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## Can ever a beam of light be trapped in an orbit of a star?why?

Phillip E.
Featured 1 month ago

A beam of light can't be trapped in an orbit of a star.

#### Explanation:

The mathematics of General Relativity are really hard, so I will try and explain in English.

Now we used to think of three dimensional space with time being an independent variable ticking at a constant rate. We now know that this is not correct.

We live in four dimensional spacetime. Time passes at different rates for different observers. An observer in a spaceship sees time passing more slowly on another spaceship which is travelling faster of is in stronger gravity.

Massive bodies such as stars cause spacetime to curve. A body orbiting a star is actually travelling in a straight line, called a geodesic, in spacetime. The close to the star the orbit the faster the body needs to travel to stay in orbit.

Light consists of photons which have zero mass. This means that they travel at the speed of light, have no mass and do not experience the passing of time.

For light to be trapped in orbit, the orbital speed must be the speed of light. It also means that gravity must be strong enough to stop time completely. This condition is only known to occur at the event horizon of a black hole. Also, the event horizon is a singularity which means that it and everything inside if are completely invisible to everything else in the universe.

To sum up nothing except for a black hole is massive enough or dense enough to capture light in orbit. Light which enters the event horizon of a black hole can't be considered to be in orbit as it is forever lost to our universe.

## How would you summarize the second Law of Thermodynamics?

Truong-Son N.
Featured 1 month ago

Levine puts it rather elegantly:

As some wit has put it: "The first law says you can't win; and the second law says you can't break even." (Physical Chemistry, Levine, pg. 79)

Or, in my own words,

The first law says energy is conserved, and the second law says the change in entropy of the universe is nonnegative.

I like Levine's wording better, honestly.

• The first law of thermodynamics states:

Energy cannot be created nor destroyed. That is, energy is conserved in the universe.

Therefore, you "can't win" - you can't obtain more energy than you put in. It all has to be accounted for (since the universe is so big, it is quasi-closed).

• The Kelvin-Planck statement of the second law of thermodynamics is (Physical Chemistry, Levine, pg. 79):

It is impossible for a system to undergo a cyclic process whose sole effects are the flow of heat into the system from a heat reservoir and the performance of an equivalent amount of work by the system on the surroundings.

In other words, the work output can be no greater than the heat input, i.e. the energy efficiency $e$ can never be 100%:

$e = | w \frac{|}{{q}_{H}}$, " "" "0% < e < 100%

where $w$ is the work output from the engine, and ${q}_{H}$ is the heat input from the hot reservoir into the engine.

Another way of saying it is therefore, "you can't break even!"

Put in yet another way - since some heat input is always lost outwards (towards the cold reservoir in the engine scheme), there is always more heat to give more motion.

Entropy, the amount of energy dispersal, increases with more motion present, so the change in entropy of the universe must be nonnegative.

## Can two atoms traveling fast enough create an explosion?

Mark C.
Featured 1 month ago

Yes, but a very, very small one.

#### Explanation:

This happens often (in any chemical reaction for example) but even two particles of the most powerful explosive we have (PETN is a good choice) will only yield billionths of millionths of a joule, and the majority of that is probably light not sound.

Two atoms are just not massive enough for their electrons rearranging (a collision / interaction / explosion) to cause anything noticeable and that may be the most interesting answer .... an atom is unimaginably tiny.

If they were expanded to the size of orange pips (5mm diameter say) the smallest visible speck of dust would become about 50 million kilometres across on the same scale. That’s from here to Mars.

A back of envelope calculation reveals that two particles of PETN totally annihilating in a nuclear reaction (way more powerful than an electronic interaction) would only yield around $100 \mu$J, and an atom is much smaller than that.

## Question details are below ?

dk_ch
Featured 3 weeks ago

Let

the mass of the cannon be $M = 1000 k g$

the mass of the shell be $m = 100 k g$

the angle of inclination of the inclined plane be $\theta = {45}^{\circ}$

the coefficient of friction between the cannon and inclined plane be $\mu = 0.5$

the velocity of the shell be $v = 180 k m \text{/"hr=(180xx1000)/3600=50m"/} s$

If the velocity of recoil of the cannon be $V$ then by conservation of momentum

$V = - \frac{m v}{M} = - \frac{100 \times 50}{1000} = - 5 m \text{/} s$
( the negative sign indicates the reverse direction of the cannon w r to the shell.)

If the height of ascent of the center of mass of the cannon over the inclined plane be $h$ m from its initial position, then the slant height covered by the cannon due to its ascent will be $L = \frac{h}{\sin} \theta$

By conservation of energy the sum of gravitational potential energy gained by the cannon at height h m and work done against the frictional force to reach at this height will be equal to the kinetic energy of the cannon due to its velocity of recