0.078 times the orbital radius r of the earth around our sun is the exoplanet's orbital radius around its sun.
Answer: Option B
<u>Explanation:</u>
Given that planet is revolving around the earth so from the statement of centrifugal force, we know that any
![\frac{G M m}{r^{2}}=m \omega^{2} r](https://tex.z-dn.net/?f=%5Cfrac%7BG%20M%20m%7D%7Br%5E%7B2%7D%7D%3Dm%20%5Comega%5E%7B2%7D%20r)
The orbit’s period is given by,
![T=\sqrt{\frac{2 \pi}{\omega r^{2}}}=\sqrt{\frac{r^{3}}{G M}}](https://tex.z-dn.net/?f=T%3D%5Csqrt%7B%5Cfrac%7B2%20%5Cpi%7D%7B%5Comega%20r%5E%7B2%7D%7D%7D%3D%5Csqrt%7B%5Cfrac%7Br%5E%7B3%7D%7D%7BG%20M%7D%7D)
Where,
= Earth’s period
= planet’s period
= sun’s mass
= earth’s radius
Now,
![T_{e}=\sqrt{\frac{r_{e}^{3}}{G M_{s}}}](https://tex.z-dn.net/?f=T_%7Be%7D%3D%5Csqrt%7B%5Cfrac%7Br_%7Be%7D%5E%7B3%7D%7D%7BG%20M_%7Bs%7D%7D%7D)
As, planet mass is equal to 0.7 times the sun mass, so
![T_{p}=\sqrt{\frac{r_{p}^{3}}{0.7 G M_{s}}}](https://tex.z-dn.net/?f=T_%7Bp%7D%3D%5Csqrt%7B%5Cfrac%7Br_%7Bp%7D%5E%7B3%7D%7D%7B0.7%20G%20M_%7Bs%7D%7D%7D)
Taking the ratios of both equation, we get,
![\frac{T_{e}}{T_{p}}=\frac{\sqrt{\frac{r_{e}^{3}}{G M_{s}}}}{\sqrt{\frac{r_{p}^{3}}{0.7 G M_{s}}}}](https://tex.z-dn.net/?f=%5Cfrac%7BT_%7Be%7D%7D%7BT_%7Bp%7D%7D%3D%5Cfrac%7B%5Csqrt%7B%5Cfrac%7Br_%7Be%7D%5E%7B3%7D%7D%7BG%20M_%7Bs%7D%7D%7D%7D%7B%5Csqrt%7B%5Cfrac%7Br_%7Bp%7D%5E%7B3%7D%7D%7B0.7%20G%20M_%7Bs%7D%7D%7D%7D)
![\frac{T_{e}}{T_{p}}=\sqrt{\frac{0.7 \times r_{e}^{3}}{r_{p}^{3}}}](https://tex.z-dn.net/?f=%5Cfrac%7BT_%7Be%7D%7D%7BT_%7Bp%7D%7D%3D%5Csqrt%7B%5Cfrac%7B0.7%20%5Ctimes%20r_%7Be%7D%5E%7B3%7D%7D%7Br_%7Bp%7D%5E%7B3%7D%7D%7D)
![\left(\frac{T_{e}}{T_{p}}\right)^{2}=\frac{0.7 \times r_{e}^{3}}{r_{p}^{3}}](https://tex.z-dn.net/?f=%5Cleft%28%5Cfrac%7BT_%7Be%7D%7D%7BT_%7Bp%7D%7D%5Cright%29%5E%7B2%7D%3D%5Cfrac%7B0.7%20%5Ctimes%20r_%7Be%7D%5E%7B3%7D%7D%7Br_%7Bp%7D%5E%7B3%7D%7D)
![\left(\frac{T_{e}}{T_{p}}\right)^{2} \times \frac{1}{0.7}=\frac{r_{e}^{3}}{r_{p}^{3}}](https://tex.z-dn.net/?f=%5Cleft%28%5Cfrac%7BT_%7Be%7D%7D%7BT_%7Bp%7D%7D%5Cright%29%5E%7B2%7D%20%5Ctimes%20%5Cfrac%7B1%7D%7B0.7%7D%3D%5Cfrac%7Br_%7Be%7D%5E%7B3%7D%7D%7Br_%7Bp%7D%5E%7B3%7D%7D)
![\frac{r_{e}}{r_{p}}=\left(\left(\frac{T_{e}}{T_{p}}\right)^{2} \times \frac{1}{0.7}\right)^{\frac{1}{3}}](https://tex.z-dn.net/?f=%5Cfrac%7Br_%7Be%7D%7D%7Br_%7Bp%7D%7D%3D%5Cleft%28%5Cleft%28%5Cfrac%7BT_%7Be%7D%7D%7BT_%7Bp%7D%7D%5Cright%29%5E%7B2%7D%20%5Ctimes%20%5Cfrac%7B1%7D%7B0.7%7D%5Cright%29%5E%7B%5Cfrac%7B1%7D%7B3%7D%7D)
Given
and ![T_{e}=365 \text { days }](https://tex.z-dn.net/?f=T_%7Be%7D%3D365%20%5Ctext%20%7B%20days%20%7D)
