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valina [46]
2 years ago
13

Which of kepler’s laws explains why the sun has a slightly larger angular diameter in january than in july?

Physics
1 answer:
Bond [772]2 years ago
7 0

Kepler’s three law is the answer. Kepler’s 3 is the amount of time it takes to orbit the sun is related to size and distance.  Kepler’s 3 is one of the planetary motion and can be stated as all planets move in elliptical orbits, having the sun sits at one of the foci.

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The diagram does not represent a real electric field because the field lines, can someone help explain this for me
Papessa [141]

electric field lines are graphical presentation of electric field intensity

It is the graphical way to represent the electric field variation

If we draw the tangent to electric field line then it will give the direction of net electric field at that point

So whenever we draw the electric field lines of a charge distribution then it will always follow this basic properties

here we will always follow these basic properties of field lines

now as we can see that here two positive charges are placed nearby so the electric field must be like it can not intersect at any point because at intersection of two lines the direction of electric field not defined

As we have two directions of tangents at that point

So here the incorrect presentation is the intersection of two field lines which is not possible


4 0
3 years ago
What is unusual about the results of mass determinations of clusters of galaxies?
Art [367]

Answer:

I think it's bigger than most galaxies

3 0
2 years ago
Read 2 more answers
A force, 10 N drags a mass 10 kg on a horizontal table with an acceleration of 0.2
kifflom [539]

Answer:

20&£+)##&843&()-_££-()&_2+0&&-£_!)

4 0
2 years ago
Round the number 14.587020 to 4 significant digits.
PilotLPTM [1.2K]
A. 14.59 is correctly rounded to 4 significant digits.
4 0
3 years ago
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9. A radioisotope has a half-life of 4.50 min and an initial decay rate of 8400 Bq. What will be
Akimi4 [234]

Answer:

525 Bq

Explanation:

The decay rate is directly proportional to the amount of radioisotope, so we can use the half-life equation:

A = A₀ (½)^(t / T)

A is the final amount

A₀ is the initial amount,

t is the time,

T is the half life

A = (8400 Bq) (½)^(18.0 min / 4.50 min)

A = (8400 Bq) (½)^4

A = (8400 Bq) (1/16)

A = 525 Bq

8 0
3 years ago
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