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likoan [24]
4 years ago
13

B. how does the separation between galaxies compare to the separation between stars? based on your answer, discuss the likelihoo

d of galactic collisions in comparison to the likelihood of stellar collisions.
Physics
1 answer:
leva [86]4 years ago
4 0
Galaxies are much further apart than stars. This is the reason why they are less likely to collide and the likelihood of galactic collision is much smaller  than the likelihood of stellar collision. Example for galaxy collision is the collision of the Milky Way galaxy with Andromeda. It is estimated that the collision will be <span>in about 4.5 billion years. </span> 
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A student decides they would like to have hands-on experience when it comes to transformers. The student takes an iron core (squ
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Answer:

a ) 3 v

b ) 400 mA

Explanation:

In case of transformer there is relation between no of turns in primary and secondary coils and voltage input and output as follows

V₁ / V₂ = n₁ / n₂

Where V₁ and V₂ are input and output voltage and n₁ and n₂ are no of turns of primary and secondary coils.

Here V₁ = 1.5 v , V₂ = ? , n₁ = 10 , n₂ = 20

Put the values in the equation above

1.5 / V₂ = 10 / 20

V₂ = 3 v.

Similarly formula for current is as follows

I₁ / I₂ = n₂ / n₁

Whereas I₁ and I₂ are current in primary and secondary coil

Put the values in the equation above

I₁ / 200 mA = 20 / 10

I₁ = 400 mA.

6 0
3 years ago
Which change to a circuit is most likely to decrease its electrical power?
Harlamova29_29 [7]

Answer:

It's D

Explanation:

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3 years ago
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The frequency of a sound wave is 457
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I think it's 1.03412969 or 1.03
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Each shot of the laser gun favored by Rosa the Closer, the intrepid vigilante of the lawless 22nd century, is powered by the dis
Alisiya [41]

Answer:

a) 4.94e9 J  b) 1.07e10 J

Explanation:

The electric potential energy stored in a capacitor, expressed in terms of the value of the capacitance C, and the voltage between its terminals V, is as follows:

U =\frac{1}{2}*C*V^{2}

a) For the original capacitor, we can find directly U as follows:

U =\frac{1}{2}*1.81F*(73.9e3)^{2} V2 = 4.94e9 J  

U = 4.94*10⁹ J

b) Prior to find the electric potential energy of the upgraded capacitor, we need to find out the value of the capacitance C of this capacitor, which is identical to the original, except that has a different dielectric constant.

As the capacitance is proportional to the dielectric constant, we can write the following proportion:

ε₂ / ε₁ = \frac{943}{435}= \frac{C2}{C1} =\frac{Cx}{1.81F}

Cx =\frac{1.81F*943}{435} = 3.92 F

Once calculated the new value of the capacitance, as V remains the same, we can find the electric potential energy for the upgraded capacitor as follows:

U =\frac{1}{2}*3.92F*(73.9e3)^{2} V2 = 1.07e10 J

⇒ U = 1.07*10¹⁰ J

8 0
3 years ago
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