Recent content by phoenix20_06

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    Tangential and Radial Acceleration problem

    Thank you very much all, I think I understand now
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    Tangential and Radial Acceleration problem

    I understand that the coordinates include Radial...I'm not sure what transverse acceleration is...(is it tangent acceleration?) What I don't understand is the geometry that has to be used in with only 2 vectors. If I had tangential acceleration, finding radial would be piece of cake... I know...
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    Tangential and Radial Acceleration problem

    What confuses me is the calculation of AC in the solutions... Ac = (22.5)cos(53.1) + (20.2)cos36.9 = 29.7 It seems that (22.5)cos(53.1) implies that 22.5 is a hypotenuse in some triangle that got formed using Atot components. The same for 20.2. I don't see how ...:( Even if this gets us Ac's x...
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    Tangential and Radial Acceleration problem

    Thank you for the reply. Unfortunately, it doesn't change anything in my progress :-(I found out that the angle between Atot and AC is 11.2 degrees and from there I don't see a solution
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    Tangential and Radial Acceleration problem

    Hello, This is my problem and the solution from the book. I don't understand how they came to the solution. I believe it's wrong and I offer my own solution(but incomplete). I would appreciate if you could explain how they arrived to their solution or if I'm right. Thanks:) Please see below...
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    A simple problem that's dirving me crazy (Perimeter and length of each side)

    What the deuce! It's wichcraft not mathematics! I just solved it the sides are c = 175, b = 168 and a = 49 49^2 + 168^2 = 175^2 Thank you Cristo, I think your presence in my posts make me think twice harder :) Those minuses messed up calculations in the first place :)
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    A simple problem that's dirving me crazy (Perimeter and length of each side)

    okay this what I get then... 4c^2 + 1582c + 159250 = 0 or 2c^2 + 791c + 79625 = 0 if I divide everything by 2 c = -b +/- sqrt of(b^2 -4ac) and everything divided by 2a however b^2 - 4ac is negative and I can't do a sqrt of a negative number. What went wrong? 791^2 - 4(2)(79625) =...
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    A simple problem that's dirving me crazy (Perimeter and length of each side)

    Homework Statement A city lot has the shape of a right triangle whose hypotehenuse is 7 ft longer that one of the other sides. Perimiter of the lot is 392 ft how long is each side. Homework Equations P = a + b +c = 392 hypothenuse is c^2 = a^2 + b^2 so if c is 7ft longer then one...
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    Finding Maximum Range of a Projectile: Velocity and Formula

    Berkeman thank you for your reply. I believe the solution would be: 242 = K60^2 x = K 70^2 so x/242 = (70^2/60^2) x = (4900/3600)* 242 = 398.38m It was very helpful :-)
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    Finding Maximum Range of a Projectile: Velocity and Formula

    Homework Statement Hello, I have a language problem that affects my Math problem :-) I hope you can assist me with this. The Maximum range of a projectile is directly proportional to the square of it's velocity. A baseball pitcher can throw a ball at 60 min/h, with a maximum range of...
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    Find slope of 2 perpendicular lines

    Whoaa you guys are the best! First of all, I'm terribly sorry for confusing you all with "I forgot to mention that k of line 1 isn't necessarily equals to k of line 2." This is my mind playing games on me after 2 days of thinking about this problem. Nowhere has it said that these 2...
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    Find slope of 2 perpendicular lines

    Cristo: thank you for your reply. It seems to me no matter how much I play with these 2 equations I always have either too many variables or get stuck with x and y square. I forgot to mention that k of line 1 isn't necessarily equals to k of line 2. So I don't see how writting them both...
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    Find slope of 2 perpendicular lines

    Homework Statement Find the values of k such that lines 3kx+8y = 5 and 6y -4kx = -1 are perpendicular. I don't need the answer, but a push in a right direction. However, feel free to solve the equation :) Homework Equations 1. Slope of a line is m = [y2-y1]/[x2-x1] where we have 2...
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