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Dovator [93]
4 years ago
7

The popular GPS devices that people use to find directions while driving use which type of satellite system?

Physics
2 answers:
inessss [21]4 years ago
4 0

The popular GPS devices that people use to find directions while driving use "Global Navigation Satellite System (GNSS)".

<u>Explanation:</u>

The umbrella term for all global satellite tracking systems is GNSS i.e Global Satellite Navigation System. This involves satellite constellations circulating over the surface of the earth and continuous signal transmission that allow users to evaluate their location.

A satellite array of 18–30 medium Earth Orbit (MEO) satellites distributed across several orbital planes typically achieves greater coverage for each network. The specific systems differ, but use > 50 ° orbital inclinations and approximately twelve hours orbital cycles.

hodyreva [135]4 years ago
3 0

GNSS(Global Navigation Satellite system) is used in GPS that are employed to find directions while driving.

Explanation:

  • "Global Navigation Satellite System" (GNSS) relates to a group of satellites giving communication from space that broadcasts "positioning and timing data" to the system receivers.
  • This system includes a variety of satellites that utilize microwave signals that are broadcasted to GPS  to give data on position, vehicle velocity, time and direction.
  • So, a GPS navigating and tracking system can provide both real-time and important exploration data on any sort of journey.
  • GNSS is also known as Augmentation systems.
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A golf ball is dropped from rest from a height of 9.50 m. It hits the pavement, then bounces back up, rising just 9.70 m before
ipn [44]
The ball is in the air for 27.70m because
9.5+9.7=19.2...
9.7-1.2=8.5...
19.2+8.5=27.70...

5 0
4 years ago
Given that the mass of the Earth is 5.972 * 10^24 kg and the radius of the Earth is
Pachacha [2.7K]

Answer:

gₓ = 6.52 m/s²

Explanation:

The value of acceleration due to gravity on the surface of earth is given as:

g = GM/R²   -------------------- equation 1

where,

g = acceleration due to gravity on surface of earth

G = Universal Gravitational Constant

M = Mass of Earth

R = Radius of Earth

Now, for the alien planet:

gₓ = GMₓ/Rₓ²

where,

gₓ = acceleration due to gravity at the surface of alien planet

Mₓ = Mass of Alien Planet = 2.4 M

Rₓ = Radius of Alien Planet = 1.9 R

Therefore,

gₓ = G(2.4 M)/(1.9 R)²

gₓ = 0.66 GM/R²

using equation 1

gₓ = 0.66 g

gₓ = (0.66)(9.81 m/s²)

<u>gₓ = 6.52 m/s²</u>

3 0
3 years ago
With "normal" gravity, we used a potential energy of mgh. Now with the gravity that is more accurate over longer distances we us
Artemon [7]

Answer:

A general solution is \Delta U=mh\frac{GM}{r^{2}}\frac{r}{r+\Delta h} and a particualr case is mgh, it is just to distance around the radius Earth.

Explanation:

We can use a general equation of the potential energy to understand the particular and general case:

The potential energy is defined as U=-\int F\cdot dx, we know that the gravitational force is F=GmM/r^{2}, so we could find the potential energy taking the integral of F.

U=-GmM/r (1)

We can find the particular case, just finding the gravitational potential energy difference:

\Delta U=U_{f}-U_{i}. Here Uf is the potential evaluated in r+Δh and Ui is the potential evaluated in r.

Using (1) we can calculate ΔU.

\Delta U=-\frac{GmM}{r+\Delta h}+\frac{GmM}{r}

Simplifying and combining terms we have a simplified expression.

\Delta U=mh\frac{GM}{r^{2}}\frac{r}{r+\Delta h} (2)

Let's call g=\frac{GM}{r^{2}}. It is the acceleration due to gravity on the Earth's surface, if r is the radius of Earth and M is the mass of the Earth and we can write (2) as ΔU=mgh, but if we have distance grader than r we should use (2), otherwise, we could get incorrect values of potential energy.

I hope i hleps you!

3 0
4 years ago
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Answer:

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Explanation:

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6 0
2 years ago
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