why does an elliptical galaxy look like one giant star

Answer:

A

Explanation:

Answer:

60 meters

Explanation:

If you are going 3 meters in a second, and you are traveling for 20 seconds, you have to multiply

3meters/second*20seconds

cross out the seconds and you have

3 meters*20

60 meters

Answer:

I think its

10 kg

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If the mass of an object is approximately 60 kg on earth then its mass remains the same on Mars which is 60 kg because mass does not change from place to place.

Mass in physics is a numerical representation of inertia, a fundamental property of all matter. What a body of matter is fundamentally is the resistance it offers to a change in its speed or location brought on by the application of a force. The more mass a body has, the less of a change a force exerts results. The kilogram is the unit of mass in the International System of Units, and it is defined by Planck's constant as being equal to 6.62607015 × 10⁻³⁴ joules second (SI).

According to the question, the weight of an object depends on gravity which is different from place to place so weight also varies from place to place but the mass is constant everywhere it does not change. So, if the mass of an object is 60 kg on earth then it remains the same on mars also.

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Answer:

[tex]V_{2.2} =2.2V[/tex]

Fast and loose: that's a classic voltage divider. the drop from the i-th resistor is [tex]R_i \over {\sum R}[/tex] of the drop across the whole series. In our situation, it's [tex]2.2\cdot 10^3 \over {2.2\cdot10^3 + 6.8\cdot 10^3[/tex] of 9 V. By plugging numbers in a calculator, it's 22/90 of 9V, or 2.2V

With Ohm's Law

The series is equivalent of a single resistor of Resistance [tex]2.2\cdot 10^3+6.8\cdot 10^3 \Omega = 9.0 k\Omega[/tex]. By Ohm's first Law ([tex]V=Ri[/tex]) the current flowing through the resistor is [tex]9V = 9*10^3\Omega i \rightarrow i=1mA[/tex]. At this point, the drop across the first resistor is, again by Ohm's law[tex]V = 2.2 \cdot 10^3 \Omega \cdot 1\cdot 10^{-3} A = 2.2V[/tex]

Since the period of the supposed planet is less than the period of the earth orbiting the sun, That means the mass of the central star is greater than the mass of the sun. The answer is option C

According to Kepler's third law of planetary motion for planets and satellites, he described circular orbit about the same center body stating that:

The squared of the period is directly proportional to the cube of the radius. That is,

[tex]T^{2} \alpha r^{3}[/tex]

The orbital period formula is expressed as :

[tex]T^{2}[/tex] = 4[tex]\pi ^{2}[/tex] [tex]r^{3}[/tex]  ÷ GM

Suppose a planet orbits a star at the same distance as Earth (1 AU) but with a period of roughly 200 Earth days. The mass of the central star will be greater than the mass of our sun based on the formula given above.

In a geostationary orbit, whatever the mass of the satellite may be, they will travel at the same speed in a particular orbit. mass M in the above formula becomes the mass of the sun or the center mass.

Since the period of the supposed planet is less than the period of the earth orbiting the sun, That means the mass of the central star is greater than the mass of the sun.

Therefore, the correct answer is option C

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Answer:

The rocket travelled 37km per minute

Explanation:

185÷5=37