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klemol [59]
4 years ago
15

Comets travel in orbits around the Sun. Some comets take less than 200 years to orbit the Sun. They are called short-period come

ts. Short-period comets are thought to originate in the Kuiper Belt, which is a belt of icy objects just beyond the orbit of the planet _____. The dwarf planet ______ is located in this belt.
Physics
2 answers:
Nina [5.8K]4 years ago
5 0

the planet neptune

The dwarf planet pluto.

The question literally has the words dwarf planet in it.  Pluto is the only dwarf in the solar system. All you had to do was look up which planet is in front of Pluto.

Elenna [48]4 years ago
5 0

Answer:

Planet: Neptune

Dwarf Planet: Pluto

Explanation:

Kuiper belt starts just after the orbit of planet Neptune at a distance of around 30 AU (1AU = 150 Million km) and extends up to 50AU from the Sun. This is home to billions of comet as well as lot of rocky bodies. Most famous of them is Pluto, which was earlier designated as a planet. In 2006, when the criteria for declaring a celestial body as a planet were reworked, Pluto was designated as a dwarf Planet.

The comet are thought to have formed at the time of the Solar system formation and resides in Kuiper Belt and Oort's cloud. The comets which reside in Kuiper belt are known as short period or periodic comets.

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In a local diner, a customer slides an empty coffee cup down the counter for a refill. The cup slides off the counter and strike
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a) t=\sqrt{\frac{2h}{g}}

b) v=\frac{d}{\sqrt{\frac{2h}{g}}}

c) v=\sqrt{d^2(\frac{g}{2h})+(2gh)}

d) \theta=tan^{-1}(\frac{2h}{d}) (radians)

Explanation:

a)

The motion of the cup sliding off the counter is the motion of a projectile, consisting of two independent motions:

- A uniform motion along the horizontal direction

- A uniformly accelerated motion (free fall) along the vertical direction

The time of flight of the cup is entirely determined by the vertical motion, therefore we can use the suvat equation:

s=ut+\frac{1}{2}at^2

where here:

s=h (the vertical displacement is the height of the counter)

u=0 (the initial vertical velocity of the cup is zero)

a=g (the vertical acceleration is the acceleration of gravity)

Solving for t, we find the time of flight of the cup:

h=0+\frac{1}{2}gt^2\\t=\sqrt{\frac{2h}{g}}

b)

To solve this part, we just analyze the horizontal motion of the cup.

Here we know that the horizontal motion of the cup is uniform: this means that is horizontal speed is constant during the whole motion, and it is actually equal to the speed at which the mug leaves the counter.

For a uniform motion, the speed is given by

v=\frac{d}{t}

where

d is the distance covered

t is the time taken

Here, the distance covered is d, the distance from the base of the counter, while the time taken is the time of flight:

t=\sqrt{\frac{2h}{g}}

Substituting into the previous equation, we find the speed of the mug as it leaves the counter:

v=\frac{d}{\sqrt{\frac{2h}{g}}}

c)

Here we want to find the speed of the cup immediately before it hits the floor.

Here we have to consider that while the mug falls, its vertical speed increases, while the horizontal speed remains constant.

Therefore, the horizontal speed of the cup before it hits the ground is:

v_x=\frac{d}{\sqrt{\frac{2h}{g}}}=d\sqrt{\frac{g}{2h}}

The vertical speed instead is given by the suvat equation:

v_y=u_y + at

where:

u_y=0 is the initial vertical velocity

a=g is the acceleration

t=\sqrt{\frac{2h}{g}} is the time of flight

Substituting,

v_y = 0 +g(\sqrt{\frac{2h}{g}})=\sqrt{2gh}

The actual speed of the cup just before it hits the floor is the resultant of the horizontal and vertical speeds, so it is:

v=\sqrt{v_x^2+v_y^2}=\sqrt{d^2(\frac{g}{2h})+(2gh)}

d)

Just before hitting the floor, the velocity of the cup has two components:

v_x=d\sqrt{\frac{g}{2h}} is the horizontal component (in the forward direction)

v_y=\sqrt{2gh} is the vertical component (in the downward direction)

Since the two components are perpendicular to each other, the angle of the direction is given by the equation

tan \theta = \frac{v_y}{v_x}

where here \theta is measured as below the horizontal direction.

Substituting the expressions for v_x,v_y, we find:

tan \theta = \frac{\sqrt{2gh}}{d\sqrt{\frac{g}{2h}}}=\frac{2h}{d}

So

\theta=tan^{-1}(\frac{2h}{d}) (radians)

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