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I would like to obtain the characteristics and properties of the Alloy Chromel. How do I download that information from the database of Wolfram Alpha and then print it in a Mathematica Notebook?


Update

enter image description here

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closed as off-topic by Bob Hanlon, m_goldberg, user6014, Henrik Schumacher, Alex Trounev Mar 17 at 19:58

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "The question is out of scope for this site. The answer to this question requires either advice from Wolfram support or the services of a professional consultant." – Bob Hanlon, m_goldberg, user6014, Henrik Schumacher, Alex Trounev
If this question can be reworded to fit the rules in the help center, please edit the question.

  • 3
    $\begingroup$ Seems obvious you should do what's suggested...You have a blocked activation key. $\endgroup$ – morbo Mar 17 at 10:37
  • $\begingroup$ Seems obvious that I don't know how do I unblock the activation key $\endgroup$ – Delfin Mar 17 at 11:18
  • $\begingroup$ Ah, you need to contact wolfram, the error suggests this. $\endgroup$ – morbo Mar 17 at 11:19
  • 2
    $\begingroup$ It does say "Please contact Wolfram" in the error message; that's not a problem anyone here on SE can take care of. $\endgroup$ – J. M. will be back soon Mar 17 at 12:05
  • 1
    $\begingroup$ Are you using an illegally licensed version of Mathematica? $\endgroup$ – user6014 Mar 17 at 15:58
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Depending on the version you have, I'm running 11.2, you can type simply

=

or

==

note that just one or two equal signs, that will ask WA directly. Here is a blog post explaining the feature from WA blog

thing

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  • 2
    $\begingroup$ Equivalently, evaluate WolframAlpha["chromel alloy properties"]. $\endgroup$ – J. M. will be back soon Mar 17 at 10:20
  • $\begingroup$ Yeah, I was doing that, but I have an error. I uploaded an imaging in my post showing it $\endgroup$ – Delfin Mar 17 at 10:37
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    $\begingroup$ @Delfin - code works fine for others. You need to contact Wolfram. $\endgroup$ – Bob Hanlon Mar 17 at 12:57

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