Recent content by Clockclocle

So you only consider the force which appear on the time interval (0,1)? I suppose the flea take 1 second to jump at the height 0.54m? What if at each t=1/n for integer N an upward force of 5N applied on the flea. Do it affect the normal calculation

I know the solution is solved by the equation $ \v^2_{x} - \v^2+{0} =2ad $. But in order to the flea can jump, It must exert a force on the ground at time t=0. Do I have to include this force to substract from earth gravity? Or since this force only appear at time t=0 only while g appear the...

After a year of thinking about instantaneous velocity. I think that this notion is no more than a mathematic coincidence when mathematician tried to find the tangent of curve. The only definition of velocity that make sense is ##\frac{\Delta x}{\Delta t}##, this proportion is a quantity that...

A uniform circular motion of a point always yield an equation u=cos(wt)i +sin(wt)j of position vector. Which we deduce the acceleration and velocity vector with constant magnitude and they are perpendicular at each instant. Can I use the information of them at one instant to predict the position...

He also use the fact that lim |##\Delta A##/##\Delta t##| = |lim ##\Delta A##/##\Delta t##|. Is that accepted?

I understand the approximation statement but he divide the |delta t| in the left but only delta t on the right. Is it true because delta phi would have the same sign as delta t ?

yes the answer is 2 (sqrt(2)+1). So is it true?

Here is my attempt. At ymax the velocity turn to zero so we get time t*=v0/g and ymax=1/2 (v0^2/g). At the height y max, since the velocity at this point is 0, i get another equation y= 1/2(v0^2/g)-(g/2)t^2, this equation could be considered as continuation of first equation. Set ymax/2=1/4...

I see the mistake, I thought that it gonna fall below me so it would be the same when I stand still. But in this case I keep moving with velocity v0.

In this situation should my free fall equation contain the v0 of the balloon or I should deny it. Because it seems to me that there is no outer force acts on the sandbag, so the scenario is just the same as I climb to the same height at time t=0 and drop the sandbag at rest.

The instantaneous velocity at time a is defined as derivative of motion function f(t). It is not similar to average velocity in an interval of time. From the Newton law. If an object is at rest, we must exert a force to make it move, assume that there is no friction. Depend on the weight of...

What? I think the last book would both contact with the surface and the top of it?

Suppose n book stack on each other. Since each book have the same weigh then the last book exert a force N=nmg on the surface so it has the biggest static friction. But if we treat the whole tower of books as one particle it also has N=nmg. This mean if we exert enough force in the last book...

When I am riding a motocycle, suppose I slowly accelerate the car(push the pedal) the net force >0 for an interval of time, when it gets maximum the car move with constant velocity and net force become zero. Does it mean friction "come after" the exert force on the car?

Does the kinetic friction monotonically deacrease to 0 as the velocity become 0? Does it make velocity become zero and disappear immediately it mean the graph of kinetic friction respect to time t after I realease the object is continuous?