![\frac{r_{e}}{r_{p}}=\left(\left(\frac{365}{9.5}\right)^{2} \times \frac{1}{0.7}\right)^{\frac{1}{3}}=\left(\frac{133225}{90.25} \times \frac{1}{0.7}\right)^{\frac{1}{3}}=(2108.82)^{\frac{1}{3}}](https://tex.z-dn.net/?f=%5Cfrac%7Br_%7Be%7D%7D%7Br_%7Bp%7D%7D%3D%5Cleft%28%5Cleft%28%5Cfrac%7B365%7D%7B9.5%7D%5Cright%29%5E%7B2%7D%20%5Ctimes%20%5Cfrac%7B1%7D%7B0.7%7D%5Cright%29%5E%7B%5Cfrac%7B1%7D%7B3%7D%7D%3D%5Cleft%28%5Cfrac%7B133225%7D%7B90.25%7D%20%5Ctimes%20%5Cfrac%7B1%7D%7B0.7%7D%5Cright%29%5E%7B%5Cfrac%7B1%7D%7B3%7D%7D%3D%282108.82%29%5E%7B%5Cfrac%7B1%7D%7B3%7D%7D)
![r_{p}=\left(\frac{1}{(2108.82)^{\frac{1}{3}}}\right) r_{e}=\left(\frac{1}{12.82}\right) r_{e}=0.078 r_{e}](https://tex.z-dn.net/?f=r_%7Bp%7D%3D%5Cleft%28%5Cfrac%7B1%7D%7B%282108.82%29%5E%7B%5Cfrac%7B1%7D%7B3%7D%7D%7D%5Cright%29%20r_%7Be%7D%3D%5Cleft%28%5Cfrac%7B1%7D%7B12.82%7D%5Cright%29%20r_%7Be%7D%3D0.078%20r_%7Be%7D)
Answer:
Explanation:
We shall use Ampere's circuital law to find magnetic field at required point.
The point is outside the circumference of two given wires so whole current will be accounted for .
Ampere's circuital law
B = ∫ Bdl = μ₀ I
line integral will be over circular path of radius r = 41 cm .
Total current I = 5A -3A = 2A .
∫ Bdl = μ₀ I
2π r B = μ₀ I
2π x .41 B = 4π x 10⁻⁷ x 2
B = 2 x 10⁻⁷ x 2 / .41
= 9.75 x 10⁻⁷ T . It will be along - ve Y - direction.
Climate feedbacks: processes that can either amplify or diminish the effects of climate forcings. A feedback that increases an initial warming is called a "positive feedback." A feedback that reduces an initial warming is a "negative feedback." So I guess it is ?
Answer:
The present day model shows a nucleus composed of protons and neutrons with cloud-like spheres of different diameters surrounding the nucleus to represent the energy levels of the electrons in the atom. Rutherford's model shows electrons orbiting the nucleus along fixed, but similar- diameter circular paths.
Both models show the composition of the nucleus at the center of the atom and the much smaller electrons at some distance from the nucleus.
Explanation:
Neither of the models does a good job of representing the relative size differences of the protons, neutrons and electrons, or the distance between the nucleus and the "electron clouds."
Answer:
b. Continental-oceanic
Explanation:
The Earth's crust has been divided into a series of plates. These plates move at a different rate or speed in different directions.
These plates are divided by the boundary between them which are of three types: convergent, transform and divergent.
The convergent boundaries are the one which moves toward each other and the based on the crust they are of three types: continent-continent, ocean-ocean and ocean-continent.
In ocean-continent boundaries, the oceanic plate is denser compared to the continental plate is lighter as a result of which during the process of subduction the oceanic plate sinks beneath the continental plate towards the mantle and melts.
Thus, Option-B is the correct answer.