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deff fn [24]
3 years ago
15

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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A man has 887.5 J of kinetic energy while running with a velocity of 5 m/s. What is his mass?
monitta

Answer:

The mass of the man is 71 kg

Explanation:

Given;

kinetic energy of the man, K.E = 887.5 J

velocity of the man, v = 5 m/s

The mass of the man is calculated as follows;

K.E = ¹/₂mv²

where;

m is the mass of the man

2K.E = mv²

m = 2K.E / v²

m = (2 x 887.5) / (5)²

m = 71 kg

Therefore, the mass of the man is 71 kg

7 0
3 years ago
If the sprinter from the previous problem accelerates at that rate for 20 m, and then maintains that velocity for the remainder
kakasveta [241]

Question:

A 63.0 kg sprinter starts a race with an acceleration of 4.20m/s square. What is the net external force on him? If the sprinter from the previous problem accelerates at that rate for 20m, and then maintains that velocity for the remainder for the 100-m dash, what will be his time for the race?

Answer:

Time for the race will be t = 9.26 s

Explanation:

Given data:

As the sprinter starts the race so initial velocity = v₁ = 0

Distance = s₁ = 20 m

Acceleration = a = 4.20 ms⁻²

Distance = s₂ = 100 m

We first need to find the final velocity (v₂) of sprinter at the end of the first 20 meters.

Using 3rd equation of motion

(v₂)² - (v₁)² = 2as₁ = 2(4.2)(20)

v₂ = 12.96 ms⁻¹

Time for 20 m distance = t₁ = (v₂ - v ₁)/a

t₁ = 12.96/4.2 = 3.09 s

He ran the rest of the race at this velocity (12.96 m/s). Since has had already covered 20 meters, he has to cover 80 meters more to complete the 100 meter dash. So the time required to cover the 80 meters will be

Time for 100 m distance = t₂ = s₂/v₂

t₂ = 80/12.96 = 6.17 s

Total time = T = t₁ + t₂ = 3.09 + 6.17 = 9.26 s

T = 9.26 s

5 0
3 years ago
In operant conditioning, many complex behaviors are learned through shaping. T or F
Snezhnost [94]

Answer:  

true

Explanation:

8 0
3 years ago
A Ping-Pong ball has a mass of 2.3 g and a terminal speed of 9.4 m/s. The drag force is of the form bv^2 What is the value of b?
julia-pushkina [17]
Drag Force = bv^2 = ma; a = g = 9.81 m/s^2

b = mg/v^2 = (0.0023×9.81)/(9.4^2)

b = 0.000255


6 0
3 years ago
Water (2510 g ) is heated until it just begins to boil. if the water absorbs 5.01×105 j of heat in the process, what was the ini
natka813 [3]
E=energy=5.09x10^5J = 509KJ 
<span>M=mass=2250g=2.25Kg </span>
<span>C=specific heat capacity of water= 4.18KJ/Kg </span>
<span>ΔT= change in temp= ? </span>
<span>E=mcΔT </span>
<span>509=(2.25)x(4.18)xΔT </span>
<span>509=9.405ΔT </span>
<span>ΔT=509/9.405=54.1degrees </span>
<span>Initial temp = 100-54 = 46 degrees </span>
<span>Hope this helps :)</span>
5 0
3 years ago
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