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Andrei [34K]
4 years ago
12

Exoplanets (planets outside our solar system) are an active area of modern research. Suppose you read an article stating that th

ere is a newly discovered planetary system with three planets. The article states that the outermost planet (Planet C) goes all the way around its star in less time than the innermost planet (Planet A). According to Kepler’s laws of planetary motion, is this possible?
Physics
1 answer:
d1i1m1o1n [39]4 years ago
5 0

Answer:

Not possible.

Explanation:

Exoplanets are an active area of modern research. The article states that the outermost planet (Planet C) goes all the way around its star in less time than the innermost planet (Planet A). Which is not possible as it will violet Kepler's third law of planetary motion.  Which says that the square of orbital period of a planet is proportional to the cube of its semi-major axis of its orbit.

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Calculate the magnitude of the electric field at one corner of a square 2.42 m on a side if the other three corners are occupied
Karolina [17]

Ans:

12500 N/C

Explanation:

Side of square,  a = 2.42 m

q = 4.25 x 10^-6 C

The formula for the electric field is given by

E = \frac{Kq}{r^2}

where, K be the constant = 9 x 10^9 Nm^2/c^2 and r be the distance between the two charges

According to the diagram

BD = \sqrt{2}\times a

where, a be the side of the square

So, Electric field at B due to charge at A

E_{A}=\frac{Kq}{a^2}

E_{A}=\frac{9\times10^{9}\times 4.25 \times 10^{-6}}{2.42^2}

EA = 6531.32 N/C

Electric field at B due to charge at C

E_{C}=\frac{Kq}{a^2}

E_{C}=\frac{9\times10^{9}\times 4.25 \times 10^{-6}}{2.42^2}

Ec = 6531.32 N/C

Electric field at B due to charge at D

E_{D}=\frac{Kq}{2a^2}

E_{D}=\frac{9\times10^{9}\times 4.25 \times 10^{-6}}{2\times 2.42^2}

ED = 3265.66 N/C

Now resolve the components along X axis and Y axis

Ex = EA + ED Cos 45 = 6531.32 + 3265.66 x 0.707 = 8840.5 N/C

Ey = Ec + ED Sin 45 = 6531.32 + 3265.66 x 0.707 = 8840.5 N/C

The resultant electric field at B is given by

E=\sqrt{E_{x}^{2}+E_{y}^{2}}

E=\sqrt{8840.5^{2}+8840.5^{2}}

E = 12500 N/C

Explanation:

8 0
4 years ago
Read 2 more answers
How do streams change as they flow from mountains down to plains? Plz help URGENT I need it today
lisov135 [29]
They start as one stream, then get curvier, then empy into a delta. A delta is a ton of streams of water. It is one, then empties out into a ton of streams.
~Deceptiøn

5 0
3 years ago
Why are the nuclei of the heavier elements radioactive and not the lighter elements of nuclei?
KatRina [158]

Answer:

becouse most of nuclear elements are heave

Explanation:

6 0
3 years ago
How can a large force result in a relatively small power?
Sav [38]

Answer:

hey mate here is your answer

So if an object has a very small velocity (not moving very far over time, even though a large force may be applied to it, the Power will remain small. ... Stepping on the gas, or "speeding up" the car, is applying a force which will increase velocity and increase power.

please mark me as a brainliest

8 0
3 years ago
A block of mass 200g is oscillating on the end of a horizontal spring of spring constant 100 N/m and natural length 12 cm. When
malfutka [58]

In order to determine the acceleration of the block, use the following formula:

F=ma

Moreover, remind that for an object attached to a spring the magnitude of the force acting over a mass is given by:

F=kx

Then, you have:

ma=kx

by solving for a, you obtain:

a=\frac{kx}{m}

In this case, you have:

k: spring constant = 100N/m

m: mass of the block = 200g = 0.2kg

x: distance related to the equilibrium position = 14cm - 12cm = 2cm = 0.02m

Replace the previous values of the parameters into the expression for a:

a=\frac{(\frac{100N}{m})(0.02m)}{0.2\operatorname{kg}}=10\frac{m}{s^2}

Hence, the acceleration of the block is 10 m/s^2

8 0
1 year ago
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