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3 years ago

A planet has two

1 answer:

7 0

Kepler's third law hypothesizes that for all the small bodies in orbit around the
same central body, the ratio of (orbital period squared) / (orbital radius cubed)
is the same number.

Moon #1: (1.262 days)² / (2.346 x 10^4 km)³

Moon #2: (orbital period)² / (9.378 x 10^3 km)³

If Kepler knew what he was talking about ... and Newton showed that he did ...
then these two fractions are equal, and may be written as a proportion.

Cross multiply the proportion:

(orbital period)² x (2.346 x 10^4)³ = (1.262 days)² x (9.378 x 10^3)³

Divide each side by (2.346 x 10^4)³:

(Orbital period)² = (1.262 days)² x (9.378 x 10^3 km)³ / (2.346 x 10^4 km)³

= 0.1017 day²

Orbital period = 0.319 Earth day = about 7.6 hours.

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May I please have some help with finding the slope in this graph?

Kisachek [45]

The slope is always

   (the change in 'y')
divided by
   (the change in 'x') .

This line is so easy ... it crosses all the corners of the boxes,
so you have tons of choices of what 2 points to pick.
But let's just pick the ends of the line.

From the beginning to the end of the line:

-- the change in 'y' is 4 meters

-- the change in 'x' is 40 seconds

Slope = (the change in 'y') divided by (the change in 'x') .

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   = 0.1 meter/second .

8 0

4 years ago

When sugar is burned, water vapor and carbon dioxide are produced. A sugar molecule has a specific number of carbon, hydrogen, a

Anuta_ua [19.1K]

Answer:

Number of carbon, hydrogen, and oxygen atoms in sugar:

C: 12
H: 22
O: 11

Explanation:

1) Given balanced chemical equation:

Sugar(s) + 12O₂(g) → 11H₂O(g) + 12CO₂(g)

2) Mass conservation law: in a chemical reaction, the atoms are not created nor destroyed; they recombine to form new substances, so the number of atoms of each kind in the reactants is equal to the number of atoms of the same kind in the products.

3) Balance.

-----------------------------------------------------------------------------------------------------

Atom Products = Reactants

-----------------------------------------------------------------------------------------------------

C 12CO₂ ⇒ 12 = 12 in sugar

-----------------------------------------------------------------------------------------------------

H 11H₂O ⇒ 11 × 2 = 22 = 22 in sugar

-----------------------------------------------------------------------------------------------------

O 12CO₂ ⇒ 12 × 2 = 24 12 × 2 = 24 in O₂ + x in sugar

11H₂O ⇒ 11

----------------------------- --------------------------------------------------

24 + 11 = 35 = 24 + x

----------------------------------------------------------------------------------------------------

Solve for x: 35 = 24 + x ⇒ x = 35 - 24 = 11 ∴ 11 O atoms in sugar

Conclusion:

Number of carbon, hydrogen, and oxygen atoms in sugar:

C: 12
H: 22
O: 11

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Veronika [31]

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