How many Planck lengths are there in a parsec?

[Bob Walance]

[Mentor's note: moved to GD because this is a thread about Alexa, not parsecs or Planck lengths]

I know it's a strange question. Here is Alexa's answer:

https://drive.google.com/open?id=1hXYFT_zBnFy-N0BotiTSDL8_S6nHK3ns

 

[phinds]

I know it's a strange question. Here is Alexa's answer:

https://drive.google.com/open?id=1hXYFT_zBnFy-N0BotiTSDL8_S6nHK3ns

I can't see the link

 

[Bob Walance]

Hmmm ... I'm able to see and follow my link even when I'm not logged in. I'm not sure why others can't

My math is fine and we're not allowed to upload mp3 files, so I guess this stops here.

Better yet -- spend $50 for an Amazon Echo Dot (you'll need a wifi network) and listen to Alexa's response for yourself. She's a lot fun and can be very educational.

 

[Nugatory]

I can't see the link

I can. I expect that this has something to do with being logged into Google Drive and/or how the sharing has been set up.

 

[Ibix]

Hm.

"Alexa, what is standard form?"

"Alexa, what is spurious precision?"

 

[phinds]

Better yet -- spend $50 for an Amazon Echo Dot (you'll need a wifi network) and listen to Alexa's response for yourself. She's a lot fun and can be very educational.

Spending $50 to learn how to multiply two numbers together doesn't seem all that appealing to me.

 

[DaveC426913]

"Alexa, what is spurious precision?"

Spurious? They are both well-defined lengths...
She could have rounded, sure, but he didn't ask for approximately...

 

[Mark44]

Ask Alexa how many electrons there are in a pint...

 

[Tom.G]

Spending $50 to learn how to multiply two numbers together doesn't seem all that appealing to me.

Uhm, maybe "...divide two numbers..."?

 

[jim mcnamara]

I get ~1.9e+49 using google calculator and my crummy typing skills. The OP is asked to consider doing the same:
i.e, lookup both values then divide Planck length in meters (really small) into the number of meters in a parsec. Thread closed.

PF is not a preferred substitute for google calculator.