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3 years ago

Do black holes have an internal structure? If so, how might the internal structure be probed?

Physics

2 answers:

oksian1 [2.3K]3 years ago

5 0

Answer:

Explanation:

it can't be probed it would be like probing a goat's butt

-BARSIC- [3]3 years ago

4 0

Technically black holes have an internal structure as they are simply a small area taking up a large mass, like squishing the entire mass of planet Earth down to the size of a pea, but I don't see probing a black hole as possible with our current technology. If an object has so much mass that objects moving at the speed of light do not have the force to escape its gravity, trying to send probes to it is impossible (also the area of a black hole is so small that probes would have to be microscopic to even be able to study it up close like a probe we'd send to the moon). But, that doesn't mean it's impossible. For example we can probe a black hole without going to it by using sensors of different types as well as telescopes. There are no nearby black holes so research on that level can only be done it miniscule amounts, but that doesn't make probing a black hole impossible.

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3 years ago

Two straight wires are in parallel and carry electrical currents in opposite directions with the same magnitude of LaTeX: 2.02.0

koban [17]

Answer:

$2.4\times 10^{-5} N$

Explanation:

We are given that

$I_1=I_2=2.0 A$

Distance between the wires, $d=5.0 mm=5\times 10^{-3} m$

1mm= $10^{-3} m$

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Length of one wire is infinite.

We have to find the magnitude of force between the two wires.

We know that magnetic force

$F=\frac{2\mu_0 I_1I_2 l}{4\pi d}$

Where $\frac{\mu_0}{4\pi}=10^{-7}$

$F=\frac{2\times 10^{-7}\times 2\times 2\times 0.3}{5\times 10^{-3}}$

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3 0

3 years ago

A truck covers 40.0 m in 7.45 s while uniformly slowing down to a final velocity of 3.50 m/s.

astra-53 [7]

Answer:

(A) Original velocity will be 7.244 m /sec

(B) Acceleration will be $-0.5026m/sec^2$

Explanation:

We have given distance covers by truck s = 40 m

Time taken by truck to cover this distance t = 7.45 sec

Final velocity v = 3.50 sec

According to second equation of motion

$S=ut+\frac{1}{2}at^2$

$40=u\times 7.45+\frac{1}{2}\times a\times 7.45^2$

$7.45u+27.751a=40$ -----eqn 1

According to first equation of motion

v = u + at

So $3.5=u+7.45a$ -----eqn2

Solving equation 1 and 2

a = $-0.5026m/sec^2$

And u = 7.244 m /sec

(A) Original velocity will be 7.244 m /sec

(B) Acceleration will be

$-0.5026m/sec^2$

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4 years ago

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valentina_108 [34]

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