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Montano1993 [528]
2 years ago
12

Does everything in a black hole contain a singularity?

Physics
2 answers:
Lapatulllka [165]2 years ago
8 0
In the real universe, no black holes contain singularities. In general, singularities are the non-physical mathematical result of a flawed physical theory.
melamori03 [73]2 years ago
4 0

Answer:

All black holes contain singularities, however not all singularities involve black holes. A neutron star may be dense, matter the size of a pinhead can weigh as much as the earth, but there seems to be a mathematical cut-off point beyond which a black hole is formed.

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As a tractor pulls a plow across a field, it exerts a force on the plow in the forward direction. What is the "equal and opposit
Karolina [17]

Answer:

the force of the plow pulling backward on the tractor

Explanation:

7 0
3 years ago
A car of mass m accelerates from rest along a level straight track, not at constant acceleration but with constant engine power,
lisabon 2012 [21]
<span>Because P = W ÷ t, and W = F*t, you can substitute (W) for (F*t). Then substitute (F) for (m*a). This will leave you with P = (m*a*d)/t. Since you need velocity, youd want to solve for a so you can use v = a*t. a = (P*t)/(m*d) therefore, substituting a in v = a*t, v = (P*t*t)/(m*d)</span>
4 0
3 years ago
A
dangina [55]

Answer:

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7 0
2 years ago
What effects are jets and magnetic fields thought to have on a protostar?
alexgriva [62]

Answer:

the effects that a jet and the magnetic fields have on a ProStar Is :

Explanation:

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7 0
3 years ago
Electrical systems are governed by Ohm’s law, which states that V = IR, where V is the voltage, I is the current, and R is the r
ella [17]

Answer:

\frac{dR(t)}{dt}=0.06\Omega

Explanation:

Since R(t)=\frac{V}{I(t)}, we calculate the resistance rate by deriving this formula with respect to time:

\frac{dR(t)}{dt}=\frac{d}{dt}(\frac{V}{I(t)})=V\frac{d}{dt}(\frac{1}{I(t)})

Deriving what is left (remember that (\frac{1}{f(x)})'=-\frac{1}{f(x)^2}f'(x)):

\frac{d}{dt}(\frac{1}{I(t)})=-\frac{1}{I(t)^2}\frac{dI(t)}{dt}

So we have:

\frac{dR(t)}{dt}=-\frac{V}{I(t)^2}\frac{dI(t)}{dt}

Which for our values is (the rate of <em>I(t)</em> is decreasing so we put a negative sign):

\frac{dR(t)}{dt}=-\frac{24V}{(56A)^2}(-8A/s)=0.06\Omega

8 0
3 years ago
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