Recent content by Sleve123

I'm trying to get my head round this. I don't see why our inability to measure the world around us means that at the quantum level things must be random. I understand that measuring momentum of a particle to a high degree of accuracy means losing accuracy in it known position. But I don't...

Your acceleration is wrong, the 'V' you worked out was the average velocity, you can work out the acceleration using s = ut +0.5at^2. Alternatively assume the lift is of constant velocity and say: mgh / time = power

Thanks that helps a lot.

When I mean point then, I mean a sphere, ie the radiation coming from all directions. I'm getting this idea from the fact that some parts of the Universe will never be observable to us on Earth because the vast distances inbetween meaning that the space is expanding faster than the speed of...

Not that it has become so small it is undetectable; but the point from which it was emitted is such a distance away that the expansion of space between Earth and that point is greater than the speed of light (like swimming upstream in a river going faster than you can swim).

Homework Statement The experiment we did was pretty basic, it was a copper box with five different metal rods (unlagged and of equal lengths and diameter) welded onto the bottom, on the bottom of each were pins stuck with petroleum jelly. Hot water was then added to the box and the order at...

I agree with your answers until the last part, where you say 0.00816/.64 = 0.01266 m I think you've just missed a zero, so it would be 0.001266 m Check it by adding the two depths and multiplying by the base area to give the original volume.

Thanks

I think my problem is assuming that the car could take off at any point as it was going up the humpback, as long as it had enough speed. For example if you take a point before the apex and instantly accelerate the car up to a large speed along the tangent - it would leave the track, you...

So Rcos(x) would just be R? the thing is with this I think is that the further from the peak of the bridge you get, the smaller the component of mg is acting towards the centre, therefore a smaller speed (v) is needed, which to me doesn't make sense (mainly because there is no mention of this...

Homework Statement A car travels over a humpback bridge of radius of curvature 45m, what is the max speed it can reach before the wheel lose contact with the road? Homework Equations mg - R = (mv^2) / r The Attempt at a Solution When the car is about to lose contact with...

Thanks, as long as I'm right. I'll make it simpler for myself next time.

So: mg - T = -(-0.1)m T = m(g - 0.1) T = 800 x 9.7 = +7760N

I think what I've done is work out the force applied to the lift to produce that outcome. In my equation: T + mg = -(-0.1)m The value of T could be positve or negative. Doesn't the negative value just tell me that the force applied on the lift from the tension in the wire is in the...

A lift has mass 800 kg. It is moving upwards and has an accerleration of -0.1 ms-2. The lift is supported by a single cable. What is the tension in the cable? I get T = -7760 N (I'm taking up as negative), just checking what anybody else gets F = ma T + mg = -(-0.1)m T = 800(0.1...