Which statements describe processes involved in the accretion of planetesimals and protoplanets? Check all that apply.

Physics

2 answers:

grin007 [14]3 years ago

5 0

Answer:

B & C

Explanation:

MakcuM [25]3 years ago

4 0

Answer:

The correct options are;

Both involve the formation of solid particles from nebular materials

Both involve the work of gravitational push on nebular materials

Explanation:

Planetesimals are thought to be the product of grains of cosmic dusts that are found in the debris and protoplanetary discs, such that hundreds of planet forming embrayos are considered to be the result of the collisions of planetesimals that collide with each other to form larger embrayos

Protoplanets is a large planetary body with a stratified interior due to internal melting that has taken place. They originate in the protoplanetary discs from the collision of planetesimals that are up to a kilometer in size.

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Three dogs (Spot, Fido, and Steinberg) are pulling on a chew toy. The chew toy is experiencing no acceleration. Spot is pulling

quester [9]

Answer:

The magnitude and direction of the force applied by Steinberg are approximately 15.192 newtons and 126.704º.

Explanation:

The chew toy is at equilibrium and experimenting three forces from three distinct dogs. The Free Body Diagram depicting the system is attached below. By Newton's Laws we construct the following equations of equilibrium: (Sp is for Spot, F is for Fido and St is for Steinberg) All forces and angles are measured in newtons and sexagesimal degrees, respectively:

$\Sigma F_{x} = F_{F}\cdot \cos \theta_{F} + F_{St,x} = 0$ (1)

$\Sigma F_{y} = F_{F}\cdot \sin \theta_{F}-F_{Sp}+F_{St,y} = 0$ (2)

If we know that $F_{F} = 20\,N$ , $F_{Sp} = 30\,N$ and $\theta_{F} = 63^{\circ}$ , then the components of the force done by Steinberg on the chewing toy is:

$F_{St,x} = -F_{F}\cdot \cos \theta_{F}$

$F_{St,x} = -(20\,N)\cdot \cos 63^{\circ}$

$F_{St, x} = -9.080\,N$

$F_{St,y} = F_{Sp}-F_{F}\cdot \sin \theta_{F}$

$F_{St,y} = 30\,N-(20\,N)\cdot \sin 63^{\circ}$

$F_{St, y} = 12.180\,N$

The magnitud of the force is determined by Pythagorean Theorem:

$F_{St} = \sqrt{F_{St,x}^{2}+F_{St,y}^{2}}$

$F_{St} =\sqrt{(-9.080\,N)^{2}+(12.180\,N)^{2}}$

$F_{St} \approx 15.192\,N$

Since the direction of this force is in the 3rd Quadrant on Cartesian plane, we determine the direction of the force with respect to the eastern semiaxis:

$\theta_{St} = 180^{\circ} + \tan^{-1} \left(\frac{F_{St,y}}{F_{St,x}}\right)$