Recent content by e to the i pi

I'm confused... Where did you get that formula from? And what is the 'frictional force'?

Oh right... 9.8m/s^2 = 98/10 = 49/5. My mistake. So 14√3 * 49/5 = 686√3 / 5 Newtons going perpendicular to the plane (downwards). And 14 * 49/5 = 686/5 Newtons going down the plane. How do I work it out from here? Would the formula F = ma help me? If I let F = 686/5 and m = 28, then: a = F/m so...

So 14√3 * 49/10 = 343√3 / 5 Newtons going perpendicular to the plane (downwards). And 14 * 49/10 = 343/5 Newtons going down the plane. How do I work it out from here? Would the formula F = ma help me? If I let F = 343/5 and m = 28, then: a = F/m so a = (343/5) / 28 a = (343/140) a =...

1. A box of mass 28 kilograms is held at rest on a rough plane inclined at an angle of 60° to the horizontal by a rope attached to a vertical pole. When the rope is cut the box moves down the plane with an acceleration of a m/s^2. Find the expression for the coefficient of friction µ, in terms...

It's been almost a month and I still haven't solved it! Please help me!

Please tell me if I'm doing it right or wrong because I am still confused!

Okay, my new attempt at the solution is to use the formula: V = Integral from a to b of [f(y)]^2 * dy By transposition we get: x = sqrt((sin(y) + 1) / 2) Since x = f(y), then [f(y)]^2 = (sin(y) + 1) / 2 Now I thought that since an arcsin graph is symmetrical, I can just simply calculate it...

If the gradient of the gradient is positive, then it is an increasing gradient. Do you know how to check for that?

Please someone help me! It's been literally weeks and I still can't solve this problem :(

You said I should cut it in half and then calculate half of that. Does that mean I am only going from 0 to +pi/2 or from -pi/2 to 0, rather than the full way? So would the integral be like this: Integral from 0 to +pi/2 of: (sqrt((sin(x) + 1)/2)) * dx Is that the answer?

So once I have it in the form x = sqrt((sin(y) + 1)/2), do I then have to change the numbers on the top and bottom of the integral? So instead of going from -1 to 1, I am now going from -pi/2 to +pi/2? And by doing that, I am now wrapping it around the x-axis? So I can change it to become: y =...

1. Consider the function f(x) = arcsin(2x^2 - 1). Write down, but do not attempt to solve, a definite integral in terms of y, which when evaluated will give the volume of the solid of revolution formed by rotating the graph about the y-axis. 2. Endpoints are at (-1, pi/2) and (1, pi/2)...

Thanks a lot for your help, dynamicsolo and SteamKing. Now I know how to solve these types of problems for the future. I let x = the value that is being multiplied by i and y = the value that is being multiplied by j and then ignore the r. Then I express y in terms of the value being multiplied...

cos(2t) = 1 - 2(sin(t))^2 So r = sin(t)*i + (1 - 2(sin(t))^2)*j But I'm still confused because I don't know how to get rid of the i's and j's... Wait I think I get it: x = sin(t) y = 1 - 2(sin(t))^2 y = 1 - 2x^2 Is that right?

1. The position vector of a particle at time t ≥ 0 is given by r = sin(t)*i + cos(2t)*j. Find the cartesian equation for the path of the particle. 2. I was told that the answer is: y = 1 - 2x^2 But I don't know how to obtain that solution. 3. r = sin(t)*i + cos(2t)*j At first I...