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.
Answer:
The Total Mechanical Energy
As already mentioned, the mechanical energy of an object can be the result of its motion (i.e., kinetic energy) and/or the result of its stored energy of position (i.e., potential energy). The total amount of mechanical energy is merely the sum of the potential energy and the kinetic energy.
Explanation:
Explanation:
Using Ohm's Law and a bit of substitution, we can use voltage divided by current to solve for resistance. Doing that, we'll get 6 Ohm.
Answer:
a) Revolutions per minute = 2.33
b) Centripetal acceleration = 11649.44 m/s²
Explanation:
a) Angular velocity is the ratio of linear velocity and radius.
Here linear velocity = 72 m/s
Radius, r = 0.89 x 0. 5 = 0.445 m
Angular velocity
Frequency
Revolutions per minute = 2.33
b) Centripetal acceleration
Here linear velocity = 72 m/s
Radius, r = 0.445 m
Substituting
Centripetal acceleration = 11649.44m/s²
