Recent content by Quantum Mind

OK, I got it. For a projectile motion, the distance traveled in x-axis is x = ut + at^2/2 But since 'a' is zero along the x axis, x = ucos\theta.t distance traveled in y-axis is y = u sin\theta.t - 1/2 gt^2. Now I can substitute the value of t = x/u cos \theta and I get the...

time of flight t = 2u sin\theta/g x (horizontal range) = u^2 sin2\theta / g How is this related to gx^2 / 2u^2cos^2\Theta ? I used the equation S = ut - 1/2gt^2 and tried to substitute the value of 't', but I ended up with something else.

Homework Statement This is a problem about an aircraft flying above a cannon which is capable of firing in any direction. The plane's height, velocity and the speed of the projectile fired by the cannon are given (I am not trying to get help in solving the problem, so the numbers are not...

This is what I found on the Net, is there another way to do it without invoking the Young-Laplace equation? Assume that the air inside the bubble is an ideal gas. Because the temperature is constant, ideal gas law states that: P∙V/n = R∙T = constant (Pi pressure inside the bubble) So...

The answer is p0(R3 - 2r3)/4(2r2 - R2). From this I gather that the volume of the bubbles has something to do with this problem and that the volume of the bigger bubble will be more than the combined volume of the two smaller ones. The area of the bigger bubble will however, be smaller than...

Any hints ?

Here's a similar thread: https://www.physicsforums.com/showthread.php?t=274670

Homework Statement Two soap bubbles of equal radius 'r' coalesce isothermally to form a bigger bubble of radius 'R'. If the atmospheric pressure is P0, find the surface tension of the soap solution. Homework Equations Surface energy = surface tension x increase in area (W = T.ΔA)...

I think the problem involves three stages: When the aircraft is accelerating, when the velocity is constant and then when it is braking. Using the Suvat equations, I got an answer, but the rules forbid me from disclosing. Waiting to see if I am correct. "Colin19", if you want to learn, a...

Thank You, it is clear now.

All that talk about centripetal accelaration, radius, arc and action_reaction etc threw me off track. Thanks for your patience.

OK, so the vectors for the input and the output forces are at right angles. On adding, they give a vector which bisects the angle at 45°? In other words, it is the diagonal of a square. In that case, the resultant vector would be sqrt of 2.ρAv2 (Sorry, but LaTEX is a pain, I am unable to use it...

So this means that ρAv2 = - ρAv2? What other forces act on the tube? I still don't get it.

OK, so force = pressure in this scenario. Since the cross sectional area is not given, I can assume it as A. Velocity is given as 'v'. So the force exerted by the liquid on the tube is ρAv2. But what is the force exerted by the tube on the liquid? Is it -ρAv2? But this cannot be right, as...

No. But I think centripetal acceleration should be R3∏/2.