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Virty [35]
4 years ago
7

Match the following rocks with the correct type.

Physics
2 answers:
slamgirl [31]4 years ago
6 0
I\B
2/A
3\C
4\E
5\D
6\F
I think this is right

Mila [183]4 years ago
4 0

The correct matches are:

1. obsidian - B) extrusive igneous;

2. granite - A) intrusive igneous;

3. conglomerate - C) clastic sedimentary;

4. evaporite - E) chemical sedimentary;

5. coal - D) organic sedimentary;

6. marble - F) metamorphic;

There are three big groups of rocks, igneous, sedimentary, and metamorphic. They further diversify into multiple types of rocks in accordance to their structure, chemical composition, crystallization, formation etc. The igneous can simple be divided into intrusive (ones that formed on the surface or very close to it) and extrusive (ones that formed deep into the ground). The sedimentary rocks can be divided into multiple groups in accordance to their composition, as some can be made of multiple sediments, some from organic material, some from mixture of sediments and organic material etc. And the metamorphic can simply be seen as rocks that come to be as a result of the transformation of the other two types if they are under big pressure and temperatures for longer periods of time.

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tekilochka [14]

Answer:

2 meter/ sec

Explanation:

Speed= distance/ time

s= 10/5

=2 meter/ sec

6 0
3 years ago
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Electrons are ejected from sodium metal by any light that has a wavelength shorter than 544 nm. What is the kinetic energy of an
wel

Answer:

KE=2.3 x 10⁻¹⁹ J

Explanation:

Given that

λ = 544 nm

λ' = 485 nm

The kinetic energy KE given as

KE= E - Ф

Where

E=\dfrac{hC}{\lambda'}

\phi=\dfrac{hC}{\lambda}

h= 6.626 x 10⁻³⁴

C=3 x 10⁸ m/s

Now by putting the values

KE=\dfrac{hC}{\lambda'}-\dfrac{hC}{\lambda}

KE=\dfrac{34.34\times 10^{-34}\times 3\times 10^8}{485\times 10^{-9}}- \dfrac{34.34\times 10^{-34}\times 3\times 10^8}{544\times 10^{-9}}

KE=2.3 x 10⁻¹⁹ J

This is kinetic energy.

6 0
3 years ago
Two spherical conductors are separated by a distance much larger than either of their radii. Sphere A has a radius of 11.5 cm an
bonufazy [111]

Explanation:

As the given spheres are connected by a thin wire so, the potential on the spheres are the same.

          \frac{q_{1}}{r_{1}} = \frac{q_{2}}{r_{2}} ......... (1)

Hence, total charge will be as follows.

              q_{1} + q_{2} = Q = -95.5 nC .......... (2)

Using the above two equations, the final equation will be as follows.

          q_{2} = \frac{Qr_{2}}{r_{1} + r_{2}}

and,    q_{1} = \frac{Qr_{1}}{r_{1} + r_{2}}

Hence, we will calculate the charge on sphere B after the equilibrium is reached as follows.

          q_{2} = \frac{Qr_{2}}{r_{1} + r_{2}}

                     = \frac{-95.5 \times 74.4 cm}{(11.5 + 74.4) cm}

                     = 82.714 nC

Thus, we can conclude that the charge on sphere B after equilibrium has been reached is 82.714 nC.

                       

5 0
3 years ago
A balloon contains 0.075 m^3 of
inessss [21]

Answer:

600 KPa.

Explanation:

From the question given above, the following data were obtained:

Initial volume (V1) = 0.075 m³

Final volume (V2) = 0.45 m³

Final pressure (P2) = 100 KPa

Initial pressure (P1) =?

Temperature = constant

The initial pressure can be obtained by using the Boyle's law equation as shown below:

P1V1 = P2V2

P1 × 0.075 = 100 × 0.45

P1 × 0.075 = 45

Divide both side by 0.075

P1 = 45 / 0.075

P1 = 600 KPa.

Thus, the initial pressure in the balloon is 600 KPa.

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3 years ago
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if you on UsaTestPrep its A

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4 years ago
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