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Olin [163]
3 years ago
14

Which statement about the minerals plagioclase feldspar, gypsum, biotite mica, and talc can best be inferred from the chart? (1)

These minerals have the same
chemical and physical properties. (2) These minerals have different chemical properties, but they have similar physical properties. (3) These minerals have different physical and chemical properties, but they have identical uses. (4) The physical and chemical properties of these minerals determine how humans use them.
Physics
1 answer:
madreJ [45]3 years ago
7 0

Answer:

(4) The physical and chemical properties of these minerals determine how humans use them.

Explanation:

All the materials and the metals found on earth shows certain characteristics naturally or in their physical state. These physical characteristics can be their look, their structure, their color, strength, melting point, boiling point, density, etc.

And chemical properties of the metals are defined as those characteristics or features of the metals that it exhibits when these metals reacts chemically.

The physical properties as well as the chemical properties distinguishes each metal from each other. These properties determines how people use these metals in their life.

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Answer:

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Oil having a density of 926 kg/m3 floats on water. A rectangular block of wood 3.69 cm high and with a density of 974 kg/m3 floa
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Answer:

the position of the wood below the interface of the two liquids is 2.39 cm.

Explanation:

Given;

density of oil, \rho _o = 926 kg/m³

density of the wood, \rho _{wood} = 974 kg/m³

density of water, \rho _w = 1000 kg/m³

height of the wood, h = 3.69 cm

Based on the density of the wood, it will position across the two liquids.

let the position of the wood below the interface of the two liquids = x

Let the wood be in equilibrium position;

F_{wood} - F_{oil} - F_{water} = 0\\\\\rho _{wood} .gh - \rho _o .g(h-x) - \rho_w .gx = 0\\\\\rho _{wood} .h - \rho _o (h-x) - \rho_w .x = 0\\\\\rho _{wood} .h -\rho _o h + \rho _o x - \rho_w .x =0\\\\h (\rho _{wood}  -\rho _o ) = x( \rho_w - \rho _o)\\\\x =h[\frac{ \rho _{wood}  -\rho _o }{\rho_w - \rho _o} ]\\\\x = 3.69\ cm \times [\frac{974 - 926}{1000-926} ]\\\\x = 2.39 \ cm

Therefore, the position of the wood below the interface of the two liquids is 2.39 cm.

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