Answer:
Resultant force, R = 10 N
Explanation:
It is given that,
Force acting along +x direction,
Force acting along +y direction,
Both the forces are acting on a point object located at the origin. Let the resultant force of the object is given by R. So,
Here
R = 10 N
So, the resultant force on the object is 10 N. Hence, this is the required solution.
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:
As an object’s temperature increases, the Rate at which it radiates energy increases.
<span>The sparse area surrounding a spiral galaxy is called a
Halo
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<span>they have to be selective permeable to filter out the waste products but allow the nutrients and blood cells to pass through. </span>