described the entire meter stick is in balance, then the center of its mass/gravity would be where the support is, at the 22.4cm mark\n\naha ... here we go then,\n\n142.3 * (22.4-9)  = 134 * (x - 22.4)\n\nsolve this for x to get the answer\n( think of the entire stick as a ball placed at position x)
902316
Let y be required distance from the far end.\n\nThe weight is acting at the C.G i.e., at 2 m from the far end.\n\n2/3 of the weight w is acting upward at y and \n\n1/3 of the weight w is acting upward at 4m from the far end.\n\nTaking moments about the far end, \n\n2/3 W y + 1/3 W 4 = W x 2. \n\n Dividing both sides by 2W and multiplying by 3 on both sides\n\ny + 2 = 3 \n\ny = 1m.
105220
From Outdoor-Links.com:\n\nHooks are sized by the number that represents the distance between the point and the inside of the shank or gap.\n\nLarge hooks (1/0- 100/0 increase in size as the number increases.\nSmall hooks (1-14) decrease in size as the number increases.
1154907
since P1.V1 / T1 = P2.V2 / T2 ... u don't actually care about the altitude. so V2 = 8.88 cm^3 (btw those degrees are °C, °F ?? cuz u gotta take them to Kevin before using them)\n\n1*10^5 Pa * 2cm^3 / 270 °K = 2*10^4 Pa * V2 / 240 °K\nV2 = 8.88 cm^3
604983
Thats simple. The formula is: Y2-Y1/X2-X1\n\nIt would be: 7-3/6-2 which is 4/4 which is 1. The slops is 1.
416914
Try to make it at 3 feet.  It will probably be easier.  I'd probably use balsa wood.
436114
well, you could be an architect and be using a 30-60-90 triangle and you needed to know a particular side of it. you can use x times the square root of 3, x being the shorter leg, to find the length of the longer leg.