All categories

3 years ago

Why is Pluto not considered a planet?

2 answers:

4 0

Pluto is considerd a dwarf planet because it was too small in size to be an acutal planet and its location in space. Pluto and planets like it are now called Dwarf planets. Pluto is also called a Plutoid, a plutoid is a dwarf planet that is farther out in space then the planet Neptune.

I hope that helped☺️

ASHA 777 [7]3 years ago

3 0

Pluto is actually considered to be a dwarf planet and was not considered a planet due to the fact that it was too small in size and it's location is space. Pluto and others similar to it are called dwarf planets. Pluto is also referred to as a plutoid, which is any dwarf planet farther out than the planet Neptune.

You might be interested in

the statement that current is equal to the voltage difference divided by the resistance is known as what

dangina [55]

Answer:it is known as (Ohm's law)

Explanation:

3 0

3 years ago

Aiden has basketball practice after school on Mondays, Wednesdays, and Fridays. He has practiced really hard this year to get be

Jlenok [28]

Ok I’ll help you I think 20 -192 = 1928172 then ur divided • by 181$1 then u get 244141551611671718181919191827337533535352526

6 0

3 years ago

A horizontal 745 N merry-go-round of radius

Arturiano [62]

Answer:

The kinetic energy of the merry-goround after 3.62 s is 544J

Explanation:

Given :

Weight w = 745 N

Radius r = 1.45 m

Force = 56.3 N

To Find:

The kinetic energy of the merry-go round after 3.62 = ?

Solution:

Step 1: Finding the Mass of merry-go-round

$m = \frac{ weight}{g}$

$m = \frac{745}{9.81 }$

m = 76.02 kg

Step 2: Finding the Moment of Inertia of solid cylinder

Moment of Inertia of solid cylinder I = $0.5 \times m \times r^2$

Substituting the values

Moment of Inertia of solid cylinder I

=> $0.5 \times 76.02 \times (1.45)^2$

=> $0.5 \times 76.02\times 2.1025$

=> $79.91 kg.m^2$

Step 3: Finding the Torque applied T

Torque applied T = $F \times r$

Substituting the values

T = $56.3 \times 1.45$

T = 81.635 N.m

Step 4: Finding the Angular acceleration

Angular acceleration , $\alpha = \frac{Torque}{Inertia}$

Substituting the values,

$\alpha = \frac{81.635}{79.91}$

$\alpha = 1.021 rad/s^2$

Step 4: Finding the Final angular velocity

Final angular velocity , $\omega = \alpha \times t$

Substituting the values,

$\omega = 1.021 \times 3.62$

$\omega = 3.69 rad/s$

Now KE (100% rotational) after 3.62s is:

KE = $0.5 \times I \times \omega^2$

KE = $0.5 \times 79.91 \times 3.69^2$

KE = 544J

6 0

3 years ago

A 20N net force acts upon an object with a mass of 4. 0kg. What is the object's acceleration?

galina1969 [7]

Answer:

5ms^-2

Explanation:

F = 20N, m = 4.0kg, a = ?

Using the formula for Force

F = ma

Then making acceleration the subject of the formula

a = F/m

a = 20N/4.0kg

a = 5ms^-2

8 0

2 years ago

15 POINTS

zhenek [66]

Answer:

Acceleration of the bullet while moving into the clay is -312500 ms⁻²

Explanation:

Given data

Initial velocity (v₁) = 250 m/s Final velocity (v₂) = 0 m/s

Distance (s)= 0.1 m acceleration (a) = ?

Using 3rd equation of motion

2as = v₂²- v₁²

2 × a × 0.1 = 0² - 250²

0.2 × a = -62500

a = -62500/0.2

a = -312500 m/s²

Here negative sign indicate that bullet slow down and it is deceleration i.e negative acceleration.

3 0

3 years ago