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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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The wave nature of light, due to the experiment having bright and dark bands corresponding to places where you have constructive and destructive interference.
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Transcranial magnetic stimulation (TMS) is a noninvasive technique used to stimulate regions of the human brain. A small coil is
attashe74 [19]

Answer:

0.125 volts

Explanation:

The induced emf can be sufficient to stimulate neuronal activity.

One such device generates a magnetic field within the brain that rises from zero to 1.5 T in 120 ms.

We need to find the induced emf within a circle of tissue of radius 1.6 mm and that is perpendicular to the direction of the field. The formula for the induced emf is given by :

\epsilon=-\dfrac{d\phi}{dt}

Where

\phi is magnetic flux

So,

\epsilon=-\dfrac{d(BA)}{dt}\\\\=2\pi r\times \dfrac{dB}{dt}\\\\=2\pi \times 1.6\times 10^{-3}\times \dfrac{1.5-0}{120\times 10^{-3}}\\\\=0.125\ V

So, the induced emf is equal to 0.125 volts.

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Where are alkaline earth metals found on the periodic table?
mojhsa [17]

Answer:

B

Explanation:

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3. When encountering low visibility from rain or fog, you
zimovet [89]

Answer:

c. low beams and fog lights

Explanation:

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i just got the answer wrong and the drivers ed gave me this explanation !!

8 0
3 years ago
Read 2 more answers
A charged paint is spread in a very thin uniform layer over the surface of a plastic sphere of diameter 13.0 cm , giving it a ch
Leokris [45]

a) Electric field inside the paint layer: zero

b) Electric field just outside the paint layer: -3.62\cdot 10^7 N/C

c) Electric field 8.00 cm outside the paint layer: -7.27\cdot 10^7 N/C

Explanation:

a)

We can find the electric field inside the paint layer by applying Gauss Law: the total flux of the electric field through a gaussian surface is equal to the charge contained within the surface divided by the vacuum permittivity, mathematically:

\int EdS = \frac{q}{\epsilon_0}

where

E is the electric field

dS is the element of surface

q is the charge within the gaussian surface

\epsilon_0 = 8.85\cdot 10^{-12}F/m is the vacuum permittivity

Here we want to find the electric field just inside the paint layer, so we take a sphere of radius r as Gaussian surface, where

R = 6.5 cm = 0.065 m is the radius of the plastic sphere (half the diameter)

By taking the sphere of radius r, we note that the net charge inside this sphere is zero, therefore

q=0

So we have

\int E dS=0

which means that the electric field inside the paint layer is zero.

b)

Now we want to find the electric field just outside the paint layer: therefore, we take a Gaussian sphere of radius

r=R=0.065 m

The area of the surface is

A=4\pi R^2

And since the electric field is perpendicular to the surface at any point, Gauss Law becomes

E\cdot 4\pi R^2 = \frac{q}{\epsilon_0}

The charge included within the sphere in this case is the charge on the paint layer, therefore

q=-17.0\mu C=-17.0\cdot 10^{-6}C

So, the electric field is:

E=\frac{q}{4\pi \epsilon_0 R^2}=\frac{-17.0\cdot 10^{-6}}{4\pi(8.85\cdot 10^{-12})(0.065)^2}=-3.62\cdot 10^7 N/C

where the negative sign means the direction of the field is inward, since the charge is negative.

c)

Here we want to calculate the electric field 8.00 cm outside the surface of the paint layer.

Therefore, we have to take a Gaussian sphere of radius:

r=8.00 cm + R = 8.00 + 6.50 = 14.5 cm = 0.145 m

Gauss theorem this time becomes

E\cdot 4\pi r^2 = \frac{q}{\epsilon_0}

And the charge included within the sphere is again the charge on the paint layer,

q=-17.0\mu C=-17.0\cdot 10^{-6}C

Therefore, the electric field is

E=\frac{q}{4\pi \epsilon_0 r^2}=\frac{-17.0\cdot 10^{-6}}{4\pi(8.85\cdot 10^{-12})(0.145)^2}=-7.27\cdot 10^7 N/C

Learn more about electric field:

brainly.com/question/8960054

brainly.com/question/4273177

#LearnwithBrainly

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