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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.

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(a) You short-circuit a 20 volt battery by connecting a short wire from one end of the battery to the other end. If the current

erica [24]

(a) $1.11 \Omega$

When the battery is short-cut, the only resistance in the circuit is the internal resistance of the battery. Therefore, we can apply Ohm's law:

$r=\frac{V}{I}$

where

V = 20 V is the voltage across the internal resistance of the battery

I = 18 A is the current flowing through it

Solving the equation,

$r=\frac{20 V}{18 A}=1.11\Omega$

(b) 360 W

The power generated by the battery is given by the equation

$P=VI$

where

V = 20 V is the voltage of the battery

I = 18 A is the current

Substituting into the formula,

$P=(20 V)(18 A)=360 W$

(c) 360 J

The energy dissipated by the internal resistance is given by

$E=Pt$

where

P = 360 W is the power generated

t = 1 s is the time

Solving the equation, we find

$E=(360 W)(1 s)=360 J$

(d) 1.65 A

The battery is now connected to a $R=11 \Omega$ resistor. This means that the internal resistance of the battery is now connected in series with the other resistor R: so, the total resistance of the circuit is

$R_T = r+R=1.11 \Omega +11 \Omega = 12.11 \Omega$

And so, the current flowing through the circuit is

$I=\frac{V}{R_T}=\frac{20 V}{12.11\Omega}=1.65 A$

(e) 29.9 W

The power dissipated in the external resistor is given by

$P=I^2 R$

where

I = 1.65 A is the current

$R=11 \Omega$ is the resistance

Solving the equation, we find

$P=(1.65 A)^2(11 \Omega)=29.9 W$

(f) 18.17 V

The terminals of the voltmeter are placed at the two end of the battery. The battery provides an emf of 20 V, however due to the internal resistance, some of this voltage is dropped across the internal resistance. Therefore, the actual potential difference that will be read by the voltmeter will be:

$V=\epsilon - Ir =20 V -(1.65 A)(1.11 \Omega)=18.17 V$

4 0

3 years ago

Which statement best describes scientific laws? (1 point)

erica [24]

Answer:

The best statement would be scientific laws describe phenomenon that can be observed in nature. Sometimes, laws are related to principle defining and predicting a certain happening without explaining its nature. It is different with ‘hypothesis’ or ‘theory’ as it aims to answer only the ‘what’ (observation only)...

Explanation:

hopefully this is correct

3 0

3 years ago

Suppose a 4.0 kg mass of water gained heat and increased in temperature from 10°C to 15°C. The specific heat capacity of water i

Alex73 [517]

It will be <span>Your answer </span>84 ok

8 0

3 years ago

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Please help, confused on this whole lesson

ira [324]

Answer:

1,4,3and 2 would happen hope this helps

3 0

3 years ago

Read 2 more answers

Consider two vectors A~ and B~ and their resultant A~ +B~ . The magnitudes of the vectors A~ and B~ are, respectively, 18.2 and

saw5 [17]

Answer:

15.76°

Explanation:

Hi!

Let C~ = A~ + B~

Then

C~ dot A~ = |C| |A| cos Ф

C~ dot A~ = (A~ + B~) dot A~ = |A|^2 + |B| |A| cos(120)

Therefore

cos Ф = (|A| + |B| cos(120)) / |C| = 15.65/|C|

and

|C|^2 = |A|^2 + |B|^2 + 2 |B| |A| cos(120)

|C| = 16.261

cos Ф = 15.65/16.261

Ф = 15.76°

7 0

4 years ago