The first method doesn't work. The second method does work but I can't run avira and malware bytes. Can someone help?
First method
I hold the shift button and click the restart button.
I click on troubleshoot then click on advance objects then I click on see more recovery options.
The...
Thanks for the response, why is the css not working?
Also since I have min and max characters do I need InputRequired() or datarequired() Validators? https://wtforms.readthedocs.io/en/2.3.x/validators/#built-in-validators
I think so but from your question it seems it is not. Why is it not?
Do you mean something like 17,987,547,480.00 meters/minutes? except it is = 1 y/minutes?
Where y is an unknown unit of distance.
But how do I make y/minutes = 1 for velocity of light?
I am sorry I am just confused.
time = x(min)
distance = 1(y)
y = unknown units
I think the answer should be 1(y)/ Min. This is not correct becase 1(y) is unknown. Any help?
I have the answer but am confused
The answer is tD = [D_0 - 3/4ct - 1/2ct] I just have 2 questions.
I realize for 2 vectors approaching it is negative for distance and for velocity positive. What be the rule for time? How do I find vector answers for velocity and distance and time?
I am confused why I have "td = ..." ? Can...
I don't own a external camera and rather not spend the money are there any other options? I guess I could buy an webcam and return it. I manage to roll back the drivers and same problem persist. Just wanted to add when I use skype or some other application I don't see the flashing. I rebooted...
My application is the camera app. I just push the windows button and search camera. I don't have very many settings. There is no flicker setting I will download https://www.yawcam.com/ and see if that works.
I did say I am not sure about the hacking though.
Thanks for the help so far.
I am using Avira and Malwarebytes and my firewall is on. It shows no viruses on avira or Malwarebytes.
At the very least can someone help me fix my webcam?
Here is the video. As you can see the webcam is flickering.
Thanks for the help
This picture shows the specs
assuming I am understand the question correctly Bob and Alice have the same location in both frames.
Event 0 everything is 0.
event 1
I assume both Alice and Bob are at the same location for every location. So I am just doing the calculations once for
Alice's frame for Alice and Bob's frame...
Thanks for the help and the extreme patience.
Let us start with the spaceship frame.
Event 0 ## x ~ rear = -15 ## and ##t = 0##.
Event 1 ## x ~ rear = -15 ## and ## t = -25 sec##
So there are 2 ways to solve this. ## t ~ rear = \frac { ~delta ~ x ~ rear} {v} ## or just ## delta ~ t ~ rear...
Sorry for the delay. I just calculated t_rear_1. I didn't calculate t'_rear_1 I will do that after assuming I got the correct answer.
Here is my thought process.
event 0
t = 0s
x = gamma(x' + vt')
x_rear = 1.25( -15LS + (-.6c)(0) )
x_rear = -18.75LS
x_front = 1.25(0 + -.6c(0))
x_front...
##d'rear1## is just the second event for x_rear in the prime frame.
I will post doing it your way and my way and then ask a few question and speculate why I am wrong using the example in post 18 my way and post 17 your way.
Thanks for the help.
Also you mention 2 events isn't event 1 happen at t = 0 and t' = 0?
Are there other ways to solve the answer?
Instead of t1 and t2 I am going to use t0 and t1.
Why can't I use ## {t'rear1} = ## ## \frac {d'rear1} {v} ##
Then once I have ##t'rear1## I use the Lorentz transform for time to...
greater then 15LS
Here is my reasoning. It should be -.6c. Also x'_rear = -15. I forgot the minus sign in my drawing. I am little confused why in the equation x'_rear needs to be positive.
Here is my thought process.
How do I figure out the distance for an person on the ground frame? I use the fact that moving clocks move slow. So I try the equation x′=gamma(x+vt).
But I don't have time of the rear clock.
Then I try to calculate trear=dv. trear=.6c. I know this is wrong because...
Here is my attempt.
I have velocity = .6c and d' = 15LS.
t' = d'/v = 25 Seconds
x = gamma(x'+vt')
x = 1.25 (15LS + (.6c)(25 Seconds) ) =
x = 1.25 (15LS + 15LS) =
x = 1.25 (30LS) =
x = 37.5 LS
Is this correct? Or should x' = 0?
After thinking about I think x ' = 0 let me correct the answer.
x =...
I think so. I assumed the value 15LS is already the contracted length. I guess I was wrong. I assumed L' = 15 LS was L = 15LS. The definition is moving objects are length contracted.
Here is my attempt. I first did your question The answer I first got was t = 12 seconds.
Then I did the problem above. I got t = 25 seconds. I am not sure if I need to use relativity of simultaniety for the time in the stationary person frame. Do I ? Then I calculated the time in the train...
Whoops careless mistake on the d = 3 x10^8 m/s it should be d = 3 x10^8 m
What is really confusing me is what is the initial position and time of both ends of the train and the initial position and time of the person standing still. Can you help with this question? Well I guess I can just...
Because the book doesn't have an answer. Can someone confirm if I am correct?
The question can be found on page 49 of the link below.
I added arrows to the picture just to separate the equations not to say greater then.
https://scholar.harvard.edu/files/david-morin/files/relativity_chap_1.pdf...
Special Relativity: For the Enthusiastic Beginner on page 49 1.8 Exercises I can't find the answers to the questions? I apologize if this is a dumb question.