You do not have to explicitly find the speed of C1 after it has fallen 2.5m. You just need to find the time C1 takes to fall 7.5m, as I have pointed out. Nevertheless, your method is OK. I have not checked the calculations.
Welcome to PF.
You know the time C1 takes to fall the 7.5 m height ( time taken to fall 10m minus time taken to fall 2.5m). In this time, C2 falls 10m. Now which formula do you think you should use?
Insulators are substances which impede the free flow of electrons from atom to atom and molecule to molecule. If charge is supplied to an uncharged insulator at a point, it will tend to remain there.
The presence of other charges outside a conductor will result in induced charges being...
We are considering the marble as a point mass sliding down the ramp with no friction, whereas in reality it is a solid sphere with some radius rolling down the ramp with friction. Part of the potential energy of the marble goes into its rotational motion and another part goes toward overcoming...
The fuel and air mixture exit the rocket after burning and therefore produces a reaction force on the rocket. The impulse on the existing gas has been given.The impulse is the change in momentum and the time is one second. Use the equation I have given you. The force on the exiting gas must be...
In this case , "consumes" means that the fuel and the air mixes and combustion takes place and the end product exits as a hot gas , producing thrust on the docket . That is given in the very first sentence of the problem . Do you understand now ?
Hi, welcome to PF.
Can this equation be correct? The left hand side has the dimension of force while the right hand side has the dimension of momentum.
Let's do it in a very simple way.
In one second, how much mass is exiting the rocket? That is given.
So, what is the impulse on this...
Hi, welcome to PF.
(What do you mean by the total velocity?)
Use the equation linking distance with time and acceleration. The initial speed I think you know. Do you know that acceleration of the body along the slope in terms of g and theta? Give it a try.
It is precisely when we cannot or do not need to calculate the force or the time for which it acts that we use the concept of impulse. The impulse is the integral of the force with respect to time and can be proved to be equal to the change in momentum. Here are some good sites for preliminary...
(Sorry for the delayed post.)
I believe that applies to all the laws of science. The point was that the equation for static fiction holds quite true within a narrow regime. Or does it? The video was most illuminating.
Yes, my oversight. I was thinking of something else.
Could you give a...
No, it is not the case in the regime where the laws of static and dynamic friction are valid -- otherwise those laws woudn't exist. I was not consideiing rolling friction, so the tire scenario is not very pertinent to my point..
This point is well exemplified by Pupil in post #9.
That would imply that the co-efficient of friction between the same pair of materials would vary with the area of contact, which is not the case.
This makes good sense.
(The brackets that you are using is generally used for matrices; for a determinant, vertical lines are used to enclose it.)
Multiply the first row by 'a' and get 1/a outside. The 1st row becomes |a a^2 abc|. Now think what you can multiply the 2nd and 3rd rows with.
After that, take out...