Answer:
(a). The path length is 3.09 m at 30°.
(b). The path length is 188.4 m at 30 rad.
(c). The path length is 1111.5 m at 30 rev.
Explanation:
Given that,
Radius = 5.9 m
(a). Angle 
We need to calculate the angle in radian
We need to calculate the path length
Using formula of path length
(b). Angle = 30 rad
We need to calculate the path length
(c). Angle = 30 rev
We need to calculate the angle in rad
We need to calculate the path length
Hence, (a). The path length is 3.09 m at 30°.
(b). The path length is 188.4 m at 30 rad.
(c). The path length is 1111.5 m at 30 rev.
Answer:
Increasing the mass and decreasing the distance between the two objects.
Explanation:
An increase in mass will cause them to have a stronger pull or gravity. A decrease of distance will make it easier for the objects to fall into each other because they would be further into the other objects area of influence.
D) decreasing the temperature lowers the average kinetic energy of the reactants.
Medicine to a patient. That should be calculated based on weight, strength/dosage and possibly other factors
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.
