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its basically because if you want a vector that doesn't start at the origin you do endpt - startpt since its QO and not OQ, endpt is O startpt is q

Re: Where can you get a collection of all past Maintaing a Balance Questions from the don't the success one books do this?

For the triangle APB, the base length is distance AB and the height you get using the perp dist eqn using P and x-y+6=0. Using area=bh/2, you can get an equation for the area in terms of p, which you can differentiate and find the max value of. Evaluate the given integral and show that the max...

Let the length PA = x. Therefore QA = p-x. If you sub in all your values, you get a quadratic equation in x with p as a constant. The use quadratic equation.

The reaction force is always in the opposite direction of the applied force. The normal force is perpendicular to the surface that the object is in contact with.

no you don't have to but it is the safe option you can mix and match colours if you know what you are doing btw why is this in chemistry?

oh ok so for escape velocity if you were to throw an object away from a planet at escape velocity, it would break free from the gravitational attraction. This implies that it would never come back down. It keeps going until it reaches the "peak" of its motion where it final velocity is 0...

did you mean gravitational potential energy? because otherwise i have never heard of this

the SI unit for density is kg/m^3

lol beat me

if the range for theta os 0 to 360, then the range for 2theta is 0 to 720 you should get 4 answers

same thing. multiply both sides by 3 then move the y over. the form they gave you is the general form and the answer you got is the gradient-intercept form.

what is the answer supposed to be?

the tangent must pass through 2 points: (2, -12) and (a, a^2). it must also have a gradient of 2a. try putting these two conditions together

for q1 first find sec(45) then determine whether it would be +ve or -ve based on quadrants. same for q2

(x+2)^2 = (2x-3)^2 + (x+3)^2 - 2(2x-3)(x+3)cos(90-q)

it isnt, the markings on the diagram are wrong. i repeat cos rule anyone?

cant you just use the cos rule since cos(90-q) = sin q

you're basically just finding the equation of the tangent at x=1/4

assuming there is no initial velocity (it was dropped not thrown downwards) you can use s=0.5at^2