Ask question

Ask question

All categories

4 years ago

6

All of the planets in Earth's solar system must follow the same plane of motion. What's this plane called? A. Plane of the eclip

tic B. Plane of spherical motion C. Plane of triangulation D. Plane of orbital rotation

2 answers:

Katena32 [7]4 years ago

7 0

Answer:

D. Plane of orbital rotation

Explanation:

The orbital plane of an object orbiting around another is the geometric plane that contains the orbit in question.

By definition, the inclination of a planet in the solar system is the angle between the orbital plane of the solar system and the orbital plane of the Earth.

Feliz [49]4 years ago

5 0

Answer:

D. Plane of orbital rotation

Explanation:

As per Kepler's law we know that all planets revolve in same plane around the position of sun in such a way that the path of all planets are elliptical in shape and position of sun is at one of its focii.

So here since the plane of rotation of all planets are same so this is also known as orbital plane of the planets.

So here we can say that the plane is known as

D. Plane of orbital rotation

So this is given by the help of kepler's law

You might be interested in

Mark and Paul are in a race. Mark is 20 meters from the finish line and running at a constant 3.5 m/s. Paul is 5 meters behind h

USPshnik [31]

Answer:

$25 m= 2.7 m/s * (5.71 s)+ \frac{1}{2} a (5.71s)^2$

And solving for a we got:

$9.583 m = \frac{1}{2} a (5.71s)^2$

$a = 0.588 \frac{m}{s^2}$

Explanation:

For this case we have an illustration for the problem on the figure attached.

And we can solve this problem analyzing each one of the runner. Let's begin with Mark

Mark

For this case we know that $V_M = 3.5 m/s$ and th velocity is constant. The distance from Mark and the finish line is $D_M = 20 m$

Since the velocity is constant we can create the following relation:

$D_M = V_M t_M$

And solving for $t_M$ we got:

$t_m = \frac{D_M}{V_M}= \frac{20m}{3.5 m/s}= 5.71 s$

So then Mark will nd the race after 5.71 seconds

Paul

We know that the initial velocity for Paul is given $V_{iP}= 2.7 m/s$ we also know that the total distance between Paul and the finish line is 25 m and we want to find the acceleration that Paul needs to apply in order to tie the race, and Paul have 5.71 sconds in order to reach the finish line.

We can use this formula in order to find the acceleration (because we assume that the acceleration is constant) that he needs to apply:

$x_f = x_i + v_i t + \frac{1}{2} a t^2$

And since $\Delta x = x_f - x_i$ we have this:

$\Delta x= v_i t + \frac{1}{2} a t^2$

And if we replace we have this:

$25 m= 2.7 m/s * (5.71 s)+ \frac{1}{2} a (5.71s)^2$

And solving for a we got:

$9.583 m = \frac{1}{2} a (5.71s)^2$

$a = 0.588 \frac{m}{s^2}$

And the final velocity for Paul using this acceleration would be:

$V_{fP}= V_{iP}+ a_P t = 2.7m/s + 0.588 m/s^2 (5.71s)= 6.057 m/s$

3 0

4 years ago

Why is oxygen more chemically reactive than nitrogen?

grin007 [14]

Answer:

O2 has two more electrons compared to N2, with extra 2 electrons in the higher energy anti-bonding orbitals known as Diradical. These electrons have higher energy and are unpaired; therefore, O2 is more reactive

Explanation:

5 0

3 years ago

Read 2 more answers

At some instant, a particle traveling in a horizontal circular path of radius 7.90 m has a total acceleration with a magnitude o

DochEvi [55]

Answer:

a) Speed of the particle at this instant

v = 8.43 m/s

b) Speed of the particle at (1/8) revolution later

v = 14.83 m/s

Explanation:

We apply the equations of circular motion uniformly accelerated :

$(a_{T}) ^{2} = (a_{n} )^{2} +(a_{t} )^{2}$ Formula (1)

