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Allisa [31]
2 years ago
11

If the mass of the sun is 2x, at least one planet will fall into the habitable zone if I place a planet in orbits___, ____, ____

, and ____, and all planets will orbit the sun successfully.
Pleas help this is Flvs comprehensive science class
6.01 please please help
Physics
1 answer:
nalin [4]2 years ago
6 0

So, the complete sentence is If the mass of the sun is 2x, at least one planet will fall into the habitable zone if I place a planet in orbits<u> 84, 1, </u>and <u>5</u>, and all planets will orbit the sun successfully for the best conditions.

When the mass of the sun is larger, Earth moves around the sun at a faster pace and When the mass of the sun is smaller, Earth moves around the sun at a slower pace.

When Earth is closer to the sun, its orbit becomes faster and When Earth is farther from the sun, its orbit becomes slower.

When Earth is closer to the sun, there will be a hotter climate. A little movement that takes one closer to the sun could lead to a huge impact, as the sun is very hot.

So, it can be concluded that If the mass of the sun is 2x, at least one planet will fall into the habitable zone if I place a planet in orbits<u> 84, 1, </u>and <u>5</u>, and all planets will orbit the sun successfully for the best conditions.

Learn more about Sun here:

brainly.com/question/15837114

#SPJ10

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A small boy is playing with a ball on a stationary train. If he places the ball on the floor of the train, when the train starts
expeople1 [14]

A small boy is playing with a ball on a stationary train. If he places the ball on the floor of the train, when the train starts moving the ball moves toward the back of the train. This happened due to inertia

An object at rest remains at rest, or if in motion, remains in motion unless a net external force acts on it .  

When a train starts moving forward, the ball placed on the floor  tends to fall backward is an example of inertia of rest. Due to the reason that the lower part of the ball is in contact with the surface and rest of the part is not . As the train starts moving, its lower part gets the motion as the floor starts moving but the upper part will remain as it is as it is not in contact with the floor , hence do not attain any motion due to the inertia of rest simultaneously i.e. it tends to remain at the same place.

To learn more about inertia here :

brainly.com/question/11049261

#SPJ1

8 0
2 years ago
Explain what is happening when roller coaster is at each point. **The roller coaster has started at A and goes to D.
yaroslaw [1]

Answer:

At point A, the cart has high potential energy. At point b, the cart is pulled down by gravity. At point c, the cart gains its highest kinetic energy. At point d, the cart returns back to the same state but with lower potential energy.

3 0
2 years ago
A 150-N box is being pulled horizontally in a wagon accelerating uniformly at 3.00 m/s2. The box does not move relative to the w
Zepler [3.9K]

Answer:

Frictional force, F = 45.9 N

Explanation:

It is given that,

Weight of the box, W = 150 N

Acceleration, a=3\ m/s^2

The coefficient of static friction between the box and the wagon's surface is 0.6 and the coefficient of kinetic friction is 0.4.  

It is mentioned that the box does not move relative to the wagon. The force of friction is equal to the applied force. Let a is the acceleration. So,

m=\dfrac{W}{g}

m=\dfrac{150}{9.8}

m=15.3\ kg

Frictional force is given by :

F=ma

F=15.3\times 3

F = 45.9 N

So, the friction force on this box is closest to 45.9 N. Hence, this is the required solution.

8 0
4 years ago
During a goal-line stand, a 112-kg fullback moving eastward with a speed of 6 m/s
11111nata11111 [884]

Answer:

-1.24 m/s

Explanation:

Total momentum before collision = total momentum after collision

Total momentum before collision = (mass of full back * velocity of fullback) + (mass of lineman * velocity of line man).

Mass of full back = 112 kg, mass of line bag = 120 kg, velocity of full back  6 m/s (east), velocity of line back = -8 m/s (west). Hence:

Total momentum before collision = (112 * 6) + (120 * -8) = 672 - 960 = -288 kgm/s

The total momentum after collision = (mass of full back + mass of line back) * velocity after collision.

Let velocity after collision be v, hence:

The total momentum after collision = (112 + 120)v = 232v

Total momentum before collision = total momentum after collision

-288 = 232v

v = -288 / 232

v = -1.24 m/s

Therefore after collision, the two players would move at a velocity 1.24 m/s west (the same direction as the lineman).

7 0
3 years ago
12) A horizontal force of 200 N is applied to move a 55-kg cart (initially at rest) across a 10 m level surface. What is the fin
WINSTONCH [101]

Hi there!

We can use the following:

W = ΔKE = F · d

Find the work done on the cart:

W = 200 · 10 = 2000 J

Now, this is equal to the change in kinetic energy of the object. Its initial kinetic energy is 0 J since it starts from rest, so:

2000J = KEf - KEi

KE is given as:

KE = \frac{1}{2}mv^2

2000J = 1/2(55)v²

4000 = 55v²

√(4000/55) = 8.53 m/s

3 0
3 years ago
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