Recent content by Dr. Jekyll

You just need to look at the net force on each string. For example, you have two forces on the first string (weights of the hanging bodies). Other one is even easier. Remember the 1st Newton's law. That's when the elevator is stationary. The formula F=ma can be used in part (iv) and (v). It's...

Maybe. I'm not sure too, but it's probably very difficult to bring the atoms of these gases together, because the electrons repel each other (and they won't make any molecules because they already have maximum number of valence electrons).

Of course, the book is right. Now, you don't need the energies here. It's all about forces. Since the crate moves at constant velocity, the net force on it should be zero (1st Newton's law). You ignored the y component of the force that the crate is being pushed by. It effects the force...

What happens with the air (wavelength) when you drink more soda? Mass of the soda doesn't have anything to do with it.

Let's say that meteor hits the Earth (and doesn't make much damage). Then both meteor and the Earth will act on us and pull us towards them, now with greater force than it was when the meteor was far away in the space. We'll do the same thing to them (according to Newton's 3rd law). So, the...

I would say that it increases because you need to do some work in order to bring the proton near the positively charged sphere.

You have some help https://www.physicsforums.com/showthread.php?t=227924". For the (a) part you have vertical shot (y component of the velocity) and then free fall from the height of 150 m at initial velocity of v_{y}=v_{0}sin\alpha. For the (b) part you need the x component. You have the...

The initial velocity can be divided in two components: \vec{v_{0}}=\vec{v_{x}}+\vec{v_{y}}. So you get these realtions: v_{x}=v_{0}cos\alpha v_{y}=v_{0}sin\alpha. v_{y} is a vertical shot, so the time which it needs to reach the ground again will be: g=\frac{\Delta v}{\Delta t}...

Seems correct to me.

Maybe this has something to do with the Bernoulli's principle: \frac{\rho v^{2}}{2}+\rho g h + p = const. But then you don't need the surface...

You're on the right way - conservation of energy: E_{potential}=E_{translation}+E_{rotation} \Rightarrow mg\Delta h = \frac{mv^{2}}{2}+\frac{I \omega^{2}}{2}, \Delta h = r(1-cos\alpha). My result differs a bit from yours. I don't have the a parameter. Everything else is the same...

We assume that the particles within the ideal fluid do not interact and that their volume is negligible. It's not the same in reality, but there is not much deviation (of course, there are some limits). So, many non-ideal fluids act almost the same as the ideal model. So, your conclusion is not...

My answer doesn't match too. I did it like this: t=\frac{v_{1}}{a_{1}}=\frac{v_{2}}{a_{2}}, s_{1}=v_{1}t-\frac{a_{1}}{2}t^{2}=\frac{v_{1}^{2}}{a_{1}}-\frac{v_{1}^2}{2a_{1}}=\frac{v_{1}^{2}}{2a_{1}}...

The maximum current flows through the resistor on the left. So, if you calculate that current (from the power), you can get the voltage by using Ohm's law. I'm not sure if this is correct.

It's easy to say: "Not a clue..." :smile: Maybe https://www.physicsforums.com/showthread.php?t=225490" will help (as it already did). It's almost the same. You need the velocity at which man must leave the ground to reach the height of 0.8 m (you know the g). Therefore you can calculate...