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Lynna [10]
3 years ago
14

Determine the distance above Earth's surface to a satellite that completes four orbits per day.

Physics
1 answer:
serg [7]3 years ago
8 0

This question involves the concepts of the time period, orbital radius, and gravitational constant.  

The distance of the satellite above the Earth's Surface is "10400 km ".

The theoretical time period of the satellite around the earth can be found using the following formula:

\frac{T^2}{R^3}=\frac{4\pi^2}{GM}\\\\

where,

T = Time Period of Satellite = \frac{1}{frequency} = \frac{1}{4\ orbits/day}\frac{3600*24\ s}{1\ day} = 21600\ s

R = Orbital Radius = ?

G = Gravitational Constant = 6.67 x 10⁻¹¹ N.m²/kg²

M = Mass of Earth = 5.97 x 10²⁴ kg

Therefore,

\frac{(21600\ s)^2}{R^3}=\frac{4\pi^2}{(6.67\ x\ 10^{-11}\ N.m^2/kg^2)(5.97\ x\ 10^{24}\ kg)}\\\\R = \sqrt[3]{\frac{4.66\ x\ 10^8\ s^2}{9.91\ x\ 10^{-14}\ s^2/m^3}} \\R = 1.675\ x\ 10^7\ m = 1.68\ x\ 10^4\ km

Now, this orbital radius is the sum of the radius of the earth (r) and the distance of satellite above earth's (h) surface:

R = r + h

1.68 x 10⁴ km = 0.64 x 10⁴ km + h

h = 1.68 x 10⁴ km - 0.64 x 10⁴ km

<u>h = 1.04 x 10⁴ km = 10400 km</u>

Learn more about the orbital time period here:

brainly.com/question/14494804?referrer=searchResults

The attached picture shows the derivation of the formula for orbital speed.

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Rudik [331]

Answer:

I think is 2.

Explanation:

(The entire range of wavelengths or frequencies of electromagnetic radiation extending from gamma rays to the longest radio waves and including visible light)

7 0
3 years ago
What additional information do you need to prove ∆ABC ≅ ∆DEF by the SAS Postulate?
miv72 [106K]

Answer:

Option A

You need a Angle C congruent to angle F

Explanation:

EX) Side angle Side = sas

6 0
3 years ago
Learning Goal:
enot [183]

Answer:

A. U_0 = \dfrac{\epsilon_0 A V^2}{2d}

B. U_1 = \dfrac{\epsilon_0 A V^2}{6d}

C. U_2 = \dfrac{K\epsilon_0 A V^2}{2d}

Explanation:

The capacitance of a capacitor is its ability to store charges. For parallel-plate capacitors, this ability depends the material between the plates, the common plate area and the plate separation. The relationship is

C=\dfrac{\epsilon A}{d}

C is the capacitance, A is the common plate area, d is the plate separation and \epsilon is the permittivity of the material between the plates.

For air or free space, \epsilon is \epsilon_0 called the permittivity of free space. In general, \epsilon=\epsilon_r \epsilon_0 where \epsilon_r is the relative permittivity or dielectric constant of the material between the plates. It is a factor that determines the strength of the material compared to air. In fact, for air or vacuum, \epsilon_r=1.

The energy stored in a capacitor is the average of the product of its charge and voltage.

U = \dfrac{QV}{2}

Its charge, Q, is related to its capacitance by Q=CV (this is the electrical definition of capacitance, a ratio of the charge to its voltage; the previous formula is the geometric definition). Substituting this in the formula for U,

U = \dfrac{CV^2}{2}

A. Substituting for C in U,

U_0 = \dfrac{\epsilon_0 A V^2}{2d}

B. When the distance is 3d,

U_1 = \dfrac{\epsilon_0 A V^2}{2\times3d}

U_1 = \dfrac{\epsilon_0 A V^2}{6d}

C. When the distance is restored but with a dielectric material of dielectric constant, K, inserted, we have

U_2 = \dfrac{K\epsilon_0 A V^2}{2d}

6 0
3 years ago
You have 1.2kg of (AU)gold (that's quite a bit of money at $1839/ounce). Using a heat source you apply 3096 J to the gold and re
melamori03 [73]

Answer:

129 J/Kg°C

Explanation:

Given :

Mass of gold, m = 1.2kg

Quantity of heat applied, Q = 3096 J

Temperature, t2 = 40°C

Temperature, t1 = 20°C

Change in temperature, dt = (40-20)°C = 20°C

Using the relation :

Q = mcdt

Where, C = specific heat capacity of gold

3096 = 1.2kg * C * 20°C

3096 J = 24kg°C * C

C = 3096 J / 24 kg°C

C = 129 J/Kg°C

7 0
3 years ago
A hand-held video game is powered by batteries. After playing the game for several minutes, a student notices that the game feel
scoundrel [369]
I would say your answer is B- Some of the chemical energy from the batteries is converted into heat energy. 
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