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lions [1.4K]
3 years ago
6

Why isn't pluto considered a planet anymore?

Physics
1 answer:
aivan3 [116]3 years ago
3 0

The answer is. It did not meet the three criteria the IAU uses to define a full-sized planet.

1. It is in orbit around the Sun

2. It has sufficient mass to assume hydrostatic equilibrium

3. It has 'cleared the neighborhood' around its orbit

Pluto has not 'cleared its neighborhood'

You might be interested in
supose at 20 degree celsius the resistance of Tungsten thermometer is 154.9. WHen placed in a particular solution , the resistan
saw5 [17]

Answer:

T₂ = 95.56°C

Explanation:

The final resistance of a material after being heated is given by the relation:

R' = R(1 + αΔT)

where,

R' = Final Resistance = 207.4 Ω

R = Initial Resistance = 154.9 Ω

α = Temperature Coefficient of Resistance of Tungsten = 0.0045 °C⁻¹

ΔT = Change in Temperature = ?

Therefore,

207.4 Ω = 154.9 Ω[1 + (0.0045°C⁻¹)ΔT]

207.4 Ω/154.9 Ω = 1 + (0.0045°C⁻¹)ΔT

1.34 - 1 = (0.0045°C⁻¹)ΔT

ΔT = 0.34/0.0045°C⁻¹

ΔT = 75.56°C

but,

ΔT = Final Temperature - Initial Temperature

ΔT = T₂ - T₁ = T₂ - 20°C

T₂ - 20°C = 75.56°C

T₂ = 75.56°C + 20°C

<u>T₂ = 95.56°C</u>

7 0
3 years ago
A trip is taken that passes through the following points in order
riadik2000 [5.3K]

Answer:

35, I got you bro, i got you

8 0
3 years ago
Suppose you exert a 25-N force to lift a ball 0.4 m in 2 s. How much work is done?
Sergio [31]

work is force x distance = 25 x 0.4

= 2.5x4 = 10joules

pwer would be 10j/2s watts .... 5 watts

3 0
3 years ago
A spherical ball of charged particles has a uniform charge density. In terms of the ball's radius R, at what radial distances in
eimsori [14]

Answer:

Electric field at radius r inside the solid sphere is

Electric field at radius r between inner radius and outer radius of the shell is

E=0 N/C

Explanation:Given

The radius of the solid sphere is

The charge on the solid sphere is

The inner radius of the shell is

The outer radius of the shell is

The total charge on the shell is

PART(A)

The magnitude of electric field at radius r where  \\The volumetric charge density of the solid sphere will be

The charge enclosed by the radius r inside the solid sphere is

rho=

According to gauss law

PART(B)

The electric field at radius r where

The shell is conducting so due to induction of charge

The charge induced on the inner surface of the shell is equal in magnitude of the total charge on the solid sphere but polarity will be changed because a conducting shell has no net electric field inside the shell

So,

The charge on the inner surface of the shell is

Due to conservation of the charge on the shell

The charge accumulated on the outer surface of the shell is

The charge enclosed by the radius r where

According to gauss law

Read more on Brainly.com - brainly.com/question/13242041#readmore

3 0
3 years ago
Convertir 340,5 grados Fahrenheit a centígrados
Hunter-Best [27]

Answer:

171.38889

Explanation:

not really any i just looked it up,

6 0
3 years ago
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