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frosja888 [35]
3 years ago
12

In order from lowest urgency to highest, which sequence properly ranks the product categories issued by the National Weather Ser

vice? Note that not all product categories are issued for all hazard types.
A. Watch, Advisory, Outlook, Warning
B. Outlook, Watch, Advisory, Warning
C. Outlook, Advisory, Watch, Warning
D. Advisory, Outlook, Watch, Warning
Physics
1 answer:
matrenka [14]3 years ago
4 0

<u>Answer:</u>

In order from lowest urgency to highest, the sequence which properly ranks the product categories issued by the National Weather Service are as follows:

B- Outlook, Watch, Advisory, Warning.

<u>Explanation:</u>

An outlook for a hazardous weather describes the potential hazardous weather of concern in day 1 through 7. There are total two segments of the outlook, one for the marine zones and the second for the land based zones.  

A watch is issued when there is the possibility of hazardous weather within 48 hours, it does not guarantee that a hazardous weather is going to come, it just reminds the possibilities of any such weather to come.

Advisory comes third in the urgency ranking, we can explain this through an example: a winter weather advisory may be issued for amount of freezing rain or when there are chance of 2 to 4 inches of snow. And, in winter weather, an warning may be issued when there is one fourth inch or more of ice accumulation.

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We have

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\vec v(t) = \left(5\dfrac{\rm m}{\rm s}\right)\,\vec\imath + \displaystyle \int_0^t \left(3\dfrac{\rm m}{\mathrm s^2}\right)\,\vec\jmath\,\mathrm du \\\\ \vec v(t) = \left(5\dfrac{\rm m}{\rm s}\right)\,\vec\imath + \left(3\dfrac{\rm m}{\mathrm s^2}\right)t\,\vec\jmath \\\\ \boxed{\vec v(t) = \left(5\dfrac{\rm m}{\rm s}\right)\,\vec\imath + \left(3\dfrac{\rm m}{\mathrm s^2}\right)t\,\vec\jmath}

• position at time <em>t</em> :

\vec r(t) = \displaystyle \int_0^t \left(\left(5\dfrac{\rm m}{\rm s}\right)\,\vec\imath + \left(3\dfrac{\rm m}{\mathrm s^2}\right)u\,\vec\jmath\right) \,\mathrm du \\\\ \boxed{\vec r(t) = \left(5\dfrac{\rm m}{\rm s}\right)t\,\vec\imath + \frac12 \left(3\frac{\rm m}{\mathrm s^2}\right)t^2\,\vec\jmath}

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Answer:

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