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Soloha48 [4]
3 years ago
8

Planet C has a tilt of zero degrees. What seasonal changes would be expected on this planet?

Physics
2 answers:
marshall27 [118]3 years ago
8 0
The second statement is your answer:

 No change in temperatures between seasons. Since tilt is the angle between the direction of positive pole and the normal to the orbital plane, seasons occur due to the tilt of axis. If there were to be no, or zero degree tilt, then there will be no season because of the lack in temperature changes. The amount of sunlight received remains same throughout the year which results in no temperature change 

harina [27]3 years ago
4 0
"Tilt of zero degrees" means the planet's rotation axis is "straight up and down".

...  No place on the planet has any seasons.
...  The climate at every place is the same all year around.
...  No place on the planet has any changes in the length of its days.
...  Every place on the planet has 50% light and 50% dark, every day.
...  At the poles, the sun goes around and around and around the horizon. 
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Answer:

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Explanation:

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2 years ago
Gravity causes objects to be attracted to one another. This attraction keeps our feet firmly planted on the ground and causes th
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The units of G must be C. m³ / ( kg s² )

<h3>Further explanation</h3>

Newton's gravitational law states that the force of attraction between two objects can be formulated as follows:

\large {\boxed {F = G \frac{m_1 ~ m_2}{R^2}} }

<em>F = Gravitational Force ( Newton )</em>

<em>G = Gravitational Constant ( 6.67 × 10⁻¹¹ Nm² / kg² )</em>

<em>m = Object's Mass ( kg )</em>

<em>R = Distance Between Objects ( m )</em>

Let us now tackle the problem !

To find unit of Gravitational Constant can be carried out in the following way:

F = G \frac{m_1 ~ m_2}{R^2}

{[N]}= G\frac{{[kg]}{[kg]}}{{[m^2]}}

{[kg ~ m / s^2]}= G \frac{{[kg^2]}}{{[m^2]}}

G = \frac{{[kg ~ m / s^2]}{[m^2]}} {{[kg^2]} }

G = \frac{{[kg ~ m^3 / s^2]}} {{[kg^2]} }

G = \frac{{[m^3 / s^2]}} {{[kg]} }

\boxed {G = \frac{{[m^3]}} {{[kg ~ s^2]} }}

The unit of G must be \large {\boxed {\frac{m^3} {kg ~ s^2 }}}

<h3>Learn more</h3>
  • Impacts of Gravity : brainly.com/question/5330244
  • Effect of Earth’s Gravity on Objects : brainly.com/question/8844454
  • The Acceleration Due To Gravity : brainly.com/question/4189441

<h3>Answer details</h3>

Grade: High School

Subject: Physics

Chapter: Gravitational Fields

Keywords: Gravity , Unit , Magnitude , Attraction , Distance , Mass , Newton , Law , Gravitational , Constant

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Explanation:

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