GREPhysics.NET
GR | # Login | Register
   
  GR9677 #50
Problem
GREPhysics.NET Official Solution    Alternate Solutions
This problem is still being typed.
Special Relativity}Spacetime Interval

The spacetime interval is defined by the metric that negates spatial and time variables as dS^2 = (cdt)^2 - (dx)^2. dS is invariant. One has thus dS^2 = dS'^2 \Rightarrow (3c)^2=(5c)^2-(ct)^2\Rightarrow ct = 4 c minutes, as in choice (C).

See below for user comments and alternate solutions! See below for user comments and alternate solutions!
Alternate Solutions
There are no Alternate Solutions for this problem. Be the first to post one!
Comments
TypT
2018-10-26 22:10:37
Quick, dumb way I did it without having to figure out what exactly the question was asking:\r\n\r\nSpace time intervals have a Pythagorean relationship, the two numbers we are given are 3 and 5, to complete the Pythagorean triple we need a 4 so we get (C). It even has the trap answer of 4^2 = 16 (E), to give a little more confidence in (C).NEC
mpdude8
2012-04-20 15:12:22
Man, I was kind of surprised that only 1/3 of test-takers got this one correct.NEC
pam d
2011-09-16 14:40:56
<3 spacetime intervalNEC

Post A Comment!
You are replying to:
Man, I was kind of surprised that only 1/3 of test-takers got this one correct.

Username:
Password:
Click here to register.
This comment is best classified as a: (mouseover)
 
Mouseover the respective type above for an explanation of each type.

Bare Basic LaTeX Rosetta Stone

LaTeX syntax supported through dollar sign wrappers $, ex., $\alpha^2_0$ produces .
type this... to get...
$\int_0^\infty$
$\partial$
$\Rightarrow$
$\ddot{x},\dot{x}$
$\sqrt{z}$
$\langle my \rangle$
$\left( abacadabra \right)_{me}$
$\vec{E}$
$\frac{a}{b}$
 
The Sidebar Chatbox...
Scroll to see it, or resize your browser to ignore it...