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Hunter-Best [27]
2 years ago
15

What is the planet's rotational speed from slowest to fastest?​

Physics
1 answer:
larisa86 [58]2 years ago
5 0

Answer:

Jupiter is the fastest whereas Venus is the slowest

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4
tensa zangetsu [6.8K]

Answer:

C. amount of charge on the source charge.

Explanation:

Electric field lines can be defined as a graphical representation of the vector field or electric field.

Basically, it was first introduced by Michael Faraday and it is typically a curve drawn to the tangent of a point is in the direction of the net field acting on each point.

The number, or density, of field lines on a source charge indicate the amount of charge on the source charge. Therefore, the density of field lines on a source charge is directly proportional to quantity of charge on the source.

8 0
2 years ago
In what way would a digital thermometer be preferable to those from a liquid-based thermometer?
garik1379 [7]
Reading a digital thermometer is more preferable because you can just easily read the measurement from a screen, a number will just show in a screen as compared to a liquid-based thermometer where you still have to read the measurement by counting the markers in the thermometer. Hope this helps! Mark brainly please!
5 0
3 years ago
What is the weight of an object with a mass of 6.0 kg on Earth?
gregori [183]
<h2>since weight is measured in newtons, convert the 6 kg to newtons</h2><h3>the formula to convert is kg x 9.807 = N</h3>
  • 6 x 9.807
  • = 58.842 N

hope that helps :))

3 0
3 years ago
Read 2 more answers
A hydrogen atom in a galaxy moving with a speed of 6.65×106 m/???? away from the Earth emits light with a wavelength of 5.13×10−
Mumz [18]

Answer:

The observed wavelength on Earth from that hydrogen atom is 5.24\times 10^{-7}\ m.

Explanation:

Given that,

The actual wavelength of the hydrogen atom, \lambda_a=5.13\times 10^{-7}\ m

A hydrogen atom in a galaxy moving with a speed of, v=6.65\times 10^6\ m/s

We need to find the observed wavelength on Earth from that hydrogen atom. The speed of galaxy is given by :

v=c\times \dfrac{\lambda_o-\lambda_a}{\lambda_a}

\lambda_o is the observed wavelength

\lambda_o=\dfrac{v\lambda_a}{c}+\lambda_a\\\\\lambda_o=\dfrac{6.65\times 10^6\times 5.13\times 10^{-7}}{3\times 10^8}+5.13\times 10^{-7}\\\\\lambda_o=5.24\times 10^{-7}\ m

So, the observed wavelength on Earth from that hydrogen atom is 5.24\times 10^{-7}\ m. Hence, this is the required solution.

8 0
3 years ago
30 points for all 5 answers please
Lynna [10]

Answer:

he is doing science

he is doing science

he is doing science

he is doing science

he is doing science

Explanation:

8 0
3 years ago
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