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2 years ago

Give me four reasons pluto is a cool planet / dwarf planet

2 answers:

6 0

Pluto is categorized as a dwarf planet. In 2006, Pluto was categorized with three other objects in the solar system that are about the same small size as Pluto: Ceres, Makemake and Eris. These objects, along with Pluto, are much smaller than the "other" planets. Here is another one Pluto is a dwarf planet that lies in the Kuiper Belt. It's an area full of icy bodies and other dwarf planets at the edge of our solar system. Because Pluto is the biggest known object in this region, some call it "King of the Kuiper Belt."

One thing is certain. Pluto and its neighborhood are very peculiar. If scientists could unravel some of their mysteries, we would know more about how our solar system formed. Hope this helps! Mark brainly pls!

AleksandrR [38]2 years ago

5 0

Answer:

1. Just because it is small doesn't mean it needs to be excluded. If that were the case I would've been out of my friend group a while ago

2. Look at it it's fricking beautiful (see attachment)

3. It is just a great planet I don't think there needs to be any reason given I meannn you agree?

4. There's no other reason needed it's fricking gorgeous and amazing it needs no other reason. It needs a freaking

opening announcement to announce the arrival of the gorgeouness.

You might be interested in

A 36.3 kg cart has a velocity of 3 m/s. How much kinetic energy does the object have?

uysha [10]

Answer:

163.35

__________________________________________________________

<u>We are given:</u>

Mass of the object (m) = 36.3 kg

Velocity of the object (v) = 3 m/s

<u>Kinetic Energy of the object:</u>

We know that:

Kinetic Energy = 1/2(mv²)

KE = 1/2(36.3)(3)² [replacing the variables with the given values]

KE = 18.15 * 9

KE = 163.35 Joules

Hence, the cart has a Kinetic Energy of 163.35 Joules

7 0

2 years ago

A seagull flying horizontally at 8.00m/s carries a clam with a mass of 300g in its beak. Calculate the total mechanical energy o

Stells [14]

Answer:

9.6J+88.2J=97.8J

Explanation:

Here the velocity of the seagull is given,mass is given and its height.

We have to find its mechanical energy my friend.

Mechanical energy=kinetic energy + potential energy.

First we will find kinetic energy.

For calculating kinetic energy we need mass and velocity,which are given here.

So, Ek=

$1 \div 2mv {?}^{2}$

So by substituting the values we get 9.6J.

Now we find the potential energy which is mgh.

By substituting the values we get 88.2J.

Then we add both of those and get 97.8J

I hope this satisfies you and make sure you contact me if it doesn't

7 0

3 years ago

Should I apply for the Trusted Helpers Program? I want a real answer, if you don’t understand it don’t answer.

Svetradugi [14.3K]

Yes you should if you will like to. It is your opinion so follow your dreams if they are your dreams.

8 0

3 years ago

Read 2 more answers

How to do this, i'm completely lost

vaieri [72.5K]

There are two torques t1 and t2 on the beam due to the weights, one torque t3 due to the weight of the beam, and one torque t4 due to the string.

You need to figure out t4 to know the tension in the string.

Since the whole thing is not moving t1 + t2 + t3 = t4.

torque t = r * F * sinФ = distance from axis of rotation * force * sin (∡ between r and F)

t1 =3.2 * 44g
t2 = 7 * 49g
t3 = 3.5 * 24g

t4 = t1 + t2 + t3 = 5570,118

The t4 also is given by:

t4 = r * T * sin Ф

r = 7
Ф = 32°
T: tension in the string

T = t4 / (r * sinФ)

T = t4 / (7 * sin(32°))

T = 1501,6 N

8 0

3 years ago

Radar uses radio waves of a wavelength of 2.4 ${\rm m}$ . The time interval for one radiation pulse is 100 times larger than t

blondinia [14]

Answer:

120 m

Explanation:

Given:

wavelength 'λ' = 2.4m

pulse width 'τ'= 100T ('T' is the time of one oscillation)

The below inequality express the range of distances to an object that radar can detect

τc/2 < x < Tc/2 ---->eq(1)

Where, τc/2 is the shortest distance

First we'll calculate Frequency 'f' in order to determine time of one oscillation 'T'

f = c/λ (c= speed of light i.e 3 x $10^{8}$ m/s)

f= 3 x $10^{8}$ / 2.4

f=1.25 x $10^{8}$ hz.

As, T= 1/f

time of one oscillation T= 1/1.25 x $10^{8}$

T= 8 x $10^{-9}$ s

It was given that pulse width 'τ'= 100T

τ= 100 x 8 x $10^{-9}$ => 800 x $10^{-9}$ s

From eq(1), we can conclude that the shortest distance to an object that this radar can detect:

$x_{min}$ = τc/2 => (800 x $10^{-9}$ x 3 x $10^{8}$ )/2

$x_{min}$ =120m

8 0

3 years ago