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Alinara [238K]
3 years ago
9

If earth increase the distance from the sun, what will happen to the period of orbi t(the time it takes to complete one revoluti

on around the Sun)?​
Physics
1 answer:
Mandarinka [93]3 years ago
5 0

The period of the orbit would increase as well

Explanation:

We can answer this question by applying Kepler's third law, which states that:

"The square of the orbital period of a planet around the Sun is proportional to the cube of the semi-major axis of its orbit"

Mathematically,

\frac{T^2}{a^3}=const.

Where

T is the orbital period

a is the semi-major axis of the orbit

In this problem, the question asks what happens if the distance of the Earth from the Sun increases. Increasing this distance means increasing the semi-major axis of the orbit, a: but as we saw from the previous equation, the orbital period of the Earth is proportional to a, therefore as a increases, T increases as well.

Therefore, the period of the orbit would increase.

Learn more about Kepler's third law:

brainly.com/question/11168300

#LearnwithBrainly

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Please help on this one?<br> I'm guessing it's either electric or mechanical? <br> Please help.
Dmitriy789 [7]
A . Thermal

Thermal Energy, or heat, is the internal energy in substance the vibration & movement of the atoms & molecules within substances. Geothermal energy is an example of thermal energy.
3 0
3 years ago
What traction of the radioisotope<br>remains in the body after one day?​
r-ruslan [8.4K]

The fraction of radioisotope left after 1 day is (\frac{1}{2})^{\frac{1}{\tau}}, with the half-life expressed in days

Explanation:

The question is incomplete: however, we can still answer as follows.

The mass of a radioactive sample after a time t is given by the equation:

m(t)=m_0 (\frac{1}{2})^{\frac{t}{\tau}}

where:

m_0 is the mass of the radioactive sample at t = 0

\tau is the half-life of the sample

This means that the mass of the sample halves after one half-life.

We can rewrite the equation as

\frac{m(t)}{m_0}=(\frac{1}{2})^{\frac{t}{\tau}}

And the term on the left represents the fraction of the radioisotope left after a certain time t.

Therefore, after t = 1 days, the fraction of radioisotope left in the body is

\frac{m(1)}{m_0}=(\frac{1}{2})^{\frac{1}{\tau}}

where the half-life \tau must be expressed in days in order to match the units.

Learn more about radioactive decay:

brainly.com/question/4207569

brainly.com/question/1695370

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5 0
3 years ago
When converting from a smaller to a larger unit of measurement, you move the _______ to the right. A. second number B. ratio C.
viktelen [127]
The answer to your question is D
7 0
3 years ago
Read 2 more answers
Find the moment of inertia about each of the following axes for a rod that is 0.360 {cm} in diameter and 1.70 {m} long, with a m
mamaluj [8]

The complete question is;

Find the moment of inertia about each of the following axes for a rod that is 0.36 cm in diameter and 1.70m long, with a mass of 5.00 × 10 ^(−2) kg.

A) About an axis perpendicular to the rod and passing through its center in kg.m²

B) About an axis perpendicular to the rod and passing through one end in kg.m²

C) About an axis along the length of the rod in kg.m²

Answer:

A) I = 0.012 kg.m²

B) I = 0.048 kg.m²

C) I = 8.1 × 10^(-8) kg.m²

Explanation:

We are given;

Diameter = 0.36 cm = 0.36 × 10^(−2) m

Length; L = 1.7m

Mass;m = 5 × 10^(−2) kg

A) For an axis perpendicular to the rod and passing through its center, the formula for the moment of inertia is;

I = mL²/12

I = (5 × 10^(−2) × 1.7²)/12

I = 0.012 kg.m²

B) For an axis perpendicular to the rod and passing through one end, the formula for the moment of inertia is;

I = mL²/3

So,

I = (5 × 10^(−2) × 1.7²)/3

I = 0.048 kg.m²

C) For an axis along the length of the rod, the formula for the moment of inertia is; I = mr²/2

We have diameter = 0.36 × 10^(−2) m, thus radius;r = (0.36 × 10^(−2))/2 = 0.18 × 10^(−2) m

I = (5 × 10^(−2) × (0.18 × 10^(−2))^2)/2

I = 8.1 × 10^(-8) kg.m²

3 0
3 years ago
Although he did not present a mechanism, what were the key points of Alfred Wegener’s proposal for the concept of continental dr
valentinak56 [21]

Answer: Alfred Wegener provided some of the important points that supported the theory of continental drift. They are as follows-

  1. The continents were once all attached together, and this can be proved by studying the coastlines of some of the continents that perfectly matches with one another.
  2. The appearance of similar rock types and similar fossils (including both animals and plants) has also contributed much information that continents were once all together.
4 0
3 years ago
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