A piece of wire is 14m long is cut into 2 pieces. One is bent into a square. The other is bent into an equilat
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Let x be the length for the square and 14-x for the triangle.\n\nThe area of the square is (x/4)^2 = (x^2)/16.\n\nThe height of the triangle is sqrt(3)(14-x)/6 and the base of the triangle is (14-x)/3, so the area of the triangle is\n\n1/2 bh = [(14-x)/6] [sqrt(3)(14-x)/6] = (14-x)^2 sqrt(3)/36\n\nTotal area A = (x^2)/16 + (14-x)^2 sqrt(3)/36\n\nA = [9x^2 + 4(x^2 - 28x + 196) sqrt(3)] / 144\n\nA = [(9 + 4 sqrt 3) x^2 - (112 sqrt 3) x + (784 sqrt 3)] / 144\n\ndA/dx = [2(9 + 4 sqrt 3) x - (112 sqrt 3)] / 144 = 0\n\n(9 + 4 sqrt 3) x = 56 sqrt 3\n\nx = (56 sqrt 3) / (9 + 4 sqrt 3) = 6.0895 meters  (Answer)\n\nThe area of the square is 2.32 m^2.  The area of the triangle is 3.01 m^2, and the total area is 5.33 m^2.
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T1=m1*a, no matter are there m2 & m3 or not.\nT2=(m1+m2)*a, no matter whether m1 & m2 are connected with a string or are in one lump, the presence of m3 being unimportant too.
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Let x be the side of a square\nThen, area of the square is, A=x^2\nThe rate of change of area w.r.t side is dA/dx =2x\nIn the problem another parameter is given, that is time.It is given in the problem that there is a constant change in the length of the side at the rate 0.4cm /sec.Here we have to notice that corresponding change in area is not uniform throughout.Therefore we can only find instantaneous change in area or accumulated area in the given periodor interval.That means change in area at a given instant of time or interval.\nFor example, let initial area be 0.Now, to find the area between the interval a and b we have to reintegrate. ie. integration of dA between the interval a and b =integration of 2x in the interval a & b.\nie. [A]a&b= [x^2]a&b=b^2-a^2 square units\nFor instance, change in area between the interval 9th and 10th seconds is =[x^2]a&b=b^2-a^2 square units\nGiven, 0.4units/sec,considering intial length and time as 0,we have, length of the side on the 9th second is a=9*0.4=3.6units and in the same way length on the 10th second is b= 4units\nAccordingly change in area(accumulated area) at the 10th second is =b^2-a^2\n=16-12.96=3.04 square units [this is an instataneous change]\nPlease note that, this change cannot be written as units^2 /sec
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Strain is defined as the fractional change in length under an applied force.  Since at the instant the other wire is cut (i.e., at t=0), the remaining wire remains stationary (until it starts to stretch at a later time), the strain on that wire is 0 becuase it hasn't changed length yet at t=0..\n\nThe answer is 0.
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the wire is magnetized, which can be either through the action of another magnet, or by its being coiled and run through by an intensity
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For a), we are asked to assume this is a parabola, even though on a real bridge the cable does not assume such a shape.\n\nTo solve for k, put in\n\n143 = k * 640^2\nk = 143 / 640^2\n\nDo the computation yourself to at least one decimal place.\n\nFor b), we need to be able to compute the length of a section of a parabola. For that, we need the derivative of the curve,\n\nf'(x) = 2x\n\nThe formula for the length is the integral of \n\nsqrt(1 + 4x^2) \n\nover the desired range. To try to integrate this, we can use trigonometric substitution, defining an angle theta such that \n\n2x = tan(theta)\nx = tan(theta) / 2\n\nNow we take dx/d(theta) = sec^2(theta) / 2\n\nsqrt(1 + 4x^2) = sec^2(theta) / 2
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Try these answers:\n\nFor the first one, shortest is 0.0086 m, longest is 8.6 m.\nFor the second one, it should be 2.95 Hz
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2/x = 4/14\n\n4x = (2)(14)\n\n4x = 28\n\nx = 7 Meters\n\nThe answer is x = 7 meters\n\nThe Streetlight is 7 Meters high.\n\n- - - - - - -s-
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4 times the length of one side
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You could just use the old force = (G*M*m)/r^2\nWhat is missing from the diagram is that the small weights were held by a long thin wire.  The wire formed a weak torsional spring.  If there were some gravitational force, it would cause the weights to twist the wire until the torque of the twisted wire matched the force of gravitation (times the radius).  He could measure the amount of twist by shining a light at a mirror attached to the weights and seeing how far the reflection moved.
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a) 72.072 cm/sec^2\n\nb) 82.63 cm/sec^2
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One Cubit:\nFeet:1 1/2\nInches:18
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eletricity pasess through a metal wire and then pasess through filiment which not alot of electricity can pas so it is given off as heat and electricity that does make it through goes down another metal wire.
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there is insulation on the wire...........
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A hexagon has 6 sides.
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A is the answer.  The other answers doesn't make sense at all.\nI just learned this stuff 3 years ago!
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Assuming you mean as 4 shapes equivalent to the original L shape, the answer is given in the attached image.