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Anton [14]
3 years ago
7

An Earth satellite needs to have its orbit changed so the new orbit will be twice as far from the center of Earth as the origina

l orbit. The new orbital period will be twice as long as the original period. O true O false
Physics
1 answer:
horsena [70]3 years ago
5 0

Answer:

False.

Explanation:

From Kepler's Third Law of plenetary motion, we know that:

<em>"The square of the orbital period of a planet is directly proportional to the cube of the semi-major axis of its orbit."</em>

Or, as expressed in mathematical terms:

\frac{a^3}{T^2}=constant, where <em>a</em> is the semi-major axis of the orbit (the distance from the center), and <em>T </em>is the orbital period of the satellite.

From this expression we can clearly see that if the orbit's semi-major axis is doubled, orbital period will be \sqrt{8} times longer to compensate the variation.

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A 5.00-V battery charges the parallel plates in a capacitor, with a plate area of 865 mm2 and an air-filled separation of 3.00 m
Westkost [7]

Answer:

W = 3.21x10⁻¹¹ J

Explanation:

The work required to separate the plates can be calculated using the following equation:

W = U_{2} - U_{1} = \frac{1}{2}(C_{2}V_{2}^{2} - C_{1}V_{1}^{2})

<u>Where</u>:

U₂: is the final stored energy

U₁: is the initial stored energy

C₂: is the final capacitance

C₁: is the initial capacitance

V₁: is the initial potential difference = 5.00 V

V₂: is the final potential difference

The initial and final capacitance is:

C_{1} = \epsilon_{0}*\frac{A}{d_{1}}

<u>Where</u>:

ε₀: is the vacuum permittivity = 8.85x10⁻¹² C²/(N*m²)

d: is the initial distance = 3.00 mm = 3.00x10⁻³ m    

A: is the plate area = 865 mm² =  8.65x10⁻⁴ m²

C_{1} = \epsilon_{0}*\frac{A}{d_{1}} = 8.85 \cdot 10^{-12} C^{2}/(N*m^{2})*\frac{8.65 \cdot 10^{-4} m^{2}}{3.00 \cdot 10^{-3} m} = 2.55 \cdot 10^{-12} F      

Similarly, C₂ is:

C_{2} = \epsilon_{0}*\frac{A}{d_{2}} = 8.85 \cdot 10^{-12} C^{2}/(N*m^{2})*\frac{8.65 \cdot 10^{-4} m^{2}}{3.00 + 3.00 \cdot 10^{-3} m} = 1.28 \cdot 10^{-12} F

Now, V₂ can be calculated by finding the initial charge (q₁):

q_{1} = C_{1}V_{1} = 2.55 \cdot 10^{-12} F*5.00 V = 1.28 \cdot 10^{-11} C

Since, q₁ is equal to q₂, V₂ is:

V_{2} = \frac{q_{2}}{C_{2}} = \frac{1.28 \cdot 10^{-11} C}{1.28 \cdot 10^{-12} F} = 10 V

Finally, we can find the work:

W = \frac{1}{2}(C_{2}V_{2}^{2} - C_{1}V_{1}^{2}) = \frac{1}{2}(1.28 \cdot 10^{-12} F*(10 V)^{2} - 2.55 \cdot 10^{-12} F(5.00 V)^{2}) = 3.21 \cdot 10^{-11} J

Therefore, the work required to separate the plates is 3.21x10⁻¹¹ J.

I hope it helps you!

4 0
3 years ago
Which of the following is the most effective relaxation technique?
nevsk [136]

Answer:

A

Explanation:

It varies from person to person.

3 0
2 years ago
Read 2 more answers
1-Calculate Req
ArbitrLikvidat [17]

Answer:

1. 21.66 Ohms

2.  3.38 A

3. 6.7 V

Explanation:

1. Req = 6+2 = 8 Ohms (2 and 6 are in a series circuit)

  Req = 1/8 +1/4 = 3/8 = 8/3 = 2.66 Ohms (8 and 4 are parallel, so we will add them using this equation)

  Req = 2.66 + 1 + 9 + 3 + 6 = 21.66 Ohms

2. I = V/R = 9/2.66 = 3.38 A (In a series circuit, the current is the same across the resistors, so we will add them and divided them by 9 volts)

3. V = IR = 3.38 x 2 = 6.7 V (In a series circuit, the voltage is different, so each resistor will have a different voltage.)

I hope this helps.  I am not an expert in physics but its ok :)

<u><em>Note: If the answer benefited u, mark me as the brainliest answer if u can, thx.</em></u>

3 0
2 years ago
Make a rough estimate of the number of quanta emitted in one second by a 100 W light bulb. Assume that the typical wavelength em
mixas84 [53]

Answer:

#_photon = 5 10²⁰ photons / s

Explanation:

For this exercise let's calculate the energy of a single quantum of energy, use Planck's law

         E = h f

         c= λ f

         E = h c / λ

          λ= 1000 nm (1 m / 109 nm) = 1000 10⁻⁹ m

Let's calculate

          E₀ = 6.6310⁻³⁴ 3 10⁸/1000 10⁻⁹

          E₀ = 19.89 10⁻²⁰ J

This is the energy emitted by a photon let's use a proportions rule to find the number emitted in P = 100 w

                #_photon = P / E₀

               #_photon = 100 / 19.89 10⁻²⁰

              #_photon = 5 10²⁰ photons / s

6 0
3 years ago
How many suns do you think are in our galaxy? Explain your answer PLS HELP ASP BEST GETS BRAINLIEST​
dexar [7]

Answer: There is only one Sun in the galaxy … that is the thing that rises in the morning and sets at night. However, there is a use of “sun” to signify any old star … nobody knows exactly there might be trillions out there

Explanation:

3 0
3 years ago
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