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2 years ago

9

Why is Pluto considered less of an oddball planet now?

1 answer:

solmaris [256]2 years ago

4 0

Answer:

The International Astronomical Union (IAU) downgraded the status of Pluto to that of a dwarf planet because it did not meet the three criteria the IAU uses to define a full-sized planet. Essentially Pluto meets all the criteria except one—it “has not cleared its neighboring region of other objects

Explanation:

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An ion has unequal numbers of which two particles?

Scrat [10]

Protons and electrons

7 0

3 years ago

The potential at location A is 382 V. A positively charged particle is released there from rest and arrives at location B with a

jarptica [38.1K]

Answer: 247.67 V

Explanation:

Given

Potential At A $V_a=382\ V$

Potential at $V_c=785\ V$

when particle starts from A it reaches with velocity $v_b$ at Point while when it starts from C it reaches at point B with velocity $2v_b$

Suppose m is the mass of Particle

Change in Kinetic Energy of particle moving under the Potential From A to B

$q\cdot \left ( V_a-V_b\right )=0.5m\cdot (v_b)^2----1$

Change in Kinetic Energy of particle moving under the Potential From C to B

$q\cdot \left ( V_c-V_b\right )=0.5m\cdot (2v_b)^2-----2$

Divide 1 and 2 we get

$\frac{V_a-V_b}{V_c-V_b}=\frac{v_b^2}{4v_b^2}$

on solving we get

$V_b=\frac{4}{3}\cdot V_a-\frac{1}{3}\cdot V_c$

$V_b=\frac{743}{3}=247.67\ V$

4 0

3 years ago

Pls help I need this done soon 50 points

lidiya [134]

Answer:

1,3,5

Explanation:

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6 0

3 years ago

Read 2 more answers

when a electric current is passed through an insulated wire that is coiled around an iron core like a nail what is created?

Anika [276]

electromagnet

When a electric current is passed through an insulated wire that is coiled around an iron core, like a nail, an electromagnet is created.

8 0

2 years ago

A very delicate sample is placed 0.150 cm from the objective lens of a microscope. The focal length of the objective is 0.140 cm

iVinArrow [24]

Answer:

m = 14*26 = 364

Explanation:

overall magnification is given as m

$m = m_{o}* m_{e}$

mo magnification of objective lens

me magnification of EYE lens

where mo is given as

$m_{o} = \frac{v_{o}}{-u _{0}}$

and me as

$m_{e} = 1+\frac{D}{f_{e}}$

d is distant of distinct vision = 25.0 cm for normal eye

fe = focal length of eye piece

focal length of objective lense is 0.140 cm

we know that

$\frac{1}{v_{0}}-\frac{1}{u_{0}}=\frac{1}{f_{0}}$

$\frac{1}{v_{0}} = \frac{1}{u_{0}} + \frac{1}{f_{0}}$

$\frac{1}{v_{0}} = \frac{1}{0.150} + \frac{1}{0.14}$

$\frac{1}{v_{o}} = 2.1 cm$

$m_{o} = \frac{2.1}{0.150} = 14$

$m_{e} = 1+\frac{25}{1}$

$m_{e} =26$

$m = m_{o}* m_{e}$

m = 14*26 = 364

4 0

3 years ago