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3 years ago

Please help quick! 3 questions if you answer all I will give brainliest

1 answer:

4 0

Answer: 1.Spectroscopes breaks the light from a single material into its component colors the way a prism splits white light into a rainbow. It records this spectrum, which allows scientists to analyze the light and discover properties of the material interacting with it.

2.Optical Telescopes The now-indispensable optical telescope instrument was pioneered by Galileo Galilei in 1609, although others had created similar tools by then

2.A stellar spectrum can reveal many properties of stars, such as their chemical composition, temperature, density, mass, distance, luminosity, and relative motion using Doppler shift measurements.

3.Stars form from an accumulation of gas and dust, which collapses due to gravity and starts to form stars. The process of star formation takes around a million years from the time the initial gas cloud starts to collapse until the star is created and shines like the Sun.Once the pressure and the temperature inside get high enough for nuclear fusion to ignite, it creates a star.

Explanation:

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What are minerals made of a single element called?

Fed [463]

Minerals of a single element are called Native elements. These include gold, silver, copper, etc. They can be divided into three groups such as metals, semi-metals, and nonmetals. Hope this helped.

3 0

3 years ago

What is a group of two or more atoms that are held together by chemical bonds?

Bad White [126]

The answer is B.) molecule

8 0

3 years ago

Read 2 more answers

In March 2006, two small satellites were discovered orbiting Pluto, one at a distance of 48,000 km and the other at 64,000 km. P

tatiyna

Answer:

Time period for first satellites 24.46 days and for second satellites 37.67 days

Explanation:

Given :

Distance of first satellites $r_{sat1} = 48000 \times 10^{3}$ m

Distance of second satellites $r _{sat2} = 64000 \times 10^{3}$ m

Distance of charon $r_{c} = 19600 \times 10^{3}$ m

Time period of charon $T_{c} = 6.39$ days

From the kepler's third law,

Square of the time period is proportional to the cube of the semi major axis.

$T^{2} = r^{3}$

$\frac{T}{r^{\frac{3}{2} } } = constant$

For first satellites,

$\frac{T_{c} }{r_{c} ^{\frac{3}{2} } } = \frac{T_{sat1} }{r_{sat1} ^{\frac{3}{2} } }$

${T_{sat1} } = 6.39 \times \frac{(48000 \times 10^{3} )^{\frac{3}{2} } }{(19600\times 10^{3} )^{\frac{3}{2} }}$

$T_{sat1} = 24.46$ days

For second satellites,

$\frac{T_{c} }{r_{c} ^{\frac{3}{2} } } = \frac{T_{sat2} }{r_{sat2} ^{\frac{3}{2} } }$

${T_{sat2} } = 6.39 \times \frac{(64000 \times 10^{3} )^{\frac{3}{2} } }{(19600\times 10^{3} )^{\frac{3}{2} }}$

$T_{sat2} = 37.67$ days

Therefore, time period for first satellites = 24.46 days and for second satellites 37.67 days

8 0

3 years ago

How could you increase power of a wave in a spring

lora16 [44]

Alll of the above I just read it from someone and if it’s not that lmk

8 0

3 years ago

Read 2 more answers

Charge q1 = +2.00 μC is at -0.500 m along the x axis. Charge q2 = -2.00 μC is at 0.500 m along the x axis. Charge q3 = 2.00 μC i

Kobotan [32]

The magnitude of electrical force on charge $q_{3}$ due to the others is 0.102 newtons.

<h3>How to calculate the electrical force experimented on a particle</h3>

The vector position of each particle respect to origin are described below:

$\vec r_{1} = (-0.500, 0)\,[m]$

$\vec r_{2} = (+0.500, 0)\,[m]$

$\vec r_{3} = (0, +0.500)\,[m]$

Then, distances of the former two particles particles respect to the latter one are found now:

$\vec r_{13} = (+0.500, +0.500)\,[m]$

$r_{13} = \sqrt{\vec r_{13}\,\bullet\,\vec r_{13}} = \sqrt{(0.500\,m)^{2}+(0.500\,m)^{2}}$

$r_{13} =\frac{\sqrt{2}}{2}\,m$

$\vec r_{23} = (-0.500, +0.500)\,[m]$

$r_{23} = \sqrt{\vec r_{23}\,\bullet \,\vec r_{23}} = \sqrt{(-0.500\,m)^{2}+(0.500\,m)^{2}}$

$r_{23} =\frac{\sqrt{2}}{2}\,m$

The resultant force is found by Coulomb's law and principle of superposition:

$\vec R = \vec F_{13}+\vec F_{23}$ (1)

Please notice that particles with charges of same sign attract each other and particles with charges of opposite sign repeal each other.

$\vec R = \frac{k\cdot q_{1}\cdot q_{3}}{r_{13}^{2}}\cdot \vec u_{13} +\frac{k\cdot q_{2}\cdot q_{3}}{r_{23}^{2}}\cdot \vec u_{23}$ (2)

Where:

$k$ - Electrostatic constant, in newton-square meters per square Coulomb.
$q_{1}$ , $q_{2}$ , $q_{3}$ - Electric charges, in Coulombs.
$r_{13}$ , $r_{23}$ - Distances between particles, in meters.
$\vec u_{13}$ , $\vec u_{23}$ - Unit vectors, no unit.

If we know that $k = 8.988\times 10^{9}\,\frac{N\cdot m^{2}}{C^{2}}$ , $q_{1} = 2\times 10^{-6}\,C$ , $q_{2} = 2\times 10^{-6}\,C$ , $q_{3} = 2\times 10^{-6}\,C$ , $r_{13} =\frac{\sqrt{2}}{2}\,m$ , $r_{23} =\frac{\sqrt{2}}{2}\,m$ , $\vec u_{13} = \left(-\frac{\sqrt{2}}{2}, - \frac{\sqrt{2}}{2} \right)$ and $\vec u_{23} = \left(\frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{2} \right)$ , then the vector force on charge $q_{3}$ is:

$\vec R = \frac{\left(8.988\times 10^{9}\,\frac{N\cdot m^{2}}{C^{2}} \right)\cdot (2\times 10^{-6}\,C)\cdot (2\times 10^{-6}\,C)}{\left(\frac{\sqrt{2}}{2}\,m \right)^{2}} \cdot \left(-\frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{2} \right) + \frac{\left(8.988\times 10^{9}\,\frac{N\cdot m^{2}}{C^{2}} \right)\cdot (2\times 10^{-6}\,C)\cdot (2\times 10^{-6}\,C)}{\left(\frac{\sqrt{2}}{2}\,m \right)^{2}} \cdot \left(\frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{2} \right)$

$\vec R = 0.072\cdot \left(-\frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{2} \right) + 0.072\cdot \left(\frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{2} \right)\,[N]$

$\vec R = 0.072\cdot \left(0, -\sqrt{2}\right)\,[N]$

And the magnitude of the electrical force on charge $q_{3}$ ( $R$ ), in newtons, due to the others is found by Pythagorean theorem:

$R = 0.102\,N$

The magnitude of electrical force on charge $q_{3}$ due to the others is 0.102 newtons. $\blacksquare$

To learn more on Coulomb's law, we kindly invite to check this verified question: brainly.com/question/506926

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2 years ago