"risky business" aged dudes

Were yenze all that clueless back then? Holy cow leave it to beaver!

Yes. Sure. Why the hell not?

I love Blanch brain teasers

I'm having a salad for lunch.

ps. not Detroit Lions salad.

Is Blanch from Pittsburgh?

Nordic wrote:I love Blanch brain teasers

I think it's a riddle.

Nordic wrote:I love Blanch brain teasers

Since when does blanche have a fucking brain?

Frank Ricard wrote:

Since when does blanche have a fucking brain?

Cents he oners colledge brah

Travis of the Cosmos wrote:Yes. Sure. Why the hell not?

Whoa. What was it like to be that naive?

Rocinante wrote:Is Blanch from Pittsburgh?

Nope. Lived there for a yr. BTW that movie was based in chitown

Blanch32 wrote:

Whoa.  What was it like to be that naive?

I know man. It's a crazy world we live in but if we don't produce these prepackaged crab rangoons then who will?

Blanch32 wrote:

Nope. Lived there for a yr. BTW that movie was based in chitown

I was asking because of "yenz"

Consider a simple clock consisting of two mirrors A and B, between which a light pulse is bouncing. The separation of the mirrors is L and the clock ticks once each time the light pulse hits a given mirror.

In the frame where the clock is at rest, the light pulse traces out a path of length 2L and the period of the clock is 2L divided by the speed of light
\Delta t = \frac{2 L}{c}.
From the frame of reference of a moving observer traveling at the speed v relative to the rest frame of the clock, the light pulse traces out a longer, angled path. The second postulate of special relativity states that the speed of light is constant in all frames, which implies a lengthening of the period of this clock from the moving observer's perspective. That is to say, in a frame moving relative to the clock, the clock appears to be running more slowly. Straightforward application of the Pythagorean theorem leads to the well-known prediction of special relativity:

The total time for the light pulse to trace its path is given by
\Delta t' = \frac{2 D}{c}.
The length of the half path can be calculated as a function of known quantities as
D = \sqrt{\left (\frac{1}{2}v \Delta t'\right )^2+L^2}.
Substituting D from this equation into the previous and solving for Δt' gives:
\Delta t' = \frac{\frac{2L}{c}}{\sqrt{1-\frac{v^2}{c^2}}}
and thus, with the definition of Δt:
\Delta t' = \frac{\Delta t}{\sqrt{1-\frac{v^2}{c^2}}}
which expresses the fact that for the moving observer the period of the clock is longer than in the frame of the clock itself.

forty grommets to the back of wood tantalizing furnace wheels, one at a time upward towards near Fall harvest penalty most times a year.

Watch Out Pylon! wrote:Consider a simple clock consisting of two mirrors A and B, between which a light pulse is bouncing. The separation of the mirrors is L and the clock ticks once each time the light pulse hits a given mirror.

In the frame where the clock is at rest, the light pulse traces out a path of length 2L and the period of the clock is 2L divided by the speed of light
\Delta t = \frac{2 L}{c}.
From the frame of reference of a moving observer traveling at the speed v relative to the rest frame of the clock, the light pulse traces out a longer, angled path. The second postulate of special relativity states that the speed of light is constant in all frames, which implies a lengthening of the period of this clock from the moving observer's perspective. That is to say, in a frame moving relative to the clock, the clock appears to be running more slowly. Straightforward application of the Pythagorean theorem leads to the well-known prediction of special relativity:

The total time for the light pulse to trace its path is given by
\Delta t' = \frac{2 D}{c}.
The length of the half path can be calculated as a function of known quantities as
D = \sqrt{\left (\frac{1}{2}v \Delta t'\right )^2+L^2}.
Substituting D from this equation into the previous and solving for Δt' gives:
\Delta t' = \frac{\frac{2L}{c}}{\sqrt{1-\frac{v^2}{c^2}}}
and thus, with the definition of Δt:
\Delta t' = \frac{\Delta t}{\sqrt{1-\frac{v^2}{c^2}}}
which expresses the fact that for the moving observer the period of the clock is longer than in the frame of the clock itself.

Is that the same thing as that prehistoric answering machine they invented?

Blanch32 wrote:

Is that the same thing as that prehistoric answering machine they invented?

It's bascially the plot of Bill and Ted's Excellent Adventure.