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Rashid [163]
4 years ago
7

What key ingredient in the modern condensation theory was missing or unknown in the nebula theory?

Physics
1 answer:
lbvjy [14]4 years ago
5 0

Answer:

Interstellar dust

Explanation:

In modern solar system theory of condensation, interstellar dust, which was lacking in nebular theory, would be the essential component.Cosmic dust is dust that exists in outer space or that has fallen to earth, also called extra-terrestrial dust or spatial powder. The majority of the cosmic dust particle sizes range from a few molecules to 0.1 μm.

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Bob and Lily are riding on a typical carousel. Bob rides on a horse near the outer edge of the circular platform, and Lily rides
alukav5142 [94]

Answer:

Bob's angular speed is the same as that of lily

Explanation:

Because for a carousel the angular speed remains the same since velocity at center and edge are the same

6 0
4 years ago
What is the difference between folk dance and ballroom dance
disa [49]

Answer:

folk dancing is old music and ball room is young dancing

7 0
3 years ago
How do you find the letters to the science weather​
yaroslaw [1]

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ummmm

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3 0
3 years ago
Calculate the force between two objects that have masses of 70 Kilograms and 2,000 kilograms separated by a distance of 1 meter
Hunter-Best [27]

9.3 x 10⁻⁶N

Explanation:

Given parameters:

Mass 1 = 70kg

Mass 2 = 2000kg

distance = 1m

Unknown:

force between them =  

Solution:

The force between the two masses will be a gravitational force of attraction.

  F = \frac{G m_{1}m_{2}  }{r^{2} }

 G is universal gravitation constant = 6.67430×10−¹¹ N⋅m²/kg²

 r is the distance between the two masses

Substituting the parameters:

 F = \frac{6.67 x 10^{-11} x 70 x 2000}{1^{2} } = 9.3 x 10⁻⁶N

 Learn more:

Universal gravitation constant   brainly.com/question/1724648

#learnwithBrainly

5 0
4 years ago
A hockey puck moving at 0.4600 m/s collides with another puck that was at rest. The pucks have equal mass. The first puck is def
Sladkaya [172]

Answer:

Speed =0.283m/ s

Direction = 47.86°

Explanation:

Since it is a two dimensional momentum question with pucks having the same mass, we derive the momentum in xy plane

MU1 =MU2cos38 + MV2cos y ...x plane

0 = MU2sin38 - MV2sin y .....y plane

Where M= mass of puck, U1 = initial velocity of puck 1=0.46, U2 = final velocity of puck 1 =0.34, V2 = final velocity of puck 2, y= angular direction of puck2

Substitute into equation above

.46 = .34cos38 + V2cos y ...equ1

.34sin38 = V2sin y...equ2

.19=V2cos Y...x

.21=V2sin Y ...y

From x

V2 =0.19/cost

Sub V2 into y

0.21 = 0.19(Sin y/cos y)

1.1052 = tan y

y = 47.86°

Sub Y in to x plane equ

.19 = V2 cos 47.86°

V2=0.283m/s

7 0
3 years ago
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