Gareth56

So where is the higher potential of electrons in my 240V wall socket and how does a voltage of however many volts "push" electrons along a wire? Ta

What "pushes" the electrons through the wires when I switch on e.g. an electric kettle? And do the electrons travel at a certain speed in wires? Thanks G56

Exactly that's why (in this example) if Fy > Fx then the barge will tend to go towards the y direction ie the bank. As for the velocity being zero, it's being pulled at constant speed which isn't zero.

I'm sorry but I think it would make a difference. If the vertical component of the force is greater than the horizontal component then the barge will be pulled more towards the bank. Try changing the angle of pull to 60 deg and see what happens. With respect to the second part above I suggest it's equal to the x component of the force.

Because the horizontal component of the force (F x costheta) is greater than the vertical(towards the bank) component (F x sintheta)of the force. That is Fx > Fy

This was the angle the force was making with the barge.

So presumably the drag force of the water on the barge is Zero?

I did that as Fx = 2000N x cos25deg but the question asked what was the drag force of the water on the boat.

Because it's being pulled along by a horse (it's a barge really in the book) and the book only gives the force the horse provides to the barge. Sadly this is an even numbered question in my physics book

If a boat is moving with constant speed through the water are there any drag forces acting on it?

Because you're balancing the weight of the bullet with the centripetal acceleration. So the mass doesn't enter into it.

In that case I'd be very interested in seeing the method using the escape velocity figure. Yes, you're using the wrong equations. You only need the equate the gravitational force (mg) to the centripetal force (mv^2/r) the solve for V.

But the trick is to determine what velocity just prevents the bullet from flying off into space and maintains the bullet in a circular orbit. How this can be achieved using the escape velocity figure and something else escapes me.

But if you were firing the bullet horizontally (and assuming nothing would impede its path) then the escape velocity doesn't enter into the debate and only Moon's acceleration due to gravity and radius are relevant.

I got 1.7km/s by equating Fgrav to Fcent then solving for v. Is that how you did it?

If you could fire a rifle on the Moon (i assume you can) what would the muzzle velocity of a rifle need to be to send a bullet into orbit around the Moon?

Clearly the tread on tyres are very helpful when driving in wet weather but does it matter if a car tyres have any tread if you are driving on ice? In other words does a tyre's tread decrease the danger from skidding or decrease your stopping distance when driving on ice?

Thanks for the pointers JaKiri. I've written down all the equations that should be relavent ie Ux=U*Cos alpha Uy=U*Sin alpha then y = Uyt -1/2gt^2 x = Ux * t Vy = Uy -gt Vx = Ux That's as far as I get but it doesn't really matter it was just a querry regarding a question I came across in Tipler regarding projectile motion. After the above I think it's all algebra or whatever. In case certain people are wondering it is not any kind of assignment question, it's just that I have an interest in physics...I'm well past college days

This one has me puzzled:- At 1/2 of its maximum height, the speed of a projectile is 3/4 of its initial speed. What was its launch angle?

Yes, it does matter but you should have been able to see what to do. Oh sorry "Sir"

Many thanks Edtharan for your time and trouble taken to explain this.

How would you be able to "jump" (I'm not too sure what you mean here) off me if there were no friction for you to get a grip to do that? I appreciate the fact that if I were stuck in the middle of a frictionless ice rink and had a medicine ball (as you always have on you) I could get to the edge by simply throwing the ball in one direction and I would go in the opposite. You say that SM just push the masonry downwards but what is then pushing back on the masonry in accordance with Newton's 3rd Law.

Does that mean that terminal velocity is the velocity at which the air resistance force of a falling object equals the weight of the object minus the acting force due to air, which halts acceleration and causes speed to remain constant? So if SM is pushing against the block of wall to provide him with the necessary acceleration to jump off it what is the block of wall pushing against in order for the wall to supply the equal and opposite force to supply SM with the force to push off the piece of wall? Is this the same scenario as if you are in a lift (elevator) that's in free fall then just before it hits the ground you could jump up and possibly save yourself from a nasty end?

Watching Spider-Man 3, who turns out to be made of very strong stuff in order for him to survive such accelerations and forces, I noted one thing that got me thinking. During one scene as a building was collapsing around Spider-Man he leaped onto a piece of falling wall and proceeded to launch himself off it. Assuming that the piece of wall was falling at terminal velocity and as such was in free fall would Spider-Man have been able to exert at force on the piece of free-falling wall in order to gain flight?