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    (Original post by SexyNerd)
    thought so, thanks...
    So what did you deduce?
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    (Original post by uberteknik)
    So what did you deduce?
    average acceleration is 4 t = 4 .... v = u +at ... v =4x4 v =16
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    (Original post by SexyNerd)
    average acceleration is 4 t = 4 .... v = u +at ... v =4x4 v =16
    correct.

    Another way is by finding the area under the graph: (8m x s-2) x (4 s) x 1/2.
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    (Original post by uberteknik)
    correct.

    Another way is by finding the area under the graph: (8m x s-2) x (4 s) x 1/2.
    yeah thanks... when is the speed at its greatest? speed = change in distance..
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    (Original post by SexyNerd)
    yeah thanks... when is the speed at its greatest? speed = change in distance..
    When the acceleration stops. i.e. when the graph crosses the x-axis.

    Even though the slope in places is -ve, it simply means the acceleration is decreasing. The velocity is still going up because by definition the object is still undergoing +ve acceleration.
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    (Original post by uberteknik)
    When the acceleration stops. i.e. when the graph crosses the x-axis.

    Even though the slope in places is -ve, it simply means the acceleration is decreasing. The velocity is still going up because by definition the object is still undergoing +ve acceleration.
    thank you....
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    (Original post by uberteknik)
    When the acceleration stops. i.e. when the graph crosses the x-axis.

    Even though the slope in places is -ve, it simply means the acceleration is decreasing. The velocity is still going up because by definition the object is still undergoing +ve acceleration.

    for a vector, i have a -(i/x) and a +(y), I calculated the magnitude correctly, but now it wants to now the angle to x/i, how would I do that, whats the law/rules?
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    (Original post by SexyNerd)
    for a vector, i have a -(i/x) and a +(y), I calculated the magnitude correctly, but now it wants to now the angle to x/i, how would I do that, whats the law/rules?
    This is simple trig:

    If the vector is of the form z = (a + jb) then amplitude = modulus(z) = sqrt(a2 + b2) i.e. the length of the hypotenuse.

    Angle theta is given by arg(z) = tan-1(b/a)

    The i or j complex notation represents a rotation about the origin where j = pi/2 rad anticlock from the x-axis and increases with integer powers of j by pi/2 rad.
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    (Original post by uberteknik)
    This is simple trig:

    If the vector is of the form z = (a + jb) then amplitude = modulus(z) = sqrt(a2 + b2) i.e. the length of the hypotenuse.

    Angle theta is given by arg(z) = tan-1(b/a)

    The i or j complex notation represents a rotation about the origin where j = pi/2 rad anticlock from the x-axis and increases with integer powers of j by pi/2 rad.
    i understand how to do the mag.... i also calculated the angle correctly... but im assuming because its a neg i value, its was 180 - (the angle I calculated).... Could you explain what the law/rule is as simply as possible please?
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    (Original post by SexyNerd)
    i understand how to do the mag.... i also calculated the angle correctly... but im assuming because its a neg i value, its was 180 - (the angle I calculated).... Could you explain what the law/rule is as simply as possible please?

    So you need to make a quick sketch of the vector to find out which quadrant of the (x,jy) axis the direction lies.

    Then the angle theta will be self evident as either a +ve or -ve direction from the x-axis. It is normal to quote the angle such that -pi < theta < +pi rad.
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    (Original post by uberteknik)
    So you need to make a quick sketch of the vector to find out which quadrant of the (x,jy) axis the direction lies.

    Then the angle theta will be self evident as either a + ve or -ve direction from the x-axis. It is normal to quote the angle such that -pi < theta < +pi rad.
    thanks.... object hits for with velocity u, and rebounds with a velocity v, impact with floor is time t with the floor... calculate magnitude of average acceleration?
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    (Original post by uberteknik)
    So you need to make a quick sketch of the vector to find out which quadrant of the (x,jy) axis the direction lies.

    Then the angle theta will be self evident as either a +ve or -ve direction from the x-axis. It is normal to quote the angle such that -pi < theta < +pi rad.
    Can you explain this to me from the fundamentals please ... i know T (time) for complete oscillation is one second.... the question states, of point p (start from point p)?
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    (Original post by SexyNerd)
    Can you explain this to me from the fundamentals please ... i know T (time) for complete oscillation is one second.... the question states, of point p (start from point p)?
    So the question is asking you to draw the movement of point P against time as the wave passes through. It's checking to see if you understand what an amplitude vs time graph represents for a wave passing a given point.

    Start by drawing points which can be easily calculated and then interpolate (join them up):

    The wave is traveling at 0.5 m/s and as the intervals given are in convenient 0.25 m steps, the wave will take (0.25/0.5) seconds (t = distance/speed) starting from point X to reach point P.
    So the first point will be 0 amplitude at time t = 0.5 seconds. i.e. the time it takes X to get to P. Plot this point on the graph.

    Now pick the next zero. i.e. the next point when the wave crosses the x-axis. How long does it take to reach P from it's starting position? Plot that on the graph.

