Multiply by (1000 meters / 1 km).
Then multiply by (1 hour / 3600 seconds).
Both of those fractions are equal to ' 1 ', because the top
and bottom numbers are equal, so the multiplications
won't change the VALUE of the 72 km/hr. They'll only
change the units.
(72 km/hour) · (1000 meters / 1 km) · (1 hour / 3600 seconds)
= (72 · 1000 / 3600) (km·meter·hour / hour·km·second)
= 20 meter/second
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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:
Option B
Explanation:
Option A is the wrong answer because the horizontal vector is in the opposite direction.
Option C is the wrong answer as the horizontal vector is in the opposite direction and all the vectors are connected head to tail [of the arrows] [Triangle law of vector addition]
Option D is the wrong answer as the horizontal vector is in the opposite direction.
