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Luba_88 [7]
3 years ago
10

If a substance can be separated by physical means and it is not the same throughout, what is it.

Physics
2 answers:
VMariaS [17]3 years ago
3 0

Answer:

The answer is Heterogeneous (B) I did the lesson and took the quiz today.

Explanation:

V125BC [204]3 years ago
3 0

: A Heterogeneous Mixture

~Sarah Robinsen

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(b)
Leona [35]

Answer:

i wish i knew your answer.

Explanation:

6 0
2 years ago
Give three examples of objects in equilibrium found in classroom?​
Charra [1.4K]

Answer:

Book. Bottle. table are some examples of objects in equilibrium found in classroom

7 0
2 years ago
What type of energy waves are used in Solar Cells, and what is it’s position on the electromagnetic spectrum?
Artyom0805 [142]

Answer: Electromagnetic waves (Ultraviolet light, between 100 nm and 380 nm)

Explanation:

Solar cells work by the photoelectric effect, which consists of the emission of electrons (electric current) when light (electromagnetic waves) falls on a metal surface under certain conditions.  

In this sense, the portion of the electromagnetic spectrum this cells use is Ultraviolet light (UV) from the Sun, whose wavelength is approximately between 100 nm and 380 nm.

It is important to note, this is a type of electromagnetic radiation that is not visible to the human eye.

3 0
3 years ago
Why do noble gasses rarely react with other elements?
ollegr [7]
Noble gasses have an outer shell full of electrons. A full outer energy level is the most stable arrangement of electrons. As a result, noble gases cannot become more stable by reacting with other elements and gaining or losing valence electrons. Therefore, noble gases are rarely involved in chemical reactions and almost never form compounds with other elements.
3 0
2 years ago
A uniform, solid, 1800.0 kgkg sphere has a radius of 5.00 mm. Find the gravitational force this sphere exerts on a 2.30 kgkg poi
iVinArrow [24]

Answer

given,

Mass of the solid sphere = 1800 Kg

radius of the sphere,R = 5 m

mass of the small sphere, m = 2.30 Kg

when the Point is outside the sphere the Force between them is equal to

      F = \dfrac{GMm}{r^2}   when r>R

When Point is inside the Sphere

      F = \dfrac{GMm}{R^2}\ \dfrac{r}{R}  when r<R

where r is the distance where the point mass is placed form the center

Now Force calculation

a) r = 5.05 m

       F = \dfrac{GMm}{r^2}[/tex]

       F = \dfrac{6.67\times 10^{-11}\times 1800\times 2.3}{5.05^2}

               F = 1.082 x 10⁻⁸ N

b) r = 2.65 m

      F = \dfrac{GMm}{R^2}\ \dfrac{r}{R}

      F = \dfrac{6.67\times 10^{-11}\times 1800\times 2.3}{5.05^2}\ \dfrac{2.65}{5.05}

     F = 5.68\times 10^{-9}\ N

3 0
3 years ago
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