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yKpoI14uk [10]
3 years ago
9

How do you producers consumers and decomposers all live in a cycle

Physics
1 answer:
ahrayia [7]3 years ago
4 0
The producers, produce what ever they are producing for the consumers to obtain. The decomposers, decompose the things when the producers and consumers are done with them. Therefore, it is a cycle!
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An element is highly conductive, highly reactive, soft, and lustrous. The element most likely belongs to which group?(1 point)
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An element which is highly conductive, highly reactive, soft, and lustrous is most likely an alkali metal.

Alkali metals are in group 1 of the Periodic table which means that they have only a single valence electron.

This causes them to be soft and highly reactive because:

  • The single valance electron leads to weak bonds amongst the element's atoms which makes them soft
  • The elements want to lose the single valance electron so as to become stable so they will react with other elements to give away the electron.

Examples of alkali electrons include:

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In conclusion therefore, alkali metals are highly reactive and soft and so the element described above is most likely an alkali metal.

<em>Find out more at brainly.com/question/18722874.</em>

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If 2.0×10^−4 C of charge passes a point in 2.0×10^−6 s , what is the rate of current flow?1.0×10^−10 A1.0×10^2 A4.0×10^−1 A4.0×1
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This problem is about the rate of the current. It's important to know that refers to the quotient between the electric charge and the time, that's the current rate.

I=\frac{Q}{t}

Where Q = 2.0×10^−4 C and t = 2.0×10^−6 s. Let's use these values to find I.

\begin{gathered} I=\frac{2.0\times10^{-4}C}{2.0\times10^{-6}\sec } \\ I=1.0\times10^{-4-(-6)}A \\ I=1.0\times10^{-4+6}A \\ I=1.0\times10^2A \end{gathered}

<em>As you can observe above, the division of the powers was solved by just subtracting their exponents.</em>

<em />

<h2>Therefore, the rate of the current flow is 1.0×10^2 A.</h2>
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