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3 years ago

Can black wholes be an exit to the universe,why or why not? (just curious).

Physics

2 answers:

Sidana [21]3 years ago

8 0

I really doubt because the universe is supposed to be infinite

Paul [167]3 years ago

4 0

Well, it's hard to say that. Black hole is still a mysterious object with GREAT gravitational pull, object with less than the speed of light just goes into it and never back. Black hole maybe a gateway for some other place, many scientist think, there could be a "White hole" for it's other end. If so, they behave like a wormhole, (like a tunnel, which connects two position by folding them into time-space)!!

It's still in debate, if you really like cosmology, Astronomy, and anything about celestial objects, then please PM me!!!

Hope this helps!

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7. A beam of light converges at a point P. Now a lens is placed in the path of the convergent

Rudiy27

Answer:

Lens at a distance = 7.5 cm

Lens at a distance = 6.86 cm (Approx)

Explanation:

Given:

Object distance u = 12 cm

a) Focal length = 20 cm

b) Focal length = 16 cm

Computation:

a. 1/v = 1/u + 1/f

1/v = 1/20 + 1/12

v = 7.5 cm

Lens at a distance = 7.5 cm

b. 1/v = 1/u + 1/f

1/v = 1/16 + 1/12

v = 6.86 cm (Approx)

Lens at a distance = 6.86 cm (Approx)

5 0

3 years ago

A bungee cord is 30m long and when stretched a distance x itexerts a restoring force of magnitude kx. Your father-in-law (mass95

madam [21]

Answer:

The distance the bungee cord that would be stretched 0.602 m, should be selected when pulled by a force of 380 N.

Explanation:

As from the given data

the length of the rope is given as l=30 m

the stretched length is given as l'=41m

the stretched length required is give as y=l'-l=41-30=11m

the mass is m=95 kg

the force is F=380 N

the gravitational acceleration is g=9.8 m/s2

The equation of k is given by equating the energy at the equilibrium point which is given as

$U_{potential}=U_{elastic}\\mgh=\dfrac{1}{2} k y^2\\k=\dfrac{2mgh}{y^2}$

Here

m=95 kg, g=9.8 m/s2, h=41 m, y=11 m so

$k=\dfrac{2mgh}{y^2}\\k=\dfrac{2\times 95\times 9.8\times 41}{11^2}\\k=630.92 N/m$

Now the force is

$F=kx\\$ or

$x=\dfrac{F}{k}\\$

So here F=380 N, k=630.92 N/m

$x=\dfrac{F}{k}\\x=\dfrac{380}{630.92}\\x=0.602 m$

So the distance is 0.602 m

6 0

3 years ago

A well-insulated cup of water is too hot to make tea, so you mixed it with an equal amount of cool water to lower the temperatur

KonstantinChe [14]

Answer:The change in entropy of the total amount of water is negative as a result of the mixing.

Explanation:If you increase temperature, you increase entropy

Also More energy gives you greater entropy and randomness of the atoms.

4 0

3 years ago

A vertical spring stretches 3.4 cm when a 12-g object is hung from it. The object is replaced with a block of mass 26 g that osc

Mariulka [41]

Answer:

Period of motion is approximately 0.5447 seconds

Explanation:

We start by calculating the constant "k" of the spring which can be derived from the fact that an object of mass 12 g produced a stretch of 3.4 cm: (we write everything in SI units)

F = k * x

0.012 kg * 9.8 m/s^2 = k 0.034 m

k = 0.012 kg * 9.8 m/s^2 / (0.034 m)

k = 3.46 N/m

now we use the formula for the period (T) of a spring of constant k with a hanging mass 'm':

$T=2\pi\,\sqrt{\frac{m}{k} }$

which in our case becomes:

$T=2\pi\,\sqrt{\frac{0.026}{3.46} } \approx 0.5447\,\,sec$

4 0

3 years ago

What engine thrust is required for a rocket of mass 35 kg to leave the launching pad? 3.5 N. 35 N. 351 N. 3,500 N. 35,000 N.

soldi70 [24.7K]

In order for the object to move upward, it needs an upward force
that's at least equal to its own weight.

Weight = (mass) x (gravity) = (35 kg) x (9.8 m/s²) = 343 N.

The engine thrust has to be more than 343 N.

7 0

3 years ago