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labwork [276]
2 years ago
10

You set a sensor to transmit the wind speed on top of a mountain. Which of these languages are you most likely to have used?

Engineering
1 answer:
kifflom [539]2 years ago
8 0

The programming language that is most likely used to transmit the wind speed is: B. SQL.

<h3>What is SQL?</h3>

SQL is an acronym for structured query language and it can be defined as a domain-specific programming language that is designed and developed for the management of various data that are saved in a relational or structured database.

This ultimately implies that, a structured query language (SQL) can be used to communicate with a database in accordance with the American National Standards Institute (ANSI) standards.

In conclusion, the programming language that is most likely used to transmit the wind speed is SQL.

Read more on SQL here:

brainly.com/question/25266787

#SPJ1

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An automobile weighing 2500 lbf increases its gravitational potential energy by a magnitude of 2.25 × 104 Btu in going from an e
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Answer:

The elevation at the high point of the road is 12186.5 in ft.

Explanation:

The automobile weight is 2500 lbf.

The automobile increases its gravitational potential energy in 2.25 * 10^4 BTU. It means the mobile has increased its elevation.

The initial elevation is of 5183 ft.  

The first step is to convert Btu of potential energy to adequate units to work with data previously presented.

British Thermal Unit - 1 BTU = 778.17  lbf*ft

2.25 * 10^4 BTU (\frac{778.17 lbf*ft}{1BTU} ) = 1.75 * 10^7 lbf * ft

Now we have the gravitational potential energy in lbf*ft. Weight of the mobile is in lbf and the elevation is in ft. We can evaluate the expression for gravitational potential energy as follows:  

Ep = m*g*(h_2 - h_1)\\ W = m*g  

Where m is the mass of the automobile, g is the gravity, W is the weight of the automobile showed in the problem.  

h_2 is the final elevation and h_1 is the initial elevation.

Replacing W in the Ep equation

Ep = W*(h_2 -h_1)\\(h_2 -h_1) = \frac{Ep}{W} \\h_2 = h_1 + \frac{Ep}{W}\\\\

Finally, the next step is to replace the variables of the problem.  

h_2 = 5183 ft + \frac{1.75 * 10^7 lbf*ft}{2500 lbf}\\h_2 = 5183 ft + 70003.5 ft\\h_2 = 12186.5 ft

The elevation at the high point of the road is 12186.5 in ft.  

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