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3 years ago

15

The sun and a planet with a period of 5 years would have a distance of

1 answer:

Galina-37 [17]3 years ago

8 0

To solve the problem shown above you must apply the proccedure shown below:

1. You must apply the Kepler's third law, as below:

(T1/T2)²=(R1/R2)³

T1 and T2 are the orbital periods. (T1=1 year; T2=5 years)
R1 and R2 are the orbital radius. (R1=1 Astronomical unit)

2. Now, you must substitute the values shown above into the formula and clear R2:

(1/5)²=(1/R2)³
1/25=1/R2³
R2³/25=1
R2³=25
R2=∛25
R2=2.92 AU

Therefore, the answer is: 2.92 AU

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2 years ago

In the circle below, suppose m VUX = 152° and mZUVW = 77º. Find the following.

bezimeni [28]

Given:

In the circle, $m(\overarc{VUX})=152^\circ$ and $m(\angle MUV)=77^\circ$ .

To find:

The following measures:

(a) $m\angle VUX$

(b) $m\angle UXW$

Solution:

According to the central angle theorem, the central angle is always twice of the subtended angle intercepted on the same same arc.

$m(VUX)=2\times m\angle VWX$

$152^\circ=2\times m\angle VWX$

$\dfrac{152^\circ}{2}=m\angle VWX$

$76^\circ=m\angle VWX$

In a cyclic quadrilateral, the opposite angles are supplementary angles.

UVWX is a cyclic quadrilateral. So,

$m\angle VUX+m\angle VWX=180^\circ$ [Opposite angles of a cyclic quadrilateral]

$m\angle VUX+76^\circ=180^\circ$

$m\angle VUX=180^\circ-76^\circ$

$m\angle VUX=104^\circ$

Now,

$m\angle UXW+m\angle UVW=180^\circ$ [Opposite angles of a cyclic quadrilateral]

$m\angle UXW+77^\circ =180^\circ$

$m\angle UXW=180^\circ-77^\circ$

$m\angle UXW=103^\circ$

Therefore, $m\angle VUX=104^\circ$ and $m\angle UXW=103^\circ$ .

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2 years ago

What the questions are asking help ASAP please

Rudiy27

Answer:

Step-by-step explanation:

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e.g.

In number 1 the set up would look like this

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3 years ago

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goblinko [34]

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