    Do these same steps for all the zero crossing points.

    Now start with the maximum and then minimum amplitude points: How long will each one take to reach point P from their starting positions? What does the energy in the wave do to point P? i.e. in which direction will point P move and what amplitude will it reach as the wave passes through? Plot these points.

    This gives you enough points and the insight to now sketch the resulting curve. Be careful to include all of the line from 0 to 2.5s as the question states.



    How did you get on?
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    (Original post by uberteknik)

    Do these same steps for all the zero crossing points.
    i understood how to cacl the time for each point....

    Now start with the maximum and then minimum amplitude points: How long will each one take to reach point P from their starting positions?
    i dont understand?

    What does the energy in the wave do to point P? i.e. in which direction will point P move and what amplitude will it reach as the wave passes through? Plot these points.
    yes, here i just do what a transverse wave does?

    http://www.youtube.com/watch?v=2Wlh3M2a10U so its above x, then below and the final is above?

    This gives you enough points and the insight to now sketch the resulting curve. Be careful to include all of the line from 0 to 2.5s as the question states.



    How did you get on?
    thanks, im sure i get it .... i will practice more similar questions though...
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    (Original post by uberteknik)
    So the question is asking you to draw the movement of point P against time as the wave passes through. It's checking to see if you understand what an amplitude vs time graph represents for a wave passing a given point.

    Start by drawing points which can be easily calculated and then interpolate (join them up):

    The wave is traveling at 0.5 m/s and as the intervals given are in convenient 0.25 m steps, the wave will take (0.25/0.5) seconds (t = distance/speed) starting from point X to reach point P.
    So the first point will be 0 amplitude at time t = 0.5 seconds. i.e. the time it takes X to get to P. Plot this point on the graph.

    Now pick the next zero. i.e. the next point when the wave crosses the x-axis. How long does it take to reach P from it's starting position? Plot that on the graph.

    Do these same steps for all the zero crossing points.

    Now start with the maximum and then minimum amplitude points: How long will each one take to reach point P from their starting positions? What does the energy in the wave do to point P? i.e. in which direction will point P move and what amplitude will it reach as the wave passes through? Plot these points.

    This gives you enough points and the insight to now sketch the resulting curve. Be careful to include all of the line from 0 to 2.5s as the question states.



    How did you get on?
    ke = 8.4 x 10^16

    x = 200

    v = 2 x 10^4

    mass = 3 x 10^8

    calculate average force acting object, when x = 200....
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    (Original post by uberteknik)
    So the question is asking you to draw the movement of point P against time as the wave passes through. It's checking to see if you understand what an amplitude vs time graph represents for a wave passing a given point.

    Start by drawing points which can be easily calculated and then interpolate (join them up):

    The wave is traveling at 0.5 m/s and as the intervals given are in convenient 0.25 m steps, the wave will take (0.25/0.5) seconds (t = distance/speed) starting from point X to reach point P.
    So the first point will be 0 amplitude at time t = 0.5 seconds. i.e. the time it takes X to get to P. Plot this point on the graph.

    Now pick the next zero. i.e. the next point when the wave crosses the x-axis. How long does it take to reach P from it's starting position? Plot that on the graph.

    Do these same steps for all the zero crossing points.

    Now start with the maximum and then minimum amplitude points: How long will each one take to reach point P from their starting positions? What does the energy in the wave do to point P? i.e. in which direction will point P move and what amplitude will it reach as the wave passes through? Plot these points.

    This gives you enough points and the insight to now sketch the resulting curve. Be careful to include all of the line from 0 to 2.5s as the question states.



    How did you get on?
    stopping d proportional to v^2

    v = 32


    PLEASE HELP!!!!!!!
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    (Original post by SexyNerd)
    stopping d proportional to v^2

    v = 32


    PLEASE HELP!!!!!!!
    Not enough information to understand the problem.

    What is the question?
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    (Original post by uberteknik)
    Not enough information to understand the problem.

    What is the question?
    DW but thanks... How can a potential divider circuit be used to produce a variable p.d?
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    (Original post by SexyNerd)
    DW but thanks... How can a potential divider circuit be used to produce a variable p.d?
    Use a variable resistor (potentiometer).

    If you think about the physical construction of a pot', you will note a single resistor made in the shape of a thin flat surface over which a conductive slider can move. A voltage is applied to terminals which themselves connect to either end of that flat resistor.

    A 'wiper' connected to a third middle terminal, is able to move along the entire length of the flat surface between either end of the resistor.

    This effectively creates a potential divider between one terminal of the pot' and the wiper terminal. i.e. one part of the potential divider resistance is between the wiper and one of the end terminals. The other part of the divider is between the wiper and the other end terminal.

    As the wiper moves along the flat surface, the ratio of the resistances between the wiper and either of the end terminals is continually changed.

    In this way a variable pd can be picked-off between either end of the pot (acting as a reference) and the wiper terminal.
 
 
 
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