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4 years ago

Do you get this ?????

Physics

1 answer:

Jet001 [13]4 years ago

8 0

The first one is actually 10 times as big as the second one.

Because of their places, the first one means 6000, and the second one means 600.

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ANSWER THIS AND YOU WILL BE THE BRAINLIEST

VladimirAG [237]

Answer:

Explanation:

it is correct

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3 years ago

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Firdavs [7]

Initial velocity (u) = 2 m/s

Acceleration (a) = 10 m/s^2

Time taken (t) = 4 s

Let the final velocity be v.

By using the equation,

v = u + at, we get

or, v = 2 + 10 × 4

or, v = 2 + 40

or, v = 42

The final velocity is 42 m/s.

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3 years ago

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Where is the epicenter (it's about earthquakes)

sweet [91]

Answer:

The epicenter is the point on the earth's surface vertically above the hypocenter (or focus), point in the crust where a seismic rupture begins.

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3 years ago

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Hows does the world spin and we cant feel it

Debora [2.8K]

Because it spin so slow over the course of a whole day.

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3 years ago

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

KATRIN_1 [288]

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

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

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

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

=> $T_R = 11.8 \ days$

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3 years ago