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lidiya [134]
2 years ago
3

Wich statements accurately describes accurately describe satellite motion )

Physics
2 answers:
Ann [662]2 years ago
8 0
Here are some examples of statements that accurately describe satellite motion:
Gravity is the only force acting on a satellite.
Circular orbits result from the interaction between gravity and inertia.
Gravity provides the centripetal force for satellites.Satellites are in free fall around Earth or other central objects.
Hope this helps!
Leya [2.2K]2 years ago
4 0

Hello. This question is incomplete. The full question

"Which statements accurately describe satellite motion? Check all that apply: Gravity is the only force acting on a satellite.  Inertia is the only cause of a satellite’s circular motion.  Air resistance prevents satellites from staying in orbit for very long.  Circular orbits result from the interaction between gravity and inertia.  Gravity provides the centripetal force for satellites.  Satellites are in free fall around Earth or other central objects."

Answer:

  • Gravity is the only force acting on a satellite.
  • Circular orbits result from the interaction between gravity and inertia.
  • Gravity provides the centripetal force for satellites.
  • Satellites are in free fall around Earth or other central objects.

Explanation:

Satellites are bodies that orbit around a celestial body. They can be classified as natural (celestial stars that orbit around a planet, for example, the Moon) and artificial (objects made by man and placed in orbit of a celestial body). There are several artificial satellites around our planet and with different functions, such as communication, meteorological, military and astronomical satellites.

The movement of satellites around a planet obeys Kepler's Laws and Universal Gravitation. About this movement we can say that:

  • Gravity is the only force acting on a satellite.
  • Circular orbits result from the interaction between gravity and inertia.
  • Gravity provides the centripetal force for satellites.
  • Satellites are in free fall around Earth or other central objects.
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Answer:

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Hence, if you use a mass of 0.40kg the total mechanical energy is the same as the obtained with a mas 0.20kg

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A 5.00μF parallel-plate capacitor is connected to a 12.0 V battery. After the capacitor is fully charged, the battery is disconn
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(a) 12.0 V

In this problem, the capacitor is connected to the 12.0 V, until it is fully charged. Considering the capacity of the capacitor, C=5.00 \mu F, the charged stored on the capacitor at the end of the process is

Q=CV=(5.00 \mu F)(12.0 V)=60 \mu C

When the battery is disconnected, the charge on the capacitor remains unchanged. But the capacitance, C, also remains unchanged, since it only depends on the properties of the capacitor (area and distance between the plates), which do not change. Therefore, given the relationship

V=\frac{Q}{C}

and since neither Q nor C change, the voltage V remains the same, 12.0 V.

(b) (i) 24.0 V

In this case, the plate separation is doubled. Let's remind the formula for the capacitance of a parallel-plate capacitor:

C=\frac{\epsilon_0 \epsilon_r A}{d}

where:

\epsilon_0 is the permittivity of free space

\epsilon_r is the relative permittivity of the material inside the capacitor

A is the area of the plates

d is the separation between the plates

As we said, in this case the plate separation is doubled: d'=2d. This means that the capacitance is halved: C'=\frac{C}{2}. The new voltage across the plate is given by

V'=\frac{Q}{C'}

and since Q (the charge) does not change (the capacitor is now isolated, so the charge cannot flow anywhere), the new voltage is

V'=\frac{Q}{C'}=\frac{Q}{C/2}=2 \frac{Q}{C}=2V

So, the new voltage is

V'=2 (12.0 V)=24.0 V

(c) (ii) 3.0 V

The area of each plate of the capacitor is given by:

A=\pi r^2

where r is the radius of the plate. In this case, the radius is doubled: r'=2r. Therefore, the new area will be

A'=\pi (2r)^2 = 4 \pi r^2 = 4A

While the separation between the plate was unchanged (d); so, the new capacitance will be

C'=\frac{\epsilon_0 \epsilon_r A'}{d}=4\frac{\epsilon_0 \epsilon_r A}{d}=4C

So, the capacitance has increased by a factor 4; therefore, the new voltage is

V'=\frac{Q}{C'}=\frac{Q}{4C}=\frac{1}{4} \frac{Q}{C}=\frac{V}{4}

which means

V'=\frac{12.0 V}{4}=3.0 V

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