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Umnica [9.8K]
3 years ago
12

A(n) _____ satellite system employs many satellites, each in an orbit at an altitude of less than 1,000 miles.

Physics
2 answers:
Stells [14]3 years ago
6 0

Answer:A(n) low Earth Orbit (LEO) satellite system employs many satellites, each in an orbit at an altitude of less than 1,000 miles.

Explanation:A low Earth orbit (LEO) is an Earth-centered orbit with an altitude of 2,000 km (1,200 mi) or less (approximately one-third of the radius of Earth), or with at least 11.25 periods per day (an orbital period of 128 minutes or less) and an eccentricity less than 0.25. Most of the manmade objects in outer space are in LEO.

There is a large variety of other sources that define LEO in terms of altitude. The altitude of an object in an elliptic orbit can vary significantly along the orbit. Even for circular orbits, the altitude above ground can vary by as much as 30 km (19 mi) (especially for polar orbits) due to the oblateness of Earth's spheroid figure and local topography

vlabodo [156]3 years ago
4 0

Answer:

The correct answer is a Low earth orbit.

Explanation:

A low earth orbit can be understood as an earth orbit with an altitude of 1,000 miles or less. It is a satellite sustem that employs many satelites, in fact, most man-made objects that are currently in outer-space are part of this low earth orbit. (LEO).

The most famous LEO satellite system is the one from planet earth. Almost every space flight that human beings have ever done are done in LEO, and every spacial station is located in this zone.

In conclusion, A low earth orbit satellite system employs many satellites, each in an orbit at an altitude of less than 1,000 miles.

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I think the answer is A

Explanation:

the bus has more kg than the car so. the bus need to stop first

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2 years ago
a particle with a charge of 5.5 x 10^-8 c is 3.5 cm from a particle with a charge of -2.3 x10^-8 c. the potential energy of this
Yuri [45]

Answer:

-32.5 * 10^-5 J

Explanation:

The potential energy of this system of charges is;

Ue = kq1q2/r

Where;

k is the Coulumb's constant

q1 and q2 are the magnitudes of the charges

r is the distance of separation between the charges

Substituting values;

Ue = 9.0×10^9 N⋅m2/C2 * 5.5 x 10^-8 C *( -2.3 x10^-8) C/(3.5 * 10^-2)

Ue= -32.5 * 10^-5 J

4 0
3 years ago
Which of the following is a personal decision that involves science?
malfutka [58]
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3 years ago
A kite 40 ft above the ground moves horizontally at a constant speed of 10 ft/s, with a child, holding the ball of kite string,
Lorico [155]

Answer:

 v = 27.28 m /s, θ = 63.9º

Explanation:

For this exercise we can approximate the movement to a projectile launch, let's analyze the situation.

* We must find the horizontal speed, for this we will find the descent time and the horizontal distance

* We look for the vertical speed

At the highest point the speed is horizontal

Let's find the time it takes for the kite to reach the ground

             y = y₀ + v_{oy} t - ½ g t²

             0 =y₀ + 0 -1/2 gt²

             t = \sqrt{ \frac{2y_o}{g} }

             t = √(2 40/32)

             t = 2.5 s

to find the horizontal velocity we must know the horizontal distance, let's use trigonometry

          sin θ = y / l

          θ = sin⁻¹1 y / l

          θ = sin⁻¹ 40/50

          θ = 53.1º

therefore the horizontal distance is

          x = l cos 53.1

          x = 50cos 53.1

          x = 30 m

let's use the equation

          x = v₀ₓ t

          v₀ₓ = x / t

          v₀ₓ = 30 / 2.5

          v₀ₓ = 12 m / s

we look for the vertical component of the velocity

          v_y = v_{oy} - g t

          v_y = 0 - g t

          v_y = - 9.8 2.5

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the negative sign indicates that the speed is directed downwards, because it is the arrival point, as they indicate that there is no friction, the exit speed is the same, worse with the opposite sign

We already have the two components of the velocity, let's use the Pythagorean theorem to find the modulus

          v = \sqrt{v_x^2 + v_y^2}

          v = \sqrt{12^2 + 24.5^2}

          v = 27.28 m /s

we use trigonometry for the angle

          tan θ = v_y / vₓ

          θ = tan⁻¹ v_y / vₓ

          θ = tan⁻¹ 24.5 / 12

          θ = 63.9º

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vodomira [7]
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