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IceJOKER [234]
3 years ago
6

Communications satellites are placed in a circular orbit where they stay directly over a fixed point on the equator as the earth

rotates. these are called geosynchronous orbits. the radius of the earth is 6.37 * 106 m, and the altitude of a geosynchronous orbit is 3.58 * 107 m 1â22,000 miles2. what are (a) the speed and (b) the magnitude of the acceleration of a satellite in a geosynchronous orbit?

Physics
1 answer:
zalisa [80]3 years ago
7 0
Refer to the diagram shown below.

Given:
R = 6.37 x 10⁶ m, the radius of the earth
h = 3.58 x 10⁷ m, the height of the satellite above the earth's surface.
Therefore
R + h = 4.217 x 10⁷ m

In geosynchronous orbit, the period of rotation is 1 day.
Therefore the period is
T = (24 h)*(60 min/h)*(60 s/min) = 86400 s

The angular velocity is
ω = (2π rad)/(86400 s) = 7.2722 x 10⁻⁵ rad/s

Part (a)
The tangential speed is
v = (R+h)*ω
   = (4.217 x 10⁷ m)*(7.2722 x 10⁻⁵ rad/s) 
   = 3066.7 m/s
   = 3.067 km/s

Part (b)
The centripetal acceleration is
a = v²/(R+h)
   = (3066.7 m/s)²/(4.217 x 10⁷ m)
   = 0.223 m/s²

Answers:
(a) The speed is 3.067 km/s
(b) The acceleration is 0.223 m/s²

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Zinaida [17]
<h2>Let us find the efficiency : Ans = 0.6</h2>

Explanation:

we know :

efficiency = output/input

We also know that :

output = m x g x h

where :

m = mass of body

g = acceleration due to gravity

h = height of body from floor

Thus, output = 0.6 x 10 x 1.2 = 7.2J

Similarly ,input = 0.6 x 10 x 2 = 12J

Thus efficiency = 7.2/12 = 0.6

4 0
3 years ago
Two fully charged cylindrical capacitors are connected to two identical batteries. The capacitors are identical except that the
Leni [432]

Answer:

Part(a):  The relative capacitance is \dfrac{C_{A}}{C_{B}} = 0.33

Part(b): The relative energy stored is \dfrac{U_{A}}{U_{B}} = 0.33

Part(c): The relative charge stored is \dfrac{Q_{A}}{Q_{B}} = 0.33

Explanation:

We know the capacitance (C) of a capacitor having charge (Q) and subjected to a potential difference of (V) is given by

C = \dfrac{Q}{V}

Also, the energy (U) stored by a capacitor can be written as

U = \dfrac{1}{2}C~V^{2}

Let us assume that the inner radius of the Capacitor B, as shown in the figure, be \textbf{r_{i}^{B}}\bf{r_{i}^{B}}, the outer radius be \bf{r_{o}^{B}}, the inner radius of Capacitor A be \bf{r_{i}^{A}} and the outer radius be \bf{r_{o}^{B}}.

Given in the problem,

&& r_{o}^{B} = 2~r_{B}^{i}\\&& r_{o}^{A} = 4~r_{B}^{i}\\&& and~r_{i}^{B} = 4~r_{o}^{B} = 8~r_{B}^{i}

Now, the capacitance (C) of a cylindrical capacitor is given by,

\bf{C = \dfrac{2~\pi~\epsilon_{0}~L}{ln(\dfrac{r_{o}}{r_{i}})}}

where \epsilon_{o} is the permittivity of the free space, L is the length of the cylindrical capacitor.

Part(a):

The capacitance of capacitor A,

C_{A} = \dfrac{2~\pi~\epsilon_{0}L}{ln(\dfrac{r_{o}^{A}}{r_{i}^{A}})} = \dfrac{2~\pi~\epsilon_{0}L}{ln(\dfrac{8~r_{i}^{B}}{r_{i}^{B}})} = \dfrac{2~\pi~\epsilon_{0}L}{ln(8)}

and the capacitance of capacitor B,

C_{B} = \dfrac{2~\pi~\epsilon_{0}L}{ln(\dfrac{r_{o}^{B}}{r_{i}^{B}})} = \dfrac{2~\pi~\epsilon_{0}L}{ln(\dfrac{2~r_{i}^{B}}{r_{i}^{B}})} = \dfrac{2~\pi~\epsilon_{0}L}{ln(2)}

giving the relative capacitance of each capacitor to be

\dfrac{C_{A}}{C_{B}} = \dfrac{ln(2)}{ln(8)} = \dfrac{ln(2)}{3~\ln(2)} = \dfrac{1}{3} = 0.33

Part(b):

Energy stored by capacitor A,

U_{A} = \dfrac{1}{2}~C_{A}~V^{2}

Energy stored by capacitor B,

U_{B} = \dfrac{1}{2}~C_{B}~V^{2}

giving the relative energy stored by each capacitor to be

\dfrac{U_{A}}{U_{B}} = \dfrac{C_{A}}{C_{B}} = 0.33

Part(c):

The charge stored by capacitor A,

Q_{A} = C_{A}~V

The charge stored by capacitor B,

Q_{B} = C_{B}~V

giving the relative charge stored by each capacitor to be

\dfrac{Q_{A}}{Q_{B}} =  \dfrac{C_{A}}{C_{B}} = 0.33

8 0
3 years ago
Four forces act on a hot-air balloon, as shown
dolphi86 [110]

The magnitude of the resultant force on the balloon is 374.13 N.

The given forces from the image;

  • <em>Upward force = 514 N</em>
  • <em>Downward force = 267 N</em>
  • <em>Eastward force = 678 N</em>
  • <em>Westward force = 397 N</em>

The net vertical force on the balloon is calculated as follows;

F_y = 514 \ N \ \ - \ \ 267 \ N\\\\F_y = 247 \ N

The net horizontal force on the balloon is calculated as follows;

F_x = 678 \ N \ - \ 397 \ N\\\\F_x = 281 \ N

The magnitude of the resultant force on the balloon is calculated as follows;

F = \sqrt{F_y^2 + F_x^2} \\\\F = \sqrt{(247)^2 + (281)^2} \\\\F= 374.13 \ N

Thus, the magnitude of the resultant force on the balloon is 374.13 N.

Learn more here:brainly.com/question/4404327

5 0
2 years ago
If the mass of a material is 50 grams and the volume of the material is 12 cm^3, what would the density of the material be?
AlekseyPX
The density of the material would be 
25/6 grams per cm^3.
to obtain the result above this is what we do:
density is calculated as: (the mass of the given material or object) / volume of the material 
which leads us to 50grams /12cm^3
 
4 0
3 years ago
Temperature is a measure of _ of the particles in an object
djverab [1.8K]

energy is the correct answer to fill the blank bb :)

3 0
3 years ago
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