Answer:
a) a geostationary satellite is that it is always at the same point with respect to the planet,
b) f = 2.7777 10⁻⁵ Hz
c) d) w = 1.745 10⁻⁴ rad / s
Explanation:
a) The definition of a geostationary satellite is that it is always at the same point with respect to the planet, that is, its period of revolutions is the same as the period of the planet
- T = 10 h (3600 s / 1h) = 3.6 104 s
b) the period the frequency are related
T = 1 / f
f = 1 / T
f = 1 / 3.6 104
f = 2.7777 10⁻⁵ Hz
c) the distance traveled by the satellite in 1 day
The distance traveled is equal to the length of the circumference
d = 2pi (R + r)
d = 2pi (69 911 103 + 120 106)
d = 1193.24 m
d) the angular velocity is the angle traveled between the time used.
.w = 2pi /t
w = 2pi / 3.6 10⁴
w = 1.745 10⁻⁴ rad / s
how fast is
v = w r
v = 1.75 10-4 (69.911 106 + 120 106)
v = 190017 m / s
Answer:
2C
Explanation:
The equivalent capacitance of a parallel combination of capacitors is the sum of their capacitance.
So, if the capacitance of each capacitor is half the previous one, we have a geometric series with first term = C and rate = 0.5.
Using the formula for the sum of the infinite terms of a geometric series, we have:
Sum = First term / (1 - rate)
Sum = C / (1 - 0.5)
Sum = C / 0.5 = 2C
So the equivalent capacitance of this parallel connection is 2C.
5.972 × 10^24 kg
it is the weight of earth
hope it is helpful to you
Answer:
R = 5.28 103 km
Explanation:
The definition of density is
ρ = m / V
V = m /ρ
Where m is the mass and V the volume of the body
The volume of a sphere is
V = 4/3 π r³
Let's replace
4/3 π r³ = m / ρ
R =∛ ¾ m / ρ π
The mass of the planet is
M = 5.5 Me
R = ∛ ¾ 5.5 Me /ρ π
Let's reduce the density to SI units
ρ = 1.76 g / cm³ (1 kg / 10³ g) (10² cm / 1 m)³
ρ = 1.76 10³ kg / m³
Let's calculate
R = ∛ ¾ 5.5 5.97 10²⁴ / (1.76 10³ pi)
R = ∛ 0.14723 10²¹
R = 0.528 10⁷ m
R = 0.528 104 km
R = 5.28 103 km
