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3 years ago

What is the orbital period in years of a planet with a semi major axis of 35 au

Physics

1 answer:

mezya [45]3 years ago

6 0

Answer:

Orbital period of the planet will be 207.06 year

Explanation:

We have given the planet have the semi major axis as 35 au

We have to find the orbital period of the planet

From Keplar's third law there is relation between the orbital period and semi major axis which is t $T^2=R^3$

So $T^2=35^3$

$T^2=42875$

$T=207.06year$

So orbital period of the planet will be 207.06 year

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Answer:

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3 years ago

A 120 kg tackler moving at 3.0 m/s meets head-on (and tackles) a 91 kg halfback moving at 7.5 m/s. What will be their mutual vel

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Explanation:

It is given that,

Mass of the tackler, m₁ = 120 kg

Velocity of tackler, u₁ = 3 m/s

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We need to find the mutual velocity immediately the collision. It is the case of inelastic collision such that,

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3 years ago

An RLRL circuit with with a 1212-ΩΩ resistor and a 0.06−H0.06−H inductor carries a current of 1A1A at t=0t=0, at which time a vo

tankabanditka [31]

Answer:

The equation of current is $I=0.86 cos \left (120 t + \frac{\pi}{2} \right )$

Explanation:

Resistance, R = 12 ohm

Inductance, L = 0.06 H

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Compare with the standard equation,

$E=E_{0}cos (2\pi ft)$

$2\pi ft = 120 t \\\\\\w = 2\pi f = 120 rad/s$

So, the inductive reactance is

XL = w L = 120 x 0.06 = 7.2 ohm

The impedance of the circuit is

$Z =\sqrt{12^2+7.2^2}\\\\Z = 14 ohm$

The current leads by 90degree so the equation of current is

$I=\frac{Eo}{Z} cos \left (120 t + \frac{\pi}{2} \right )\\\\I=\frac{12}{14} cos \left (120 t + \frac{\pi}{2} \right )\\\\I=0.86 cos \left (120 t + \frac{\pi}{2} \right )$