A satellite has a mass of 6463 kg and is in a circular orbit 4.82 × 105 m above the surface of a planet. The period of the orbit
1 answer:
Answer:
The weight of the planet is 29083.5 N .
Explanation:
mass of satellite, m = 6463 kg
height of orbit, h = 4.82 x 10^5 m
period, T = 2 h
radius of planet, R = 4.29 x 10^6 m
Let the acceleration due to gravity at the planet is g.
The weight of the satellite at the surface of the planet is
W = m g = 6463 x 4.5 = 29083.5 N
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Answer:
The final displacement from the origin is <u>1.6</u> km to the <u>NE</u>
Explanation:
The directions in which the delivery truck travels are;
1) 2.8 km North = 2.8·, in vector form
2) 1.0 km East = 1.0·, in vector form
3) 1.6 km South = -1.6·, in vector form
Therefore, to find the final displacement, Δx, of the delivery truck, we add the individual displacements as follows;
Final displacement, Δd = 2.8· + 1.0·
+(-1.6·
) = 1.2·
+ 1.0·
Final displacement, = 1.0· + 1.2·
Where;
Δx = The displacement in the x-direction = 1.0·
Δy = The displacement in the y-direction = 1.2·
The magnitude of the resultant displacement vector is given as follows
= √((Δx)² + (Δy)²) = √(1² + 1.2²) ≈ 1.6 (To the nearest tenth)
The magnitude of the resultant displacement vector ≈ 1.6 km
The direction of the resultant vector is positive for both the east and north direction, therefore, the direction of the resultant vector = NE
Therefore, the resultant displacement of the delivery truck is approximately 1.6 km, NE from the origin.