Can you make predictions about the Moon's appearance?

What is a loose icy body with a long narrow orbit?

Bumek [7]

A comet is the loose, icy body with a long, narrow orbit.
Comets are very small solar system body made mainly of ices mixed with smaller amounts of dust and rock. Most comets are not larger than a few kilometers across. The main body of the comet is called the nucleus, and it can contain water, methane, nitrogen and other ices. Their speeds vary depending on their orbits and where they are in it. The closer they are to the sun, the faster they are going.

4 0

3 years ago

A 755 N force is used to push a 15 Kg box across the floor. What is the acceleration of the box?

cluponka [151]

Answer:

good question!

Explanation:

3 0

2 years ago

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Please help will mark brainliest

andreyandreev [35.5K]

Answer:

b

Explanation:

it melts yes but it's still water [same element] so there fore it's a physical change

7 0

3 years ago

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An object islaunched at velocity of 20 m/s in a direction making an angle of 25 degree upward with the horizontal. what is the m

Aneli [31]

Answer:

$\displaystyle y_m=3.65m$

Explanation:

<u>Motion in The Plane</u>

When an object is launched in free air with some angle respect to the horizontal, it describes a known parabolic path, comes to a maximum height and finally drops back to the ground level at a certain distance from the launching place.

The movement is split into two components: the horizontal component with constant speed and the vertical component with variable speed, modified by the acceleration of gravity. If we are given the values of $v_o$ and $\theta\\$ as the initial speed and angle, then we have

$\displaystyle v_x=v_o\ cos\theta$

$\displaystyle v_y=v_o\ sin\theta-gt$

$\displaystyle x=v_o\ cos\theta\ t$

$\displaystyle y=v_o\ sin\theta\ t -\frac{gt^2}{2}$

If we want to know the maximum height reached by the object, we find the value of t when $v_y$ becomes zero, because the object stops going up and starts going down

$\displaystyle v_y=o\Rightarrow v_o\ sin\theta =gt$

Solving for t

$\displaystyle t=\frac{v_o\ sin\theta }{g}$

Then we replace that value into y, to find the maximum height

$\displaystyle y_m=v_o\ sin\theta \ \frac{v_o\ sin\theta }{g}-\frac{g}{2}\left (\frac{v_o\ sin\theta }{g}\right )^2$

Operating and simplifying

$\displaystyle y_m=\frac{v_o^2\ sin^2\theta }{2g}$

We have

$\displaystyle v_o=20\ m/s,\ \theta=25^o$

The maximum height is

$\displaystyle y_m=\frac{(20)^2(sin25^o)^2}{2(9.8)}=\frac{71.44}{19.6}$

$\displaystyle y_m=3.65m$

7 0

2 years ago

50POINTS! Find the orbital speed of a satellite in a circular orbit 1700km above the surface of the Earth. M_earth 5.97e24kg, r_

ivolga24 [154]

<h2>Answer: 7020.117 m/s</h2>

Explanation:

The velocity of a satellite describing a circular orbit is<u> constant</u> and defined by the following expression:

$V=\sqrt{G\frac{M}{R}}$ (1)

Where:

$G=6.674({10}^{-11})\frac{N{m}^{2}}{{kg}^{2}}$ is the gravity constant

$M_{Earth}=5.97{10}^{24}kg$ the mass of the massive body around which the satellite is orbiting, in this case, the Earth .

$R=r_{Earth}+h=8080000m$ the radius of the orbit (measured from the center of the planet to the satellite).

This means the radius of the orbit is equal to <u>the sum</u> of the average radius of the Earth $r_{Earth}$ and the altitude of the satellite above the Earth's surface $h$ .

Note this orbital speed, as well as orbital period, does not depend on the mass of the satellite. It depends on the mass of the massive body (the Earth).

Now, rewriting equation (1) with the known values:

$V=\sqrt{(6.674({10}^{-11})\frac{N{m}^{2}}{{kg}^{2}})\frac{5.97{10}^{24}kg}{8080000m}}$

$V=7020.117\frac{m}{s}$

6 0

3 years ago

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