Recent content by mlazos

So how is possible the moving object to have a higher reading that the stationary. Since time slows down then we should have a higher reading that her. Right?

So what is this value? how much? Is it 0.5sfor the O while is 0.3 for the O'? Or its 0.3/gamma for the O while its 0.3 for the O'?

is the thread number 10 of this topic and question c. This is really strange. The problem itself is not difficult its just the solution strange for me. I will be back in 2:30h. thank you for your time.

I see, makes sense. Though for the last question, the part c) is the answer 0.5 seconds? Again i say that in the solutions they say "When the racer reads 0.3 the stationary observer reads 0.3/gamma".Is there any logical explanation for this or its just a mistake?

Yes but in order all those observers to synchronize their clocks still use information and this takes time.So there is always a delay this way. Why in the book of Young and Freedman the time measured by the stationary observer is 0.3/gamma ?They substitute the 0.3 in the space vehicle as the...

Yes but how do we know someone arrived to the destination?We have to observe it somehow. So when we see someone is at the point B then we count the time to go there plus the light to get to our eyes. Since the distance is big we have to count also the time that light traveled from B to A. right?

The exact problem says As you pilot your space utility vehicle at a constant speed towards the moon, a race pilot flies past you in her spaceracer with a constant speed of 0.8c relative to you.At the instant the spaceracer passes you,both of you start timers at 0. a) At the instant when you...

0.3 seconds is the clock inside the object which moves. According to the stationary observer, the moving object needs at least 0.5 seconds from A to be plus the time the signal needs to get from B to A which is 0.25s.Thats why i said 0.75s. I have a drawing so you could check there too. In the...

Now the tricky question is: When the Observer O' reads 0.3 sec what do we read to our clock? This is the answer I need actually. I calculated that to be 0.75s and I want to verify the answer. Thank you for reading me.

Yes but in order to know that the asteroid is at point B we have somehow to observe him. Therefore we count also the delay of light since the speed is finite.Thats why when the observer is at the point B we count the time it took to get there plus the delay of light from B to our instruments. Is...

1. I give the problem and the solution. Can someone tell me if and where I am wrong? Thanks Homework Statement We have an observer O', who travels with constant speed U=0.8c, and another observer O with speed 0. The Observer O' is traveling from point A to point B (1.2 \cdot 10^8m)...

I am trying to find a program for Windows in order to draw forces, vectors and problems for physics. I want to illustrate angles, rotations etc. thank you

Homework Statement Along a certain section of road a car (m=1600) will coast in neutral at a constant speed of 72km/h if there is no wind.Examination of a topological map shows that for this section of the road the elevation decreases by 200m for each 6000m of road.What is the resistive force...

What I understand by enclosed means for the value of t where x=0. So the values are x=-\sqrt{5} , 0 , \sqrt{5} So I guess you have to find two integrals, for x>0 and for x<0 because one will get negative and the other positive and to add them.

Well we only know the relation between a and b! We can't know that a is very large, we know that a is much much greater than b.