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Sergeeva-Olga [200]
3 years ago
10

An artificial satellite circles Earth in a circular orbit at a location where the acceleration due to gravity is 9.00 m/s2. Dete

rmine the orbital period of the satellite.
Physics
1 answer:
Ksju [112]3 years ago
6 0
 <span>g = GMe/Re^2, where Re = Radius of earth (6360km), G = 6.67x10^-11 Nm^2/kg^2, and Me = Mass of earth. On the earth's surface, g = 9.81 m/s^2, so the radius of your orbit is:


R = Re * sqrt (9.81 m/s^2 / 9.00 m/s^2) = 6640km 

here, the speed of the satellite is:

v = sqrt(R*9.00m/s^2) = 7730 m/s 

  the time it would take the satellite to complete one full rotation is:

T = 2*pi*R/v = 5397 s * 1h/3600s = 1.50 h 

Hope it help i know it's long and may be confusing but if you have any more questions regarding this topic just hmu!  :)</span>
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vector u has a magnitude of 20 and direction of 0°.vector v has amagnitude of 40and a direction of 60°.find the magnitude and di
pantera1 [17]

Addition of vectors:

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has a magnitude of 20 and a direction of 0º with respect to the horizontal, vector v has a magnitude of 40 and a direction of 60º with respect to the horizontal.

a) Find the magnitude and direction of the resultant to the nearest whole

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Vector Sum:

The resultant of two vectors is simply the vector sum of the vectors. There are a handful of ways to present the resultant factor; the notation that shows the vector magnitude and direction is called the polar vector notation. An example of a vector presented in polar vector notation is

a∠θ where a is the magnitude and θis the angle that the vector makes with the horizontal axis.

Answer and Explanation:

Let's first present the vectors in rectangular vector notation.

For the vector →u of magnitude 20 and direction 0∘ to the horizontal axis, the vector is →u=^i20.

For the vector →v

of magnitude 40 and direction 60∘ to the horizontal axis, the vector is →v=^i40cos60∘+^j40sin60∘.

The resultant vector →w is the vector sum of the vectors, i.e.

→w=→u+→v

=^i20+^i40cos60∘+^j40sin60∘

=^i(20+40cos60∘)+^j(40sin60∘)

=^i40+^j20√3

For a vector ^ix+^jy, the magnitude of the vector is √x2+y2 and the direction above the horizontal axis is θ=tan−1(yx).

Let's use the formulas:

|→w|∠θ=√(40)2+(20√3)2∠tan−1(20√340)≈52.9∠40.9∘

The magnitude of the vector is about 53 units in the direction 41-degrees above the horizontal axis.

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