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Sav [38]
3 years ago
8

A student is taking an animal science and an astronomy class. In animal sciences, she learns that pumas are called mountain lion

s and cougars in different parts of the world, but they are all the same cat. She makes an analogy, or comparison, to something she learned in her astronomy class. She says that a certain space object is like a puma because it is the same object but has different names depending on where it is. What space object is she thinking of?
comet
meteoroid
asteroid
meteoride
Physics
2 answers:
Bond [772]3 years ago
6 0

Answer: Im pretty sure its meteorite.

Explanation:

nexus9112 [7]3 years ago
5 0

Answer: meteoroid

Explanation: a meteoroid becomes a meteor when it enters earths atmosphere  

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How do burrowing rodents affect soil fertility? A) They feed on the nutrients in the soil leaving little for plants to absorb. B
jek_recluse [69]

How do burrowing rodents affect soil fertility? A) They feed on the nutrients in the soil leaving little for plants to absorb. B) They loosen the soil and make it difficult for plant roots to take a firm grip. C) They decompose the remains of plants and animals and add nutrients to the soil. D) They provide means by which air can enter the soil and reach the roots of plants.

the answer is a

7 0
3 years ago
Read 2 more answers
You happen to know that the coefficient of static friction between your patio table and the ground is 0.42. You decide you want
Deffense [45]

Answer:

If the force applied is larger than 185.2 N, yes.

Explanation:

In order to move the table, the pushing force must be larger than the frictional force. The frictional force is given by:

F_f = \mu mg

where

\mu=0.42 is the coefficient of static friction

m=45 kg is the mass of the table

g=9.8 m/s^2 is the gravitational acceleration

Substituting,

F_f=(0.42)(45 kg)(9.8 m/s^2)=185.2 N

So, we are able to move the table if we push with a force larger than 185.2 N.

4 0
3 years ago
What are 3 things that can affect gravity?
den301095 [7]
The force of gravity the masses exert on each other. If one of the masses is doubled , the force of gravity between the objects is doubled. Increases , the force of gravity decreases.
5 0
2 years ago
1. Which mathematical representation correctly identifies impulse?
horsena [70]

Answer:

1. B. Impulse = Force × Time

2. A. The momentum of each ball changes, and the total momentum stays the same

3. -55 kg·m/s

4. B. 3.5 kg

5. C. 6.3 m/s

Explanation:

1. The impulse is the momentum change of an object due to a force applied for a given period

2. Given that the objects collide, and the force of the 3 kg mass moving with 24 kg·m/s acts on the 1 kg mass while the total momentum is conserved;

The stationary ball of mass 1 kg begins to moves at certain velocity after collision and therefore changes momentum, while the velocity of the ball of mass 3.0 kg reduces and the total combined momentum of the two balls in the closed system remains the same

3. By the principle of conservation of linear momentum, we have;

The sum of the momentum before the collision = The sum of the momentum after collision

Given that the objects move together after the collision, the total momentum is therefore;

Total momentum = 110 kg·m/s + -65 kg·m/s + -100 kg·m/s = 110 kg·m/s - 65 kg·m/s - 100 kg·m/s  = -55kg·m/s

4. Given that the final velocity of the two objects (m₁ + m₂) combined = 50 m/s

Where;

m₁ = The mass of the first object

m₂ = The mass of the second object

The total momentum of the system = 250 kg·m/s

From momentum = Mass × Velocity, we have;

Mass = Momentum/Velocity = 250 kg·m/s/(50 m/s) = 5.0 kg

The mass (m₁ + m₂) = 5.0 kg

Given that m₁ = 1.5 kg, we have;

m₂ = 5.0 kg - m₁ = 5.0 kg - 1.5 kg = 3.5 kg

The mass of the second object = 3.5 kg

5. The mass of the cue stick = 0.5 kg

The velocity of the cue stick = 2.5 m/s

The mass of the ball = 0.2 kg

The initial velocity of the ball = 0 m/s

Given that total initial momentum = Total final momentum, we have;

0.5 kg × 2.5 m/s + 0.2 kg × 0 = 0.2 kg × v + 0.5 kg × 0

0.5 kg × 2.5 m/s = 0.2 kg × v

v = (0.5 kg × 2.5 m/s)/(0.2 kg) = 6.25  m/s ≈ 6.3 m/s

3 0
3 years ago
The box resting on the inclined plane above has a mass of 20kg. The incline sits at a 30o angle. Find the friction force between
tekilochka [14]

The friction force between the box and the incline if the box does not slide down the incline will be 0.577

The force preventing sliding against one another of solid surfaces, fluid layers, and material components is known as friction. There are several kinds of friction: Two solid surfaces in touch are opposed to one another's relative lateral motion by dry friction.

Given the box resting on the inclined plane above has a mass of 20kg and the The incline sits at a 30 degree angle

We have to find the friction force between the box and the incline if the box does not slide down the incline

Since the frictional force F₁ must equal or exceed gravitational force F₂ down the incline:

F₁ = F₂

μmgcosΘ = mgsinΘ

μ = (mgsinΘ)/(mgcosΘ)

μ = tanΘ

μ = 0.577

Hence the friction force between the box and the incline if the box does not slide down the incline will be 0.577

Learn more about friction force here:

brainly.com/question/24386803

#SPJ4

3 0
1 year ago
Read 2 more answers
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