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tekilochka [14]
3 years ago
8

Saturn has an orbital period of 29.46 years. In two or more complete sentences, explain how to calculate the average distance fr

om Saturn to the sun and then calculate it.
Physics
2 answers:
d1i1m1o1n [39]3 years ago
3 0
For astronomical objects, the time period can be calculated using:
T² = (4π²a³)/GM
where T is time in Earth years, a is distance in Astronomical units, M is solar mass (1 for the sun)
Thus,
T² = a³
a = ∛(29.46²)
a = 0.67 AU
1 AU = 1.496 × 10⁸ Km
0.67 * 1.496 × 10⁸ Km
= 1.43 × 10⁹ Km
kari74 [83]3 years ago
3 0

Answer:

For astronomical objects, the time period can be calculated using:

T² = (4π²a³)/GM

where T is time in Earth years, a is distance in Astronomical units, M is solar mass (1 for the sun)

Thus,

T² = a³

a = ∛(29.46²)

a = 0.67 AU

1 AU = 1.496 × 10⁸ Km

0.67 * 1.496 × 10⁸ Km

= 1.43 × 10⁹ Km

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A tube with sealed ends has some sand at one end. When the tube is turned upside down the sand falls 0.6m then settles down agai
barxatty [35]

Answer:

2.943 J

Explanation:

Data:

acceleration due to gravity, g = 9.81 m/s²

mass of sand  = 0.5 kg

distance  = 0.6

Thermal energy  = the energy possessed by the falling sand

The force of sand = ma

                              = 0.5 * 9.81\\= 4.905 N

However, the energy is given as the force multiplied by the distance, so

E =fd

   4.905 * 0.6\\= 2.943 J

Therefore, the thermal energy is 2.943 joules

 

4 0
3 years ago
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The colors of pigments is it a) are cyan,yellow,and magenta b) are the same as the secondary c) combine in equal amounts to prov
fredd [130]

The answer would be A or D. The most appropriate answer would be D. Hope I helped :)

8 0
4 years ago
Agility refers to a person's level of flexibility.
tekilochka [14]
False, agility is more about how fast you are
7 0
3 years ago
Identify the kind of simple machine represented by each of these examples: a. A drill bit b. A skateboard ramp c. A boat oar
marusya05 [52]

Answer:

a skateboard ramp

Explanation:

6 0
3 years ago
Un pintor de 75.0 kg sube por una escalera de 2.75 m que está inclinada contra una pared vertical. La escalera forma un ángulo d
dezoksy [38]

Answer:

Work done, W = 1786.17J

Explanation:

The question says "A 75.0-kg painter climbs a 2.75-m ladder that is leaning against a vertical wall. The ladder makes an angle of 30.0 ° with the wall. How much work (in Joules) does gravity do on the painter? "

Mass of a painter, m = 75 kg

He climbs 2.75-m ladder that is leaning against a vertical wall.

The ladder makes an angle of 30 degrees with the wall.

We need to find the work done by the gravity on the painter.

The angle between the weight of the painter and the displacement is :

θ = 180 - 30

= 150°

The work done by the gravity is given by :

W=Fd\cos\theta\\\\=75\times 10\times 2.75\times \cos30\\\\=1786.17\ J

Hence, the required work done is 1786.17 J.

6 0
2 years ago
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