$a_{n} = \frac{v^{2} }{r}$ Formula (2)

$a_{t} = \alpha *r$ Formula (3)

v= ω*r Formula (4)

ω² = ω₀² + 2*α*θ Formula (5)

Where:

$a_{T}$ : total acceleration, (m/s²)

$a_{n}$ : normal acceleration, (m/s²)

$a_{t}$ : tangential acceleration, (m/s²)

$\alpha$ : angular acceleration (rad/s²)

r : radius of the circular path (m)

v : tangential velocity (m/s)

ω : angular speed ( rad/s)

ω₀: initial angular speed ( rad/s)

θ : angle that the particle travels (rad)

Data:

$a_{T}$ = 15 m/s²

$a_{t}$ = 12 m/s²

r=7.90 m :radius of the circular path

Problem development

In the formula (1) :

$a_{n} = \sqrt{(a_{T})^{2} -(a_{t})^{2} }$

We replace the data

$a_{n} = \sqrt{(15)^{2} -(12)^{2}}$

$a_{n} = 9 \frac{m}{s^{2} }$

We use formula (2) to calculate v:

$9 = \frac{v^{2} }{7.9}$ Equation (1)

a)Speed of the particle at this instant

in the equation (1):

$v=\sqrt{9*7.9} = 8.43 \frac{m}{s}$

b)Speed of the particle at (1/8) revolution later

We know the following data:

θ =(1/8) revolution=( 1/8) *2π= π/4

$a_{t}$ = 12 m/s²

v₀= 8.43 m/s

r=7.9 m

We use formula (3) to calculate α

$12 = \alpha *7.90$

$\alpha =\frac{12}{7.9} = 1.52 \frac{rad}{s^{2} }$

We use formula (4) to calculate ω₀

v₀= ω₀ *r

8.43 = ω₀*7.9

ω₀ = 8.43/7.9 = 1.067 rad/s

We use formula (5) to calculate ω

ω² = ω₀² + 2*α*θ

ω²= (1.067)² + 2*1.52*π/4

ω² =3.526

ω = 1.87 rad/s

We use formula (4) to calculate v

v= 1.87 rad/s * 7.9m

v = 14.83 m/s : speed of the particle at (1/8) revolution later

5 0

3 years ago

A 763 kg car moving at 26 m/s brakes to a stop. The brakes contain about 15 kg of iron that absorb the energy. What is increase

Neko [114]

Answer:

$\Delta T=38.20^{\circ}$

Explanation:

It is given that,

Mass of the car, m = 763 kg

Speed of the car, v = 26 m/s

Mass of the iron, m' = 15 kg

Specific heat of iron, c = 450 J/kg

When the car is in motion, it will possess kinetic energy. It is given by :

$K=\dfrac{1}{2}mv^2$

$K=\dfrac{1}{2}\times 763\times (26)^2$

K = 257894 J

Since, energy is absorbed by the brakes. The kinetic energy of the car is absorbed by the brakes. So,

$K=mc\Delta T$

$\Delta T$ is the increase in temperature of the brakes

$\Delta T=\dfrac{K}{m'c}$

$\Delta T=\dfrac{257894}{15\times 450}$

$\Delta T=38.20^{\circ}$

So, the increase in temperature of the brakes is 38.20 degrees Celsius. Hence, this is the required solution.

3 0

3 years ago

A car has a speed of 12m/s. The mass of the car and its passengers is 1250 kg. What is the total momentum of the car/passengers?

Temka [501]

Answer:

15000kgm/s

Explanation:

Given parameters:

Speed of the car = 12m/s

Mass of car and passengers = 1250Kg

Unknown:

Momentum of the car = ?

Solution:

Momentum is the quantity of motion a body posses;

Momentum = mass x velocity

Now insert the given parameters and solve;

Momentum = 12 x 1250 = 15000kgm/s

5 0

3 years ago