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bonufazy [111]
3 years ago
7

The distance between Pluto and the Sun is 39.1 times more than the distance between the Sun and Earth. Calculate the time taken

by Pluto to orbit the Sun in Earth days.
Physics
1 answer:
german3 years ago
7 0

Answer:

Explanation:

Given

Distance between Pluto and sun is 39.1 times more than the distance between earth and sun

According to Kepler's Law

T^2=kR^3

where k=constant

T=time period

R=Radius of orbit

Suppose R_1 is the radius of orbit of earth and sun

so Distance between Pluto and sun is R_2=39.1\cdot R_1

T_1 and T_2 is the time period corresponding to R_1 and R_2[/tex]

(T_1)^2=k(R_1)^3---1

(T_2)^2=k(R_2)^3---2

divide 1 and 2

(\frac{365}{T_2})^2=(\frac{R_1}{39.1})^3

T_2^2=365^2\times 39.1^3

T_2=89239.67\ Earth\ days                      

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When a 360 nF air capacitor is connected to a power supply, the energy stored in the capacitor is 1.85 x 10-5 J. While the capac
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Answer:

(a) Approximately 10.1\; {\rm V}.

Explanation:

Let C denote the capacitance of a capacitor. Let V be the potential difference (voltage) between the two plates of this capacitor. The energy E stored in this capacitor would be:

\displaystyle E = \frac{1}{2}\, C\, (V^{2}).

Rearrange this equation to find an expression for the potential difference V in terms of capacitance C and energy E:

\begin{aligned}V^{2} &= \frac{2\, E}{C} \end{aligned}.

\begin{aligned}V &= \sqrt{\frac{2\, E}{C}} \end{aligned}

The capacitance C of this capacitor is given in nanofarads. Convert that unit to standard unit (farads):

\begin{aligned}C &= 360\; {\rm nF} \\ &= 360\; {\rm nF} \times \frac{1\; {\rm F}}{10^{9}\; {\rm nF}} \\ &= 3.60 \times 10^{-7}\; {\rm F}\end{aligned}.

Given that the energy stored in this capacitor is E = 1.85 \times 10^{-5}\; {\rm J}, the potential difference across the capacitor plates would be:

\begin{aligned}V &= \sqrt{\frac{2 \times 1.85 \times 10^{-5}\; {\rm J}}{3.60 \times 10^{-7}\; {\rm F}}} \\ &\approx 10.1\; {\rm V}\end{aligned}.

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Answer:

The part that completes the nuclear equation is:

           ^{221}_{87}Fr

Explanation:

<h2>A) Preliminar explanation</h2>

The <em>nuclear equation</em> represents a nuclear reaction: the change of the nucleus of an atom.

The given equation represents an actinium atom releasing an alpha particle.

This is the meaning of each part of the equation:

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  • The superscript to the left of the chemical symbol is the mass number of the atom (number of protons plus number of neutrons). The mass number is 225.
  • The subscript to the left of the chemical symbol is the atomic number of the atom (number of protons). The atomic number is 89.

  • ^4_2He is the symbol of the alpha particle. It is an atom of helium
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<h2>B) Solution</h2>

To <em>complete the nuclear equation </em>you must do two balances: mass number balance and atomic number balance.

<u>i) Mass number balance</u>

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<u>ii) Atomic number balance</u>

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Therefore, the mass number of the unknown atom is 221, and the atomic number is 87.

From a periodic table, the element with atomic number 87 is francium, Fr.

Now, you have the chemical symbol, the atomic number, and the mass number of the unknown atom, which lets you to write the atom that completes the <em>nuclear equation</em>.

           ^{221}_{87}Fr

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