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faltersainse [42]
4 years ago
5

According to the condensation theory, the most important factor for the formation of our planets was _____. the gravitational pu

ll of the Sun the interstellar dust attracting heat away from the protosun the process of nuclear fusion the nebular cloud condensing
Physics
2 answers:
mart [117]4 years ago
5 0

According to the condensation theory, the most important factor for the formation of our planets was "the interstellar dust attracting heat away from the protosun".


Condensation is the procedure by which water particles noticeable all around bunch together and shape fluid water. This is regularly observed outwardly of cold glasses. This idea additionally identifies with the solar system.

The condensation theory of the solar system expresses that our solar system, and perhaps all other galaxies, were shaped from a cloud of residue and gas that consolidated into strong issue. Space experts trust that the littlest grains of residue in our cloud applied a draw on the gas about it, 'consolidating' into bigger and bigger bits of issue, similarly as a snowball moving downhill will become bigger and bigger. In the long run, the gravitational draw of these residue atoms was sufficiently solid that they started to pull in each other, developing into greater and greater clusters that had more grounded gravitational pulls. In the long run, these bunches of residue and gas from the cloud frame a star, and potentially planets, space rocks, and comets turning about the star.

Mandarinka [93]4 years ago
3 0
It was the gravitational pull of the sun
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A 5.00 g object moving to the right at 20.0 cm/s makes an elastic head-on collision with a 10.0 g object that is initially at re
marusya05 [52]

Answer: a) 6.67cm/s b) 1/2

Explanation:

According to law of conservation of momentum, the momentum of the bodies before collision is equal to the momentum of the bodies after collision. Since the second body was initially at rest this means the initial velocity of the body is "zero".

Let m1 and m2 be the masses of the bodies

u1 and u2 be their velocities respectively

m1 = 5.0g m2 = 10.0g u1 = 20.0cm/s u2 = 0cm/s

Since momentum = mass × velocity

The conservation of momentum of the body will be

m1u1 + m2u2 = (m1+m2)v

Note that the body will move with a common velocity (v) after collision which will serve as the velocity of each object after collision.

5(20) + 10(0) = (5+10)v

100 + 0 = 15v

v = 100/15

v = 6.67cm/s

Therefore the velocity of each object after the collision is 6.67cm/s

b) kinectic energy of the 10.0g object will be 1/2MV²

= 1/2×10×6.67²

= 222.44Joules

kinectic energy of the 5.0g object will be 1/2MV²

= 1/2×5×6.67²

= 222.44Joules

= 111.22Joules

Fraction of the initial kinetic transferred to the 10g object will be

111.22/222.44

= 1/2

3 0
4 years ago
Which color of light refracts at the greatest angle when white light is incident on a prism?
antiseptic1488 [7]
Red is refracted the least and violet is refracted the most
5 0
3 years ago
Read 2 more answers
A wire has resistivity of 0.12 homes per metre​.
gulaghasi [49]

Answer:

(a) The resistance of 25m of wire is 3 ohms

(b) the length of this wire that has resistance 22 ohms is 183.33 m

Explanation:

Given;

resistivity of the wire, ρ = 0.12 ohms per meter

(a) The resistance of 25m of wire is calculated as follows;

R = \rho L\\\\R = 0.12 \ \frac{ohms}{m} \times 25\ m\\\\R = 3 \ ohms

(b) the length of this wire that has resistance 22 ohms is calculated as;

L = \frac{R}{\rho} \\\\L = \frac{22 \ ohms }{0.12 \ ohms/m} = 183.33 \ m

3 0
3 years ago
A(n) 0.2 kg object is swung in a vertical circular path on a string 0.1 m long. The acceleration of gravity is 9.8 m/s2 . If a c
Leya [2.2K]

Answer:

T=83.37N

Explanation:

Since the object is under a circular motion, according to Newton's second law, when the object is at the top of the circle we have:

\sum F_y: T-mg=F_c

Where F_c is the centripetal force and is given by:

F_c=ma_c=m\frac{v^2}{r}

Replacing and solving for T:

T=m\frac{v^2}{r}+mg\\T=0.2kg\frac{(6.38\frac{m}{s})^2}{0.1m}+0.2kg(9.8\frac{m}{s^2})\\T=83.37N

8 0
4 years ago
What type of lever has the fulcrum between the resistance arm and the effort arm?
crimeas [40]
I’m not really sure but I think it’s D type 1 lever
4 0
3 years ago
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