Answer:
23376 days
Explanation:
The problem can be solved using Kepler's third law of planetary motion which states that the square of the period T of a planet round the sun is directly proportional to the cube of its mean distance R from the sun.

where k is a constant.
From equation (1) we can deduce that the ratio of the square of the period of a planet to the cube of its mean distance from the sun is a constant.

Let the orbital period of the earth be
and its mean distance of from the sun be
.
Also let the orbital period of the planet be
and its mean distance from the sun be
.
Equation (2) therefore implies the following;

We make the period of the planet
the subject of formula as follows;

But recall that from the problem stated, the mean distance of the planet from the sun is 16 times that of the earth, so therefore

Substituting equation (5) into (4), we obtain the following;

cancels out and we are left with the following;

Recall that the orbital period of the earth is about 365.25 days, hence;

Explanation:
physical exertion.
lack of physical activity.
lack of sleep.
being overweight or obese.
periods of emotional stress.
boredom.
grief.
taking certain medications, such as antidepressants or sedatives.
Answer:
Explanation:
A grounded wire is sometimes strung along the tops of the towers to provide lightning protection.
In areas where the neutral is grounded or earthed, it is essential to endure that the neutral and the live or hot wires are not confused for each other.
When this happens, the fuses on the transformer will not operate unless the fault is very close to the transformer. The fuses in the consumer's intake box, will not operate.
The value of parameter C for the function in the figure is 2.
<h3>What is amplitude of a wave?</h3>
The amplitude of a wave is the maximum displacement of the wave. It can also be described at the maximum upward displacement of a wave curve.
f(x) = Acos(x - C)
where;
- A is amplitude of the wave
- C is phase difference of the wave
<h3>What is angular frequency of a wave?</h3>
Angular frequency is the angular displacement of any element of the wave per unit time.
From the blue colored graph; at y = 1, x = -2 cm
1 = cos(2 - C)
(2 - C) = cos^(1)
(2 - C) = 0
C = 2
Thus, the value of parameter C for the function in the figure is 2.
Learn more about phase angle here: brainly.com/question/16222725
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