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3 years ago

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Current evidence suggests that many massive jovian planets orbit at very close orbital distances to their stars. How do we think

these planets ended up on these close orbits?

1 answer:

ivanzaharov [21]3 years ago

4 0

Answer:

In the Solar system, the Jovian planets are farther from the Sun. Majority of the extrasolar Jovian planets are closer to their stars. These are known as "Hot Jupiters". From the studies, the reason for the existence of massive Jovian planets to be closer to their star is found to be the gravitational interaction of these planets with other massive planets which pushes them closer to their stars. These planets are formed beyond the frost line initially but later on migrate inwards.

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In the image below, astronauts detect approaching space junk and become

Wewaii [24]

Answer:

A

Explanation:

7 0

3 years ago

If you pull on one end of a skateboard with a force of 10 N, and your friend pulls on the other end with a force of 10 N. What i

Colt1911 [192]

Answer: Imma say balanced

Explanation:

Because if u pull one side of a skate board at 10n and your friend pulls the other side of the skate board at 10n it will be balanced because it doesn't move.

8 0

3 years ago

Particle A of charge 2.76 10-4 C is at the origin, particle B of charge -6.54 10-4 C is at (4.00 m, 0), and particle C of charge

Vanyuwa [196]

Answer:

a) F_net = 30.47 N , θ = 10.6º

b) Fₓ = 29.95 N

Explanation:

For this exercise we use coulomb's law

F₁₂ = k $k \frac{ q_{1} \ q_{2} }{ r^{2} }$

the direction of the force is on the line between the two charges and the sense is repulsive if the charges are equal and attractive if the charges are different.

As we have several charges, the easiest way to solve the problem is to add the components of the force in each axis, see attached for a diagram of the forces

X axis

Fₓ = $F_{bc x}$

Y axis

$F_{y}$ Fy = $F_{ab} - F_{bc y}$

let's find the magnitude of each force

$F_{ab}$ = 9 10⁹ 2.76 10⁻⁴ 1.02 10⁻⁴ / 3²

F_{ab} = 2.82 10¹ N

F_{ab} = 28.2 N

$F_{bc}$ = 9 10⁹ 6.54 10⁻⁴ 1.02 10⁻⁴ / 4²

F_{bc} = 3.75 10¹ N

F_{bc} = 37.5 N

let's use trigonometry to decompose this force

tan θ = y / x

θ = tan⁻¹ and x

θ= tan⁻¹ ¾

θ = 37º

let's break down the force

sin 37 = F_{bcy} / F_{bc}

F_{bcy} = F_{bc} sin 37

F_{bcy} = 37.5 sin 37

F_{bcy} = 22.57 N

cos 37 = F_{bcx} /F_{bc}

F_{bcx} = F_{bc} cos 37

F_{bcx} = 37.5 cos 37

F_{bcx} = 29.95 N

let's do the sum to find the net force

X axis

Fₓ = 29.95 N

Axis y

Fy = 28.2 -22.57

Fy = 5.63 N

we can give the result in two ways

a) F_net = Fₓ i ^ + $F_{y}$ j ^

F_net = 29.95 i ^ + 5.63 j ^

b) in the form of module and angle

let's use the Pythagorean theorem

F_net = $\sqrt{ F_{x}^2 + F_{y}^2 }$

F_net = √(29.95² + 5.63²)

F_net = 30.47 N

we use trigonometry for the direction

tan θ= $\frac{ F_{y} }{ F_{x} }$

θ = tan⁻¹ \frac{ F_{y} }{ F_{x} }

θ = tan⁻¹ (5.63 / 29.95)

θ = 10.6º

3 0

3 years ago

A +2.00nc point charge is at the origin, and a second -5.00nc point charge is on the x-axis at x = 0.800m find the magnitude of

Damm [24]

You have to reduce 2.00 an5.00 I order to use the×that=0.800

3 0

3 years ago

A rock is projected upward from the surface of the moon, at time t = 0.0 s, w a velocity of 30 m/s. The acceleration due to grav

Vinvika [58]

<h2>Answer: 277.777 m</h2>

Explanation:

The situation described here is parabolic movement. However, as we are told that the rock was<u> projected upward from the surface</u>, we will only use the equations related to the Y axis.

In this sense, the movement equations in the Y axis are:

$y-y_{o}=V_{o}.t+\frac{1}{2}g.t^{2}$ (1)

$V=V_{o}-g.t$ (2)

Where:

$y$ is the rock's final position

$y_{o}=0$ is the rock's initial position

$V_{o}=30\frac{m}{s}$ is the rock's initial velocity

$V$ is the final velocity

$t$ is the time the parabolic movement lasts

$g=1.62\frac{m}{s^{2}}$ is the acceleration due to gravity at the surface of the moon

As we know $y_{o}=0$ , equation (2) is rewritten as:

$y=V_{o}.t+\frac{1}{2}g.t^{2}$ (3)

On the other hand, the maximum height is accomplished when $V=0$ :

$V=V_{o}-g.t=0$ (4)

$V_{o}-g.t=0$

$V_{o}=g.t$ (5)

Finding $t$ :

$t=\frac{V_{o}}{g}$ (6)

Substituting (6) in (3):

$y=V_{o}(\frac{V_{o}}{g})+\frac{1}{2}g(\frac{V_{o}}{g})^{2}$ (7)

$y_{max}=\frac{{V_{o}}^{2}}{2g}$ (8) Now we can calculate the maximum height of the rock

$y_{max}=\frac{{(30m/s)}^{2}}{(2)(1.62m/s^{2})}$ (9)

Finally:

$y_{max}=277.777m$

4 0

3 years ago