The Greek letter (delta) indicates a(n)
2 answers:
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Answer:
Mass of the planet = 1.48 × 10²⁵ Kg
Mass of the star = 5.09 × 10³⁰ kg
Explanation:
Given;
Diameter = 1.8 × 10⁷ m
Therefore,
Radius = =
or
Radius of the planet = 0.9 × 10⁷ m
Rotation period = 22.3 hours
Radius of star = 2.2 × 10¹¹ m
Orbit period = 407 earth days = 407 × 24 × 60 × 60 seconds = 35164800 s
free-fall acceleration = 12.2 m/s²
Now,
we have the relation
g =
g is the free fall acceleration
G is the gravitational force constant
M is the mass of the planet
on substituting the respective values, we get
12.2 =
or
M = 1.48 × 10²⁵ Kg
From the Kepler's law we have
T² =
on substituting the respective values, we get
35164800² =
or
= 5.09 × 10³⁰ kg
Answer:
The answer is E) 6.7m at an angle 63° north of east
Explanation:
We begin by writing out our given vectors A,B and C in their respective rectangular Cartesian coordinate form. which are given as A=8i, B=6j and C= -5j (negative because it points towards west which is towards the left which is taken as the negative direction).Now we simply add all three to obtain our final vector given by,
In the last step we just added the like components. Now we will have to calculate the magnitude and angle of V(with respect of the east direction), which is given by,
the angle alpha is always calculated with respect to the horizontal positive axis, and since both components of the vector are positive the vector lies in the first quadrant. Hence the 63° angle north of east is justified. The term north of east means the angle formed in moving from east towards the north that is from the positive horizontal axis to the positive vertical axis.