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KiRa [710]
3 years ago
11

What is the closest star that could go supernova?

Physics
1 answer:
sweet [91]3 years ago
6 0
The nearest star that can go supernova is either Spica or IK Pegasi
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Determine the vertex form of g(x) = x2 + 2x - 1. Which graph represents g(x)?<br> 5) Intro
Paladinen [302]

g(x) =  x ^{2}  + 2x - 1 =  {x}^{2}  + 2x - 1 + 2 - 2  =  \\  = (x + 1) {}^{2}  - 2

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i wanna learn a little bit of physics so if anyone could teach me a thing or two it would be much appreciated
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Answer:

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What is the frequency of a microwave of wavelength 3cm?
navik [9.2K]

Frequency = (speed) / (wavelength)

Speed = 3 x 10⁸ m/s

Wavelength = 3 cm = 0.03 m

Frequency = (3 x 10⁸  m/s) / (0.03 m)

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An object of unknown mass is initially at rest and dropped from a height h. It reaches the ground with a velocity v1 . The same
Hatshy [7]

Answer:

v_2=\sqrt{2}v_1

Explanation:

The velocity v₁ can be calculated with the kinematic formula:

v_1^{2} =v_0^{2} +2gh

Since the object is initially at rest, v₁ becomes:

v_1=\sqrt{2gh}

Where g is the acceleration due to gravity. Now, the velocity v₂ can be calculated with the same formula, but now the initial velocity is v₁:

v_2^{2}=v_1^{2} +2gh

Substituting v₁ in this expression and solving for v₂, we get:

v_2^{2}=(\sqrt{2gh} )^{2} +2gh=4gh\\\\\implies v_2=\sqrt{4gh}=2\sqrt{gh}

Now, dividing v₂ over v₁, we get the expression:

\frac{v_2}{v_1}=\frac{2\sqrt{gh} }{\sqrt{2gh}}=\sqrt{2}\\   \\\implies v_2=\sqrt{2}v_1

It means that v₂ is √2 times v₁.

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4 years ago
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