The orbital period could be obtained from the question as 8.98 * 10^7 m.
<h3>What is the orbital period?</h3>
We know that the solar system is composed of the sun and the planets. The planets are known to move around the sun in concentric circles. Following the heliocentric model of the solar system, the sun is at the center of the solar system at all times and all the other planets tend to move round the sun at a good distance that is appropriate for each.
Now we have that;
T = √4π^2r^3/GM
T = period of the orbit
r = radius of the orbit
G = gravitational constant
M = mass of the planet
Hence, we have;
R = ∛T^2GM/4π^2
R =∛ (26 * 60 * 60)^2 * 6.67 * 10^-11 * 4.9 * 10^25/ 4 * (3.142)^2
R = ∛2.86 * 10^25/39.49
R = 8.98 * 10^7 m
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"The marble moved 30 cm north in 6 seconds" is the one example among the following choices given in the question that <span>provides a complete scientific description of an object in motion. The correct option among all the options that are given in the question is the third option or option "C". I hope the answer has helped you.</span>
<h2>
Answer: electrons sometimes behave like waves
</h2>
The French physicist <em>Louis De Broglie</em> proposed the existence of matter waves, that is to say that <em>all matter has a wave associated with it</em>.
On the other hand, <em>Heisenberg</em> enunciated the uncertainty principle, which postulates that the fact that each particle has a wave associated with it, imposes restrictions on the ability to determine its position and speed at the same time.
These postulations were tested with the double slit experiment (<u>formerly applied to photons</u>) applied to electrons, and the result was: electrons (as well as the other particles different from the photons) are able to behave as waves.
Answer:
Acceleration,
Explanation:
Given that,
Mass of the planet Krypton,
Radius of the planet Krypton,
Value of gravitational constant,
To find,
The acceleration of an object in free fall near the surface of Krypton.
Solution,
The relation for the acceleration of the object is given by the below formula as :
So, the value of acceleration of an object in free fall near the surface of Krypton is