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3 years ago

5

Atmospheric pressure at sea level has a value of 100kpa. The density of sea water is 1020kg/m-3. At which depth in the sea would

the total pressure be 110kpa?

1 answer:

lisov135 [29]3 years ago

6 0

Answer:

1 m

Explanation:

P = Patm + ρgh

110,000 Pa = 100,000 Pa + (1020 kg/m³) (9.8 m/s²) h

h = 1 m

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Answer:

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Explanation:

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Magnetic field is given by

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3 years ago

A piece of rocky debris in space has a semi major axis of 45.0 AU. What is its orbital period?

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Complete Question

Planet D has a semi-major axis = 60 AU and an orbital period of 18.164 days. A piece of rocky debris in space has a semi major axis of 45.0 AU. What is its orbital period?

Answer:

The value is $T_R = 11.8 \ days$

Explanation:

From the question we are told that

The semi - major axis of the rocky debris $a_R = 45.0\ AU$

The semi - major axis of Planet D is $a_D = 60 \ AU$

The orbital period of planet D is $T_D = 18.164 \ days$

Generally from Kepler third law

$T \ \ \alpha \ \ a^{\frac{3}{2} }$

Here T is the orbital period while a is the semi major axis

So

$\frac{T_D}{T_R} = \frac{a^{\frac{3}{2} }}{a_R^{\frac{3}{2} }}$

=> $T_R = T_D * [\frac{a_R}{a_D} ]^{\frac{3}{2} }$

=> $T_R = 18.164 * [\frac{ 45}{60} ]^{\frac{3}{2} }$

=> $T_R = 11.8 \ days$

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