All categories

3 years ago

Black holes are the final stage of what star

2 answers:

8 0

“A black hole will suck in any nearby stuff it can. If a star or galaxy is rotating around a black hole this will be meta-stable with the gravity of the black hole matching the rotational acceleration of the star. Various events that produce imbalance, like interactions between different bodies and solar wind, allow the black hole to pick up the resulting debris.

Over time, the black hole will clear out it’s neighborhood. As the universe is expanding, and looks like it continue to do so indefinitely, our black hole will end up alone in a region of empty space.

Hawking radiation is a process at the “edge” of the black hole where quantum randomness results in creation of a particle and antiparticle pair on either side of the event horizon. One falls back into the black hole and the other radiates into space. This will slowly evaporate the black hole. This is a extremely slow process. It relies on the chance creation of particles pairs in a miroscopically thin surface bridging the event horizon. It has been calculated that this would take 10^67 years to evaporate a black hole with the mass of the Sun.

(Compare with the age of the universe, 10^10 years, and you can see how imperceptibly slow this process is. During your lifetime a black hole the mass of the Sun would lose way much than a microgram. By the time the black hole has radiated its mass, all stars will have long burned out, and the universe will have expanded so much that there are enormous and still increasing gaps between everything that even light will never cross.)

If the theory is right, and nothing unexpected happens in the long, long intervening time, the black hole will eventually evaporate in a final flash and all that will be left is an extremely weak radiation in apparent empty space”.

Andreyy893 years ago

4 0

Answer:

This:

Explanation:

Black holes occur when a massive star or larger reaches the final stage of it's lifespan. The star implodes and a black hole is the dying star's remains

You might be interested in

A plastic bottle is closed outdoors on a cold day when the temperature is −15.0°C and is later brought inside where the temperat

Lana71 [14]

Answer:

Pressure = 1.14 atm

Explanation:

Hello,

This question requires us to calculate the final pressure of the bottle after thermal equilibrium.

This is a direct application of pressure law which states that in a fixed mass of gas, the pressure of a given gas is directly proportional to its temperature, provided that volume remains constant.

Mathematically, what this implies is

P = kT k = P / T

P1 / T1 = P2 / T2 = P3 / T3 =........= Pn / Tn

P1 / T1 = P2 / T2

P1 = 1.0atm

T1 = -15°C = (-15 + 273.15)K = 258.15K

P2 = ?

T2 = 21.5°C = (21.5 + 273.15)K = 294.65K

P1 / T1 = P2 / T2

P2 = (P1 × T2) / T1

P2 = (1.0 × 294.65) / 258.15

P2 = 1.14atm

The pressure of the gas after attaining equilibrium is 1.14atm

3 0

3 years ago

An aircraft cannot reach flying speed on a short runway. Someone suggests reducing its cargo load. Which of Newton's laws is bei

ELEN [110]

Answer:

Second Law

Explanation:

Newton's second law states that the acceleration caused in a body is directly proportional to the force applied and inversely proportion to the mass of the body.

This is given by :

$Acceleration=\frac{Force}{Mass}$

In this case the suggestion given to reduce the aircraft's cargo load is the right move as reducing the load on the aircraft will decrease the mass of the whole aircraft. This in turn will help the aircraft to accelerate more as acceleration inversely varies with mass. Thus the aircraft will be able to reach its flying speed even on a short run way.

Hence, Newton's second law is applied.

6 0

4 years ago

Any help would be appreciated. Confused.

masya89 [10]

Answer:

q(problem 1) = 25,050 joules; q(problem 2) = 4.52 x 10⁶ joules

Explanation:

To understand these type problems one needs to go through a simple set of calculations relating to the 'HEATING CURVE OF WATER'. That is, consider the following problem ...

=> Calculate the total amount of heat needed to convert 10g ice at -10°C to steam at 110°C. Given are the following constants:

Heat of fusion (ΔHₓ) = 80 cal/gram

Heat of vaporization (ΔHv) = 540 cal/gram

specific heat of ice [c(i)] = 0.50 cal/gram·°C

specific heat of water [c(w)] = 1.00 cal/gram·°C

specific heat of steam [c(s)] = 0.48 cal/gram·°C

Now, the problem calculates the heat flow in each of five (5) phase transition regions based on the heating curve of water (see attached graph below this post) ... Note two types of regions (1) regions of increasing slopes use q = mcΔT and (2) regions of zero slopes use q = m·ΔH.

q(warming ice) = m·c(i)·ΔT = (10g)(0.50 cal/g°C)(10°C) = 50 cal

q(melting) = m·ΔHₓ = (10g)(80cal/g) 800 cal

q(warming water) = m·c(w)·ΔT = (10g)(1.00 cal/g°C)(100°C) = 1000 cal

q(evaporation of water) = m·ΔHv = (10g)(540cal/g) = 5400 cal

q(heating steam) = m·c(s)·ΔT = (10g)(0.48 cal/g°C)(10°C) = 48 cal

Q(total) = ∑q = (50 + 800 + 1000 + 5400 + 48) = 7298 cals. => to convert to joules, multiply by 4.184 j/cal => q = 7298 cals x 4.184 j/cal = 30,534 joules = 30.5 Kj.

Now, for the problems in your post ... they represent fragments of the above problem. All you need to do is decide if the problem contains a temperature change (use q = m·c·ΔT) or does NOT contain a temperature change (use q = m·ΔH).

Problem 1: Given Heat of Fusion of Water = 334 j/g, determine heat needed to melt 75g ice.

Since this is a phase transition (melting), NO temperature change occurs; use q = m·ΔHₓ = (75g)(334 j/g) = 25,050 joules.

Problem 2: Given Heat of Vaporization = 2260 j/g; determine the amount of heat needed to boil to vapor 2 Liters water ( = 2000 grams water ).

Since this is a phase transition (boiling = evaporation), NO temperature change occurs; use q = m·ΔHf = (2000g)(2260 j/g) = 4,520,000 joules = 4.52 x 10⁶ joules.

Problems containing a temperature change:

NOTE: A specific temperature change will be evident in the context of problems containing temperature change => use q = m·c·ΔT. Such is associated with the increasing slope regions of the heating curve. Good luck on your efforts. Doc :-)

5 0

3 years ago

Forces between similar molecules are said to be

podryga [215]

Answer:

Forces between similar molecules are said to be cohesive while those between different types of molecules are said to be adhesive.

Water 'beads' due to its strong cohesive forces. The meniscus of water in a glass tube is concave because the adhesive forces are strong.

Explanation:

The water in a tube has stronger adhesive forces between the water and glass molecules, so the cohesive forces between water molecules are weaker. That makes the water 'ascend' through the tube, giving a concave form of the meniscus. Another example is mercury, which is the opposite. In this case, the cohesive forces are stronger than the adhesive ones, thus the meniscus is convex.

4 0

3 years ago

What minimum energy is required to excite a vibration in HF?

Elodia [21]

Answer:

The energy of a vibrating molecule is quantized much like the energy of an electron in the hydrogen atom. The energy levels of a vibrating molecule are given by the equation: En=(n+21)hv where n is a quantum number with possible values of 1, 2, ... and v is the frequency of vibration.

Explanation:

hope it helps.

have a wonderful day!

7 0

3 years ago