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;
Answer:
E. all of these
Explanation:
The designation of a point in space all the points that necessary
- reference point
- a direction
- fundamental units
- a direction
- motion
all are necessary to designate a point in space. Hence option E is correct.
For example in simple harmonic motion we need to specify all the above factors of the object in order to designate the position of the object.
The answer is a) Teres Major Muscle
Answer:
Explanation:
Since the system is in international space station
so here we can say that net force on the system is zero here
so Force by the astronaut on the space station = Force due to space station on boy
so here we know that
mass of boy = 70 kg
acceleration of boy = 
now we know that
now for the space station will be same as above force
