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cluponka [151]
4 years ago
11

What causes the tail of a comet to become visible?

Physics
2 answers:
nlexa [21]4 years ago
5 0

Answer:

The correct answer is option C. Sun.

Explanation:

The tail of a comet as well as the comma are visible thanks to the influence of sunlight on these objects.

If there is a case that these objects cross the inner Solar System,<u> then they will be visible from Earth</u>. The reason why this occurs is that as the comet approaches the internal solar system, the materials that are inside the comet vaporize and flow out of the nucleus thanks to solar radiation, and while they do carry dust with them.

This dust is the one that reflects sunlight directly, while the <u>gases shine because of ionization. </u>

Usually we will need a telescope to see a comet, although there are some that can be seen with the naked eye.

zavuch27 [327]4 years ago
4 0

Answer:

it is the sun

Explanation:

i just did the test

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If the velocity of an airplane is doubled, then its kinetic energy<br> will increase by a factor of
ohaa [14]

Answer:4

Explanation:

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3 years ago
A car is traveling at 33.2 m/s. If the car has a kinetic energy of 688,900 J, what is the car's mass?​
aivan3 [116]

Answer:

1250kg

Explanation:

\frac{688900}{33.2\\^{2} } = 625

\frac{1}{2} m = 625

625 x 2 = 1250

5 0
3 years ago
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Find the kinetic energy of a 0.1-kilogram toy truck moving at the speed of 1.1 meters per second.
ZanzabumX [31]

Answer:

0.061 J

Explanation:

The kinetic energy of an object is given by

K=\frac{1}{2}mv^2

where

m is the mass of the object

v is its speed

For the toy truck in the problem, we have

m = 0.1 kg is its mass

v = 1.1 m/s is its speed

Putting the numbers into the equation, we find

K=\frac{1}{2}(0.1 kg)(1.1 m/s)^2=0.061 J

7 0
3 years ago
What happens at a reverse fault plane?
melamori03 [73]
<span>B. Compression causes upward vertical movement.</span>
5 0
3 years ago
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A generator with �# ' = 300 V and Zg = 50 Ω is connected to a load ZL = 75 Ω through a 50-Ω lossless line of length l = 0.15λ. (
ki77a [65]

Answer:

a. Zin = 41.25 - j 16.35 Ω

b. V₁ = 143. 6 e⁻ ¹¹ ⁴⁶

c.  Pin = 216 w

d. PL = Pin = 216 w

e. Pg = 478.4 w , Pzg = 262.4 w

Explanation:

a.

Zin = Zo * [ ZL + j Zo Tan (βl) ] / [ Zo + j ZL Tan (βl) ]  

βl = 2π / λ * 0.15 λ = 54 °

Zin = 50 * [ 75 + j 50 Tan (54) ] / [ 50 + j 75 Tan (54) ]

Zin = 41.25 - j 16.35 Ω

b.

I₁ = Vg / Zg + Zin ⇒ I₁ = 300 / 41.25 - j 16.35 = 3.24 e ¹⁰ ¹⁶

V₁ = I₁ * Zin = 3.24 e ¹⁰ ¹⁶ * ( 41.25 - j 16.35)

V₁ = 143. 6 e⁻ ¹¹ ⁴⁶

c.

Pin = ¹/₂ * Re * [V₁ * I₁]

Pin = ¹/₂ * 143.6 ⁻¹¹ ⁴⁶ * 3.24 e ⁻ ¹⁰ ¹⁶ = 143.6 * 3.24 / 2 * cos (21.62)

Pin = 216 w

d.

The power PL and Pin are the same as the line is lossless input to the line ends up in the load so

PL = Pin

PL = 216 w

e.

Pg Generator

Pg = ¹/₂ * Re * [ V₁ * I₁ ] = 486 * cos (10.16)

Pg = 478.4 w

Pzg dissipated

Pzg = ¹/₂ * I² * Zg = ¹/₂ * 3.24² * 50

Pzg = 262.4 w

4 0
4 years ago
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