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3 years ago

In what ways do observed extrasolar planetary systems differ from our own solar system?

Physics

1 answer:

statuscvo [17]3 years ago

4 0

Answer:

Extrasolar solar system differ from our solar system in many ways such as of mass, size and shape of the planet, as well as temperature or amount of heat received in each planet.

Explanation:

An extrasolar planet is a planet outside the Solar System, while the Solar System orbit around the sun as a result of the gravitational pull of the sun.

Thus, we can say that the major difference between extrasolar planetary systems and solar system is that in solar system, planets orbit around the Sun, while in extrasolar planetary systems, planets orbit around other stars.

All of the planets in our solar system orbit around the Sun. Planets that orbit around other stars are called exoplanets or extrasolar.

Extrasolar solar system differ from our solar system in many ways such as of mass, size and shape of the planet. They also differ in terms of temperature, because the temperature in each planet in solar system depends on its distance from the sun while that of the extrasolar depends on the activities of the star.

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Mamont248 [21]

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3 years ago

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Firdavs [7]

Answer:

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3 years ago

Explain how electromagnetic waves are produced. I

madreJ [45]

Answer:

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2 years ago

What is the main difference between magma and lava

Colt1911 [192]

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3 years ago

Each plate of a parallel‑plate capacitor is a square of side 4.19 cm, 4.19 cm, and the plates are separated by 0.407 mm. 0.407 m

alexandr1967 [171]

Answer:

The electric field strength inside the capacitor is 49880.77 N/C.

Explanation:

Given:

Side length of the capacitor plate (a) = 4.19 cm = 0.0419 m

Separation between the plates (d) = 0.407 mm = $0.407\times 10^{-3}\ m$

Energy stored in the capacitor (U) = $7.87\ nJ=7.87\times 10^{-9}\ J$

Assuming the medium to be air.

So, permittivity of space (ε) = $8.854\times 10^{-12}\ F/m$

Area of the square plates is given as:

$A=a^2=(0.0419\ m)^2=1.75561\times 10^{-3}\ m^2$

Capacitance of the capacitor is given as:

$C=\dfrac{\epsilon A}{d}\\\\C=\frac{8.854\times 10^{-12}\ F/m\times 1.75561\times 10^{-3}\ m^2 }{0.407\times 10^{-3}\ m}\\\\C=3.819\times 10^{-11}\ F$

Now, we know that, the energy stored in a parallel plate capacitor is given as:

$U=\frac{CE^2d^2}{2}$

Rewriting in terms of 'E', we get:

$E=\sqrt{\frac{2U}{Cd^2}}$

Now, plug in the given values and solve for 'E'. This gives,

$E=\sqrt{\frac{2\times 7.87\times 10^{-9}\ J}{3.819\times 10^{-11}\ F\times (0.407\times 10^{-3})^2\ m^2}}\\\\E=49880.77\ N/C$

Therefore, the electric field strength inside the capacitor is 49880.77 N/C

8 0

3 years ago