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3 years ago

According to Kepler's harmonic law, Neptune would have a _________ orbit around the Sun than Mercury.

Physics

2 answers:

N76 [4]3 years ago

8 0

Slower

Explanation:

Kepler's harmonic law (also called Kepler's third law) states that:

"The cube of the radius of the orbit of a planet around the sun is proportional to the square of its period of revolution"

In formula, this can be written as

$\frac{r^3}{T^2}=const.$

where

r is the radius of the orbit

T is the orbital period

const. is a constant number, which is the same for all the planets orbiting the sun

By looking at the equation, we can see that the further a planet from the sun, the larger its orbital period: this means that the further planet will have a slower orbit. In this problem, we have to compare Neptune and Mercury: since Neptune is further from the sun than Mercury, also its orbital period will be larger, and therefore Neptune would have a slower orbit around the Sun than Mercury.

chubhunter [2.5K]3 years ago

3 0

Neptune would have a slower orbit i just did this

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A ball of mass M collides with a stick with moment of inertia I = βml2 (relative to its center, which is its center of mass). Th

ZanzabumX [31]

Answer:

Part a)

$v_2 = \frac{\frac{2\beta mL^2v_o}{d}}{(md + \frac{\beta mL^2}{d}(1 + \frac{m}{M})}$

Part b)

$v_1 = v_0 - \frac{m}{M}(\frac{\frac{2\beta mL^2v_o}{d}}{(md + \frac{\beta mL^2}{d}(1 + \frac{m}{M})})$

Explanation:

Since the ball and rod is an isolated system and there is no external force on it so by momentum conservation we will have

$Mv_o = M v_1 + m v_2$

here we also use angular momentum conservation

so we have

$M v_o d = M v_1 d + \beta mL^2 \omega$

also we know that the collision is elastic collision so we have

$v_o = (v_2 + d\omega) - v_1$

so we have

$\omega = \frac{v_o + v_1 - v_2}{d}$

also we know

$M v_o d - M v_1 d = \beta mL^2(\frac{v_o + v_1 - v_2}{d})$

also we know

$v_1 = v_o - \frac{m}{M}v_2$

so we have

$M v_o d - M(v_o - \frac{m}{M}v_2)d = \beta mL^2(\frac{v_o + v_o - \frac{m}{M}v_2 - v_2}{d})$

$mv_2 d = \beta mL^2\frac{2v_o}{d} - \beta mL^2(1 + \frac{m}{M})\frac{v_2}{d}$

now we have

$(md + \frac{\beta mL^2}{d}(1 + \frac{m}{M})v_2 = \frac{2\beta mL^2v_o}{d}$

$v_2 = \frac{\frac{2\beta mL^2v_o}{d}}{(md + \frac{\beta mL^2}{d}(1 + \frac{m}{M})}$

Part b)

Now we know that speed of the ball after collision is given as

$v_1 = v_o - \frac{m}{M}v_2$

so it is given as

$v_1 = v_0 - \frac{m}{M}(\frac{\frac{2\beta mL^2v_o}{d}}{(md + \frac{\beta mL^2}{d}(1 + \frac{m}{M})})$

3 0

3 years ago

Please help me quickly!!!

vichka [17]

Hey! So referring to the data the thing we can clearly see is that in a vacuum, everything, regardless of its mass, falls at the same speed.

Acceleration is often confused with speed, or velocity, but the difference is, acceleration by definition is the rate of which an object falls with respect to its mass and time.

Every single thing in the world falls at the same acceleration, this is because of gravity. The difference is the speed of which it falls. In space, there is not any gravity, and so, the objects are able to fall at the same speed regardless of their mass.

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3 years ago

A person pushes a 10 kg box from rest and accelerates it to a speed of 4 m/s with a constant force. If the box is pushed for a t

lutik1710 [3]

The box is accelerated from rest to 4 m/s in a matter of 2.5 s, so its acceleration a is such that

4 m/s = a (2.5 s) → a = (4 m/s) / (2.5 s) = 1.6 m/s²

Then the force applied to the box has a magnitude F such that

F = (10 kg) (1.6 m/s²) = 16 N

7 0

3 years ago

The mole is 6.02 x 10 23 particles. If a person masses out the correct molar mass in grams for a substance then she would have a

Leto [7]

Answer:

1 mole of H2O is 18 grams (2 g H + 16 g Oxygen)

36 / 18 = 2

So 2 moles = 2 * 6.02E23 = 12.04E23 = 1.204E24

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2 years ago

When a solute can't be dissolved in a solvent what is it called called?

Ivenika [448]

Answer: SATURATED

Explanation:

When a solute can no longer be dissolved it is considered "saturated" as is.

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2 years ago