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Hitman42 [59]
3 years ago
14

A planet has been observed orbiting a nearby star. This star has a mass of 3.5 solar masses, and the planet is 4.2 AU from the s

tar. What is the planet’s orbital period in Earth years? Round your answer to the nearest whole number. __ Earth years
Physics
2 answers:
nadezda [96]3 years ago
7 1

Answer:

16 years just got it wrong for choosing 4.6 believe me if u want

Explanation:

kondaur [170]3 years ago
4 1

Answer:

4.6 years

Explanation:

This is solved using Kepler's third law which says:

T^2=\frac{4\pi ^2}{GM} a^2

Where

T = Orbital period of the planet (in seconds)

a = Distance from the star (in meters)

G = Gravitational constant

M = Mass of the parent star (in kg)

From the information given

M = 3.5M_{sun} = 6.96*10^{30} kg

a=4.2AU = 6.28*10^{11} meters

G = 6.67*10^{-11}m^3kg^{-1}s^{-2}

We put this into Kepler's law and get:

T=\sqrt{\frac{4\pi ^2}{6.67*10^{-11}*6.96*10^{30}} (6.28*10^{11})^3}=145,128,196 seconds.

This when converted to years is 4.6 years.

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Statement A: Area of a rectangle with measured length = 2.536 mm and width = 1.4 mm.
pochemuha

Explanation:

Formula for calculating the area of a  rectangle A = Length *width

For statement A;

Given area of a rectangle with measured length = 2.536 mm and width = 1.4 mm.

Area of the rectangle = 2.536mm * 1.4mm

Area of the rectangle = 3.5504mm²

The rule of significant figures states that we should always convert the answer to the least number of significant figure amount the given value in question. Since 1.4mm has 2 significant figure, hence we will convert our answer to 2 significant figure.

Area of the rectangle = 3.6mm² (to 2sf)

For statement B;

Given area of a rectangle with measured length = 2.536 mm and width = 1.41 mm.

Area of the rectangle = 2.536mm * 1.41mm

Area of the rectangle = 3.57576mm²

Similarly, Since 1.41mm has 3 significant figure compare to 2.536 that has 4sf, hence we will convert our answer to 3 significant figure.

Area of the rectangle = 3.58mm² (to 3sf)

Based on the conversion, it can be seen that 3.6mm²  is greater than 3.58mm², hence the area of rectangle in statement A is greater than the area of the rectangle in statement B.

7 0
3 years ago
A student carried out an experiment adding different weights to a spring and recording the results. Look at the table of results
MAXImum [283]

Answer:

0.25 m.

Explanation:

We'll begin by calculating the spring constant of the spring.

From the diagram, we shall used any of the weight with the corresponding extention to determine the spring constant. This is illustrated below:

Force (F) = 0.1 N

Extention (e) = 0.125 m

Spring constant (K) =?

F = Ke

0.1 = K x 0.125

Divide both side by 0.125

K = 0.1/0.125

K = 0.8 N/m

Therefore, the force constant, K of spring is 0.8 N/m

Now, we can obtain the number in gap 1 in the diagram above as follow:

Force (F) = 0.2 N

Spring constant (K) = 0.8 N/m

Extention (e) =..?

F = Ke

0.2 = 0.8 x e

Divide both side by 0.8

e = 0.2/0.8

e = 0.25 m

Therefore, the number that will complete gap 1is 0.25 m.

5 0
3 years ago
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