Recent content by dr3vil704

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    Hydraulic Lift.( U tube) and oil

    Homework Statement A 12,000 N car is raised using a hydraulic lift, which consist of a U - tube with arm of unequal areas, filled with oil with a density of 800 kg/m^3 and capped at both ends with tight-fitting pistons. The wider arm of the U tube had a radius of 18. cm and the narrower arm...
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    Solve Derivative of sin^2(pie*Z) with respect to Z

    hm..I believe the derivative of sin is cos. so the answer should be 2cos(pi*Z) * (Pi) or 2\picos(\piZ). the first part is power rule and derivative of the Sin which is Cos. for the inside of the ( ), since it's product of a variable and a constant, we know that Pi is a number hence it a...
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    Solve Derivative of sin^2(pie*Z) with respect to Z

    yeah, so Use the chain rule. Look at it as (sinx)^2 because is easier if you look at it like this.
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    Solve Derivative of sin^2(pie*Z) with respect to Z

    sin^2(pie*Z) is sin^{2}(\piz)? If it is, Then can't you use the chain rule? which is power rule on the whole out side of the ( ) and then it by multiply by the derivatives of the inside of the ( ). because sin^{2}(\piz) is just the same as (sin(\piz))^{2}
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    How can a periodic function with period 1/2 be evaluated at specific values?

    Homework Statement Suppose that f(x) is a periodic function with period 1/2 and that f(2)=5, f(9/4)=2, and f(11/8)=3. Evaluate f(1/4), f(-3), f(1,000) and F(x) - f(x+3) (I'm not sure on this one, the teacher never really taught us this, we are on Derivative right now, but this is just one...
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    Simple Ladder leading against a vertical wall become complicated.

    Oh, I see the mistake now. I guess I forgot to put the distance with the force So the equation should be (1/2 X mg X L X Cos\theta) = (N X Lsing\theta) +(Ff X cos\thetaX L) But I do no get why the Ff agaisnt the wall is not \muN
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    Simple Ladder leading against a vertical wall become complicated.

    yes, I did put the pivot point on the floor, which cancel out the normal and friction between floor and ladder. So the frictional force that is left is between the wall and floor. And my friction and normal is not incline so it just \mumg. The first equation is Tcw=Tccw
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    Simple Ladder leading against a vertical wall become complicated.

    Homework Statement A uniform ladder of length L and weight w is leaning against a vertical wall. The coefficient of static friction between the ladder and the floor is the same as that between the ladder and the wall. If this coefficient of static friction is μs = 0.500, determine the smallest...
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