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Nataly [62]
2 years ago
6

A student sorted mineral samples into two groups: dull and shiny. Which of the following properties did the student use to sort

the mineral samples into groups? a
streak
b
luster
c
cleavage
d
color free brainliest
Physics
2 answers:
Katyanochek1 [597]2 years ago
7 0

Answer:

luster

Explanation:

xenn [34]2 years ago
4 0
The answer would be luster
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_____ is the transfer of thermal energy between two bodies which are at different temperatures. The si unit for this is the joul
ss7ja [257]

Answer:

Heat

Explanation:

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This temperature change that is involved in this process can be measured.

Thus, in thermal energy, heat (heat energy) is transferred from one body to another at different distinct temperatures.

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3 years ago
Describe the organization of our solar system. What are the different divisions, or categories?
olasank [31]
Hello.

The solor system is organized by the smallest star to the largest and by the 8 planets. it is also organized by how much gas a planet has. But it is also orgainzed becasue of gravity.

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Have a nice day

6 0
3 years ago
Two spherical shells have a common center. A -2.1 10-6 C charge is spread uniformly over the inner shell, which has a radius of
julsineya [31]

Answer:

a) E_total = 6,525 10⁴ N /C ,field direction is radial outgoing

b)  E_total = 1.89 10⁶ N / C, field is incoming radial

c) E_total = 0

Explanetion:

For this exercise we can use that the charge in a spherical shell can be considered concentrated at its center and that the electric field inside the shell is zero, since Gauss's law is

                Ф = E .dA = q_{int} /ε₀

inside the spherical shell there are no charges

The electric field is a vector quantity, so we calculate the field created by each shell and add it vectorly.

We have two sphere shells with radii 0.050m and 0.15m respectively

a) point where you want to know the electric field d = 0.20 m

shell 1

the point is on the outside,d>ro,  therefore we can consider the charge to be concentrated in the center

            E₁ = k q₁ / d²

             

shell 2

the point is on the outside,d>ro

             E₂ = k q₂ / d²

the total camp is

              E_total = -E₁ + E₂

              E_total = k ( \frac{-q_1 + q_2}{d^2})

              E_total = 9 10⁹ (-2.1 10⁻⁶+ 5 10⁻⁶ / .2²

              E_total = 6,525 10⁵ N /C

The field direction is radial and outgoing ti the shells

b) the calculation point is d = 0.10m

shell 1

point outside the shell d> ro

                 E₁ = k q₁ / d²

shell 2

the point is inside the shell d <ro

Therefore, according to Gauss's law, since there are no charges in the interior, the electrioc field is zero

                E₂ = 0

               

                 E_total = E₁

                 E_total = k q₁ / d²

                 E_total = 9 10⁹ 2.1 10⁻⁶ / 0.1²

                 E_total = 1.89 10⁶ N / A

As the charge is negative, this field is incoming radial, that is, it is directed towards the shell 1

c) the point of interest d = 0.025 m

shell 1

point  is inside the shell d< ro

                 

as there are no charges inside

                     E₁ = 0

shell 2

point is inside the radius of the shell d <ro

                    E₂ = 0

the total field is

                    E_total = 0

3 0
2 years ago
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TiliK225 [7]

Answer:

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Explanation:

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same with every answer exce

6 0
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Kryger [21]
D=s(t) so it would be d=10(.19) d=.19 FOR BITH SNDWERS
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3 years ago
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