The orbiting speed of the satellite orbiting around the planet Glob is 60.8m/s.
To find the answer, we need to know about the orbital velocity a satellite.
<h3>What's the expression of orbital velocity of a satellite?</h3>
- Mathematically, orbital velocity= √(GM/r)
- G= gravitational constant= 6.67×10^(-11) Nm²/kg², M = mass of sun , r= radius of orbit
<h3>What's the orbital velocity of the satellite in a circular orbit with a radius of 1.45×10⁵ m around the planet Glob of mass 7.88×10¹⁸ kg?</h3>
- Here, M= 7.88×10¹⁸ kg, r= 1.45×10⁵ m
- Orbital velocity of the orbiting satellite = √(6.67×10^(-11)×7.88×10¹⁸/1.45×10⁵)
= 60.8m/s
Thus, we can conclude that the speed of the satellite orbiting the planet Glob is 60.8m/s.
Learn more about the orbital velocity here:
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Through Shannon's Theorem, we can calculate the capacity of the communications channel using the value of its bandwidth and signal-to-noise ratio. The capacity, C, can be expressed as
C = B × log₂(1 + S/N)
where B is the bandwidth of the channel and S/N is its signal-to-noise ratio.
Since the given SN ratio is in decibels, we must first express it as a ratio with no units as
SN (in decibels) = 10 × log (S/N)
30 = 10log(S/N)
log(S/N) = 3
S/N = 10³ = 1000
Now that we have S/N, we can solve for its capacity (in bits per second) as
C = 4000 × log₂(1 + 1000)
C = 39868.91 bps
Thus, the maximum capacity of the channel is 39868 bps or 40 kbps.
Answer: 40 kbps
Answer:
Explanation:
The magnitude of electric field = 8 x 10⁴ V /m
there is a potential difference of 8 x 10⁴ V on a separation of 1 m
so on a separation of .5 m , potential drop or change in potential will be equal to
.5 x 8 x 10⁴
= 4 x 10⁴ V .
The increase in kinetic energy of proton = V X Q
V is potential drop x Q is charge on proton
= 4 x 10⁴ x 1.6 x 10⁻¹⁹
= 6.4 x 10⁻¹⁵ J
potential energy of the proton-field system will be correspondingly decreased by the same amount or by an amount of
- 6.4 x 10⁻¹⁵ J .
Answer:
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Explanation:
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