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2 years ago

7

A new planet is discovered beyond Pluto at a mean distance to the sun of 4004 million miles. Using Kepler's third law, determine

an estimate for the time T to travel around the sun in an orbit.

1 answer:

AVprozaik [17]2 years ago

7 0

Answer:

103239.89 days

Explanation:

Kepler's third law states that the square of the orbital period of a planet is directly proportional to the cube of the semi-major axis of its orbit.

a³ / T² = 7.496 × 10⁻⁶ (a.u.³/days²)

where,

a is the distance of the semi-major axis in a.u

T is the orbit time in days

Converting the mean distance of the new planet to astronomical unit (a.u.)

1 a.u = 9.296 × 10⁷ miles

$\frac{4004 * 10^{6}}{9.296 * 10^{7}} = 43.07\ a.u.$

Substituting the values into Kepler's third law equation;

$\frac{(43.07)^{3}}{T^{2}} = 7.496 * 10^{-6}$

$T^{2} = \frac{(43.07)^{3}}{7.496 * 10^{-6}}$ (days)²

$T^{2} = \sqrt{\frac{(43.07)^{3}}{7.496 * 10^{-6}}}$

T = 103239.89 days

An estimate time T for the new planet to travel around the sun in an orbit is 103239.89 days

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When two resistors are wired in series with a 12 V battery, the current through the battery is 0.33 A. When they are wired in pa

MA_775_DIABLO [31]

Answer:

If R₂=25.78 ohm, then R₁=10.58 ohm

If R₂=10.57 then R₁=25.79 ohm

Explanation:

R₁ = Resistance of first resistor

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V = Voltage of battery = 12 V

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I = Current = 1.6 A (parallel)

In series

$\text{Equivalent resistance}=R_{eq}=R_1+R_2\\\text {From Ohm's law}\\V=IR_{eq}\\\Rightarrow R_{eq}=\frac{12}{0.33}\\\Rightarrow R_1+R_2=36.36\\ Also\ R_1=36.36-R_2$

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$\text {From Ohm's law}\\V=IR_{eq}\\\Rightarrow R_{eq}=\frac{12}{1.6}\\\Rightarrow \frac{R_1R_2}{R_1+R_2}=7.5\\\Rightarrow \frac{R_1R_2}{36.36}=7.5\\\Rightarrow R_1R_2=272.72\\\Rightarrow(36.36-R_2)R_2=272.72\\\Rightarrow R_2^2-36.36R_2+272.72=0$

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$\Rightarrow R_2=\frac{36.36\pm \sqrt{36.36^2-4\times 272.72}}{2}$

$\Rightarrow R_2=25.78\ or\ 10.57\\ If\ R_2=25.78\ then\ R_1=36.36-25.78=10.58\ \Omega\\ If\ R_2=10.57\ then\ R_1=36.36-10.57=25.79\Omega$

∴ If R₂=25.78 ohm, then R₁=10.58 ohm

If R₂=10.57 then R₁=25.79 ohm

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Given:

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initial volume, $V_0=4.4\times 10^{-4}\ m^3$

maximum pressure difference bearable by the container, $dP_{max}=2.26\times 10^{5}\ Pa$

density of the air, $\rho_a=1.2\ kg.m^{-3}$

density of sea water, $\rho_s=1.2\ kg.m^{-3}$

<u>The relation between the change in pressure with height is given as:</u>

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Answer:

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