I believe it is lithosphere
The water pressure on the first floor must be 455 PSI in order to push the water to the 13th floor at the given pressure.
The given parameters;
- <em>Pressure on the 13 th floor, P₁ = 35 PSI</em>
- <em>Distance between each floor, d = 10 ft</em>
The vertical pressure of the water is calculated as follows;
The vertical height of the first floor from the 13th floor = 130 ft
The vertical height of the 13 ft floor = 10 ft
Thus, the water pressure on the first floor must be 455 PSI in order to push the water to the 13th floor at the given pressure.
Learn more about vertical height and pressure here: brainly.com/question/15691554
Answer:
Tp/Te = 2
Therefore, the orbital period of the planet is twice that of the earth's orbital period.
Explanation:
The orbital period of a planet around a star can be expressed mathematically as;
T = 2π√(r^3)/(Gm)
Where;
r = radius of orbit
G = gravitational constant
m = mass of the star
Given;
Let R represent radius of earth orbit and r the radius of planet orbit,
Let M represent the mass of sun and m the mass of the star.
r = 4R
m = 16M
For earth;
Te = 2π√(R^3)/(GM)
For planet;
Tp = 2π√(r^3)/(Gm)
Substituting the given values;
Tp = 2π√((4R)^3)/(16GM) = 2π√(64R^3)/(16GM)
Tp = 2π√(4R^3)/(GM)
Tp = 2 × 2π√(R^3)/(GM)
So,
Tp/Te = (2 × 2π√(R^3)/(GM))/( 2π√(R^3)/(GM))
Tp/Te = 2
Therefore, the orbital period of the planet is twice that of the earth's orbital period.
Since there is no decimal point in the number given above, the counting for the number of the significant figures will start from the left. Then, the first zero from the left is insignificant. Therefore, in this number there are 6 significant figures.
