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ELEN [110]
3 years ago
11

Match the correct properties and characteristics of each type of rock?

Chemistry
2 answers:
Over [174]3 years ago
6 0

Sedimentary- deposition of material on Earth's surface or in bodies of water/ made of hard layers formed under high temperature and pressure

Igneous- cooled magma/ has a shiny surface throughout

Metamorphic- made of layers that are weathered relatively easily/ arises from transformation of existing rocks

Lelu [443]3 years ago
5 0
Sedimentary -<span>deposition of material on Earth’s surface or in bodies of water
</span><span>igneous- </span><span>cooled magma
</span>metamorphic-<span>made of hard layers formed under high temperature and pressure</span>
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Explanation:

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Water is poured into a conical container at the rate of 10 cm3/sec. The cone points directly down, and it has a height of 20 cm
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Answer:

\frac{dh}{dt}_{h=2cm} =\frac{40}{9\pi}\frac{cm}{2}

Explanation:

Hello,

The suitable differential equation for this case is:

\frac{dV}{dt}=10\frac{cm^3}{s}

As we're looking for the change in height with respect to the time, we need a relationship to achieve such as:

\frac{dh}{dt} = ?*\frac{dV}{dt}

Of course, ?=\frac{dh}{dV}.

Now, since the volume of a cone is V=\pi r^2h/3 and the ratio r/h=15/20=3/4 or r=3/4h, the volume becomes:

V=\pi (\frac{3}{4} h)^2h/3= \frac{3}{16}\pi h^3

We proceed to its differentiation:

\frac{dV}{dh} =\frac{9}{16} \pi h^2\\\frac{dh}{dV} =\frac{16}{9 \pi h^2}

Then, we compute \frac{dh}{dt}

\frac{dh}{dt} = \frac{16}{9 \pi h^2}*\frac{dV}{dt}\\\frac{dh}{dt} = \frac{16}{9\pi h^2}*10\frac{cm^3}{s} =\frac{160}{9 \pi h^2}

Finally, at h=2:

\frac{dh}{dt}_{h=2cm} =\frac{160}{9\pi 2^2}\\\frac{dh}{dt}_{h=2cm} =\frac{40}{9\pi}\frac{cm}{s}

Best regards.

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