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omeli [17]
3 years ago
10

Select the correct answer.

Physics
1 answer:
Murrr4er [49]3 years ago
3 0

Answer:

I will choose the answer A

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Which of these is the easiest mineral identification property to use when you want to know whether you have pyrite or actual gol
Makovka662 [10]

Answer:

c. streak

Explanation:

Pyrite is a mineral that looks like gold but actually is iron disulfide.

Pyrite and gold have comparable luster.

Pyrite and gold have different tones of yellow. This can be determined by their streak. Streak is the powdered form of a mineral. A streak of mineral can be found just by rubbing the mineral on a rough surface and comparing the colors.

Pyrite is diamagnetic which is not a strong form of magnetism. Gold is also diamagnetic

5 0
3 years ago
I NEED HELP ASAP
LiRa [457]

The object is called a meteor because it is producing Streak of light  and has not yet struck earth.

<h3><u>Explanation:</u></h3>

A meteoroid is a celestial object which is very smaller than an asteroid. These objects are produced as a collision impact from mars or moon and float freely in space without any specific orbit. When they come inside the Earth's gravitational field, they are attracted by the Earth's gravity to Earth's crust. These objects in Earth's atmosphere are called meteors. As they travel through Earth's atmosphere, they do face a huge friction from Earth's atmosphere which let them burn and that is visible as the tail of the meteor.

Most of them are so small that they are burnt away in the atmosphere. But some are bigger and they reach the Earth's surface and are called as meteorites.

4 0
3 years ago
Read 2 more answers
An object is attached to a vertical spring and bobs up and down between points a and
Anton [14]
 it would be at either A or B. 
7 0
3 years ago
A vacuum tube diode consists of concentric cylindrical electrodes, the negative cathode and the positive anode. Because of the a
DENIUS [597]

Answer:

   C = 4,174 10³ V / m^{3/4} ,  E = 7.19 10² / ∛x,    E = 1.5  10³ N/C

Explanation:

For this exercise we can calculate the value of the constant and the electric field produced,

Let's start by calculating the value of the constant C

           V = C x^{4/3}

           C = V / x^{4/3}

            C = 220 / (11 10⁻²)^{4/3}

            C = 4,174 10³ V / m^{3/4}

To calculate the electric field we use the expression

            V = E dx

             E = dx / V

             E = ∫ dx / C x^{4/3}

            E = 1 / C  x^{-1/3} / (- 1/3)

            E = 1 / C (-3 / x^{1/3})

We evaluate from the lower limit x = 0 E = E₀ = 0 to the upper limit x = x, E = E

            E = 3 / C     (0- (-1 / x^{1/3}))

            E = 3 / 4,174 10³   (1 / x^{1/3})

           E = 7.19 10² / ∛x

for x = 0.110 cm

          E = 7.19 10² /∛0.11

          E = 1.5  10³ N/C

6 0
4 years ago
How light is channelled down an optical fibre
coldgirl [10]

Explanation:

Suppose you want to shine a flashlight beam down a long, straight hallway. Just point the beam straight down the hallway -- light travels in straight lines, so it is no problem. What if the hallway has a bend in it? You could place a mirror at the bend to reflect the light beam around the corner. What if the hallway is very winding with multiple bends? You might line the walls with mirrors and angle the beam so that it bounces from side-to-side all along the hallway. This is exactly what happens in an optical fiber.

The light in a fiber-optic cable travels through the core (hallway) by constantly bouncing from the cladding (mirror-lined walls), a principle called total internal reflection. Because the cladding does not absorb any light from the core, the light wave can travel great distances.

However, some of the light signal degrades within the fiber, mostly due to impurities in the glass. The extent that the signal degrades depends on the purity of the glass and the wavelength of the transmitted light (for example, 850 nm = 60 to 75 percent/km; 1,300 nm = 50 to 60 percent/km; 1,550 nm is greater than 50 percent/km). Some premium optical fibers show much less signal degradation -- less than 10 percent/km at 1,550 nm.

1

3 0
3 years ago
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