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Ilya [14]
2 years ago
11

A migrating robin flies due north with a speed of 12 m/s relative to the air. The air moves due east with a speed of 6.8 m/s rel

ative to the ground.
Physics
1 answer:
Svetach [21]2 years ago
4 0

I'm assuming the question is what is the robin's speed relative to to the ground...

Create an equation that describes its relative motion.

rVg = rVa + aVg


Substitute values.

rVg = 12 m/s [N] + 6.8 m/s [E]


Use vector addition.

| rVg | = √ | rVa |² + | aVg |²

| rVg | = √ 144 m²/s² + 46.24 m²/s²

| rVg | = √ 19<u>0</u>.24 m²/s²

| rVg | = 1<u>3</u>.78 m/s


Find direction.

tanФ = aVg / rVa

tanФ = 6.8 m/s / 12 m/s

Ф = 29°


Therefore, the velocity of the robin relative to the ground is 14 m/s [N29°E]

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Mars has two moons, Phobos and Deimos. Phobos orbits Mars at a distance of 9380 km from Mars's center, while Deimos orbits at 23
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Answer:

The ratio is   \frac{T_1}{T_2}  = 3.965

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From the question we are told that

   The  radius of Phobos orbit is  R_2 =  9380 km

    The radius  of Deimos orbit is  R_1  =  23500 \  km

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Here M is the mass of Mars which is constant

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So we see that \frac{ 4 *  \pi^2  }{G * M  } =  constant

   

    T^2 = R^3   *  constant      

=>  [\frac{T_1}{T_2} ]^2 =  [\frac{R_1}{R_2} ]^3

Here T_1 is the period of Deimos

and  T_1 is the period of  Phobos

So

      [\frac{T_1}{T_2} ] =  [\frac{R_1}{R_2} ]^{\frac{3}{2}}

=>    \frac{T_1}{T_2}  =  [\frac{23500 }{9380} ]^{\frac{3}{2}}]

=>    \frac{T_1}{T_2}  = 3.965

   

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