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OLEGan [10]
10 months ago
13

Six artificial satellites complete one circular orbit around a space station in the same amount of time. Each satellite has mass

m and radius of orbit l. The satellites fire rockets that provide the force needed to maintain a circular orbit around the space station. The gravitational force is negligible.
Physics
1 answer:
REY [17]10 months ago
4 0

The satellites launch rockets to generate the force required to keep an orbit all around space station circular. The continuous centripetal force is maintained by the centripetal force.

<h3>What is a good illustration of gravity?</h3>

The energy holding the gases inside the sun. the power behind a ball's descent after being thrown into the air. the force that makes an automobile coast downward even when the gas is not depressed.

<h3>What makes anything gravitational?</h3>

Our term gravity and more specific derivation gravitation are derived from a Latin word gravity, from gravis, which itself is derived from a much older root word that is considered to have existed due to multiple cognates in closely related languages.

To know more about Gravitational visit:

brainly.com/question/3009841

#SPJ4

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A gas‑forming reaction produces 1.90 m 3 1.90 m3 of gas against a constant pressure of 179.0 kPa. 179.0 kPa. Calculate the work
goldfiish [28.3K]

Answer:

The work done is 3.4 × 10⁵ J.

Explanation:

Given:

Pressure of the gas produced (P) = 179 kPa

Volume of the gas produced (ΔV) = 1.90 m³

We need to find the work done in joules. For that, we don't need any conversion as the units are already in SI units which will give the result in Joules only.

Now, let us verify our results by using conversion factors and without using them.

Using conversion factors:

1 m³ = 1000 L

So, 1.90 m³ = 1.90 m³ × 1000 \frac{L}{m^3} = 1900 L

Also, 1 atm = 101.325 kPa

So, 179 kPa = 179 kPa × \frac{1\ atm}{101.325\ kPa} = 1.767 atm

Now, work done in a constant pressure process is given as:

Work = Pressure × Volume change

Work = P × ΔV

Work = 1.767 atm × 1900 L

Work = 3.36 × 10³ atm-L

Now, again using the energy conversion for work.

1 atm-L = 101.325 J

So, 3.36 × 10³ atm-L = 3.36 × 10³ atm-L × \frac{101.325\ J}{1\ atm-L} = 3.4 × 10⁵ J

Therefore, the work done is 3.4 × 10⁵ J.

Now, let us verify the above result without any conversion.

Work = P × ΔV = 179 × 1000 × 1.90 = 3.4 × 10⁵ J.

Therefore, the work done is same by both ways.

Hence the work done is 3.4 × 10⁵ J.

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The vector values are 4.0 km due East and 3.0 km due north the resultant vector is 5.0 km long and 37 north of East this values
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Answer:

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Explanation:

Given;

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