OCR A level Physics help required!!

thought so, thanks...

So what did you deduce?

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.

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.

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?

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?

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.

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)?

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?

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?

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!!!!!!!

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?