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3 years ago

Does time stop in a black hole

Physics

2 answers:

lbvjy [14]3 years ago

4 0

Answer:

In standard GR, nothing exists at the center of a black hole. The center of a black hole is a singularity, and because GR fails at that point it is simply removed from the manifold. That means that the singularity is not part of spacetime.

To answer your question more realistically, we believe that GR is an approximate theory that fails well before you reach the center. Unfortunately, we have no good alternative theory with which to answer the question in the region where GR fails. We simply don’t have any data from that regime and it is very hard to formulate a good theory without data. So there very well could be time at the center, but we simply don’t have a good way to even guess.

kifflom [539]3 years ago

4 0

Answer:

Black hole is a region of space having a gravitational field so intense that no matter or radiation can escape

Actually time can't be stop.

although black hole slower a time

Explanation:

Time never cannot be stop totally.

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Compute the dot product of the vectors u and v, and find the angle between the vectors. Bold v equals 7 Bold i minus Bold j and

OLga [1]

Answer:

$\theta = 106.3 degree$

Explanation:

As we know that

$\vec w = -\hat i + 7\hat j$

$\vec v = 7\hat i - \hat j$

also we know that

$\vec v. \vec w = -14$

it is given as

$\vec v. \vec w = (-\hat i + 7\hat j).(7\hat i - \hat j)$

$\vec v. \vec w = - 7 - 7 = -14$

also we can find the magnitude of two vectors as

$|v| = \sqrt{(-1)^2 + (7)^2}$

$|v| = \sqrt{50}$

similarly we have

$|w| = \sqrt{(7^2) + (-1)^2}$

$|w| = \sqrt{50}$

now we know the formula of dot product as

$\vec v. \vec w = |v||w| cos\theta$

$-14 = (\sqrt{50})^2cos\theta$

$\theta = cos^{-1}(\frac{-14}{50})$

$\theta = 106.3 degree$

3 0

3 years ago

A common flashlight bulb is rated at 0.23 a and 2.9 v (the values of the current and voltage under operating conditions). if the

allochka39001 [22]

The resistance at operating temperature is R = V/I = 2.9 V / 0.23A = 12.61 ohmsT from R – R0 = Roalpha (T – T0), we find that:T = T0 + 1/alpha (R/R0 -1) = 20 degrees Celsius + (1/ 4.3 x 10^-3/K) (12.61 ohms/ 1.1 ohms – 1)T = 2453.40 degrees Celsius

6 0

3 years ago

The table below describes some features of methods used to generate electricity. Name method 4.

alina1380 [7]

Answer:

Hydroelectricity

Explanation:

Because of flooding of water, we can assume that the electricity was generated by Water which is known as Hydroelectricity.

4 0

3 years ago

Calculate, for the judge, how fast you were going in miles per hour when you ran the red light because it appeared Doppler-shift

sammy [17]

Answer:

The doppler effect equation is:

$f' = \frac{v +v0}{v - vs}*f$

In the equation we have frequencies, but then we have the wavelengths of the lights, remember the relation:

v = f*λ

then:

f = v/λ

and v is the speed of light, then:

f = c/λ

where:

f' is the observed frequency, in this case, is equal to f = (3*10^17nm/s)/550 nm

f is the real frequency, in this case, is (3*10^17nm/s)/650 nm

vs is the speed of the source, in this case, the source is not moving, then vs = 0 m/s.

v is the speed of the wave, in this case, is equal to the speed of light, v = 3*10^8 m/s

v0 is your speed, this is what we want to find.

Replacing those quantities in the equation, we get:

(3*10^17nm/s)/550 = (3*10^8 m/s + v0)/(3*10^8 m/s)*(3*10^17nm/s)/650 nm

(650nm)/(550nm) = (3*10^8 m/s + v0)/(3*10^8 m/s)

1.182*(3*10^8 m/s) = (3*10^8 m/s + v0)

1.182*(3*10^8 m/s) - (3*10^8 m/s) = v0 = 54,600,000 m/s

So your speed was 54,600,000 m/s, which is a lot.

6 0

3 years ago

Which of the following statements is accurate? A. Sound waves passing through the air will do so as transverse waves, which vibr

egoroff_w [7]

The answer is, C. the wavelength is measured in parallel to the direction of the wave, at any point, under the same repetition for any type of wave.

7 0

3 years ago

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