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3 years ago

What does it mean when the orbital eccentricity of a planet is close to 1?

Physics

2 answers:

Vladimir [108]3 years ago

8 0

Answer:

The eccentricity of an orbit is a single number, between 0 and 1, which describes how stretched out the orbit is. Zero means the orbit is perfectly circular. An eccentricity close to 1 means the orbit is extremely elongated; only comets coming from the outer reaches of the solar system get close to this value.

ankoles [38]3 years ago

7 0

We have that Zero signifies a perfect circle shape and 1 shows it maximum out of order shape.

From the question we are told

What does it mean when the orbital eccentricity of a planet is close to 1

Generally

Eccentricity

This in its simplest definition means to be eccentric which means to be a bit out of order or for the given subject at hand means to be a bit out of shape

Naturally the Eccentricity that an object possess is defined by two number 0(zero) to 1(one)

Where

Zero signifies a perfect circle shape and 1 shows it maximum out of order shape

For more information on this visit

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A 2kg block has 70J of KE. It then travels 1.5 meters up a hill. As it travels up the hill friction does -12J of work on the blo

Dima020 [189]

Answer:

v = 5.34[m/s]

Explanation:

In order to solve this problem, we must use the theorem of work and energy conservation. This theorem tells us that the sum of the mechanical energy in the initial state plus the work on or performed by a body must be equal to the mechanical energy in the final state.

Mechanical energy is defined as the sum of energies, kinetic, potential, and elastic.

E₁ = mechanical energy at initial state [J]

$E_{1}=E_{pot}+E_{kin}+E_{elas}\\$

In the initial state, we only have kinetic energy, potential energy is not had since the reference point is taken below 1.5[m], and the reference point is taken as potential energy equal to zero.

In the final state, you have kinetic energy and potential since the car has climbed 1.5[m] of the hill. Elastic energy is not available since there are no springs.

E₂ = mechanical energy at final state [J]

$E_{2}=E_{kin}+E_{pot}$

Now we can use the first statement to get the first equation:

$E_{1}+W_{1-2}=E_{2}$

where:

W₁₋₂ = work from the state 1 to 2.

$E_{k}=\frac{1}{2} *m*v^{2} \\$

$E_{pot}=m*g*h$

where:

h = elevation = 1.5 [m]

g = gravity acceleration = 9.81 [m/s²]

$70 - 12 = \frac{1}{2}*2*v^{2}+2*9.81*1.5$

$58 = v^{2} +29.43\\v^{2} =28.57\\v=\sqrt{28.57}\\v=5.34[m/s]$

4 0

3 years ago

The use of force to move an object is <br> .

Ne4ueva [31]

The use of force to move an object is called work. This only applies if the object moves.

4 0

3 years ago

A concave mirror has a focal length of 30.0 CM. an object is placed 15.0 CM from the mirror. what is the radius of curvature of

Nana76 [90]

Answer:

60 cm

Explanation:

We are given;

Focal length of a concave mirror as 30.0 cm
Object distance is 15.0 cm

We are required to determine the radius of curvature.

We need to know that the radius of a curvature is the radius of a circle from which the curved mirror is part.

We also need to know that the radius of curvature is twice the focal length of a curved mirror.

Therefore;

Radius of curvature = 2 × Focal length

Therefore;

Radius of curvature = 2 × 30 cm

= 60 cm

7 0

3 years ago

Does anyone know the answer to all of these questions

Rasek [7]

Answer:

i don't understand the hw

5 0

3 years ago

An object has an acceleration of 6.0 m/s/s. If the net force was doubled and the mass was one-third the original value, then the

alexandr402 [8]

Hahahahaha. Okay.

So basically , force is equal to mass into acceleration.

F=ma

so when F=ma , we get acceleration=6m/s/s

Force is doubled.

Mass is 1/3 times original.

2F=1/3ma

Now , we rearrange , and we get 6F=ma

So , now for 6 times the original force , we get 6 times the initial acceleration.

So new acceleration = 6*6= 36m/s/s

5 0

3 years ago