Please help me with this

suppose you have a block resting on a horizontal smooth surface. th block with a mass m is attached to a horizontal spring which

Mrrafil [7]

Easy 400 m 600 m 700 m

4 0

3 years ago

Two satellites of equal mass are orbiting the planet Mars. Satellite B is closer to Mars than Satellite A. How does the gravitat

Gnoma [55]

The object in the lower orbit will move faster in its orbit, and will have a shorter orbital period. None of that depends on their masses either.

7 0

4 years ago

How did researchers go about investigating trans-sonic flight

Annette [7]

A wind tunnel is a traditional tool used for trans-sonic flight research. But, during those days, it’s not exact and consistent to use for aircraft flow conditions. The reason is that wind tunnel models will be visible to airflow that produces shock waves. So the U.S and Britain develop an instrument to get data and record flow conditions for a specified speed, these are the specialized research airplanes.

6 0

3 years ago

1. A satellite (mass = 4.44 x 109 kg) travels in orbit around the Earth at a distance of 1.9 x 10'm above

Nuetrik [128]

Answer:

The force of gravitational attraction between the satellite and Earth is $2.587\times 10^{10}$ newtons.

Explanation:

Statement is incorrect. Correct statement is:

A satellite ( $m = 4.44\times 10^{9}\,kg$ ) travels in orbit around the Earth at a distance of $1.9\times 10^{6}\,m$ above Earth's surface. What is the force of gravitational attraction between the satellite and Earth?

The gravitational force experimented by the satellite ( $F$ ), in newtons, is calculated by Newton's Law of Gravitation, whose equation is defined by following formula:

$F = \frac{G\cdot m \cdot M}{R^{2}}$ (1)

Where:

$G$ - Gravitational constant, in cubic meters per kilogram-square second.

$m$ - Mass of the satellite, in kilograms.

$M$ - Mass of the Earth, in kilograms.

$R$ - Distance of the satellite with respect to the center of the Earth, measured in meters.

If we know that $G = 6.674\times 10^{-11}\,\frac{m^{3}}{kg\cdot s^{2}}$ , $m = 4.44\times 10^{9}\,kg$ , $M = 5.972\times 10^{24}\,kg$ and $R = 8.271\times 10^{6}\,m$ , then the force of gravitational attraction between the satellite and Earth is: