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1 year ago

what is our best hypothesis for why all the planets in our solar system orbit in the same direction as the sun rotates?

Physics

1 answer:

avanturin [10]1 year ago

8 0

The conservation of angular momentum is the reason that (nearly) all of them rotate in the same direction.

<h3>Which explanation for the solar system is more plausible?</h3>

The nebular hypothesis is regarded as the most widely accepted theory of planetary creation. According to this theory, a massive molecular cloud several light-years across gravitationally collapsed to create the Solar System 4.6 billion years ago.

<h3>What is the solar system's foundational theory?</h3>

The nebular hypothesis states that our solar system arose at the same period as our Sun. The nebular hypothesis proposes that a spinning cloud of dust called a nebula, composed primarily of light elements, flattened into a protoplanetary disk and developed into a solar system made up of a star with orbiting planets .

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An infinite line charge of linear density λ = 0.30 μC/m lies along the z axis and a point charge q = 6.0 µC lies on the y axis a

ira [324]

The line charge E-field Ec = λ/(2πr*e0),
where λ = charge/length and e0 is the permittivity constant = 8.8542E-12 F/m.
<span>The point charge E-field Ep = kq/r^2 where electrostatic constant k = 1/(4π*e0) = 8.99E9 N-m^2/C^2.</span>

6 0

3 years ago

Read 2 more answers

The wavelength of red light is 7 x 10^-7 meter. Express this value in nanometers

3241004551 [841]

The wavelength of the red light in "nanometer" is 7× $10^{2}$

Wavelength is given as : 7× $10^{-7}$ meter

1 nanometer = ( $10^{-9}$ meter)

Let X= value of the wavelength in nanometer.

1 nanometer = $10^{-9}$ meter

X nanometer = 7× $10^{-7}$ meter

<em>If we Cross multiply</em>

X nanometer = ( $\frac{ 7* 10^{-7} }{ 10^{-9} }$ )

X= 7× $10^{2}$ nanometer

Therefore, the wavelength in "nanometer" is 7× $10^{2}$

Learn more at :brainly.com/question/12924624?referrer=searchResults

6 0

2 years ago

The position of a particle moving along the x-axis depends on the time according to the equation x = ct2 - bt3, where x is in me

Sav [38]

Answer:

(a): $\rm meter/ second^2.$

(b): $\rm meter/ second^3.$

(c): $\rm 2ct-3bt^2.$

(d): $\rm 2c-6bt.$

(e): $\rm t=\dfrac{2c}{3b}.$

Explanation:

Given, the position of the particle along the x axis is

$\rm x=ct^2-bt^3.$

The units of terms $\rm ct^2$ and $\rm bt^3$ should also be same as that of x, i.e., meters.

The unit of t is seconds.

(a):

Unit of $\rm ct^2=meter$

Therefore, unit of $\rm c= meter/ second^2.$

(b):

Unit of $\rm bt^3=meter$

Therefore, unit of $\rm b= meter/ second^3.$

(c):

The velocity v and the position x of a particle are related as

$\rm v=\dfrac{dx}{dt}\\=\dfrac{d}{dx}(ct^2-bt^3)\\=2ct-3bt^2.$

(d):

The acceleration a and the velocity v of the particle is related as

$\rm a = \dfrac{dv}{dt}\\=\dfrac{d}{dt}(2ct-3bt^2)\\=2c-6bt.$

(e):

The particle attains maximum x at, let's say, $\rm t_o$ , when the following two conditions are fulfilled:

$\rm \left (\dfrac{dx}{dt}\right )_{t=t_o}=0.$
$\rm \left ( \dfrac{d^2x}{dt^2}\right )_{t=t_o}$

Applying both these conditions,

$\rm \left ( \dfrac{dx}{dt}\right )_{t=t_o}=0\\2ct_o-3bt_o^2=0\\t_o(2c-3bt_o)=0\\t_o=0\ \ \ \ \ or\ \ \ \ \ 2c=3bt_o\Rightarrow t_o = \dfrac{2c}{3b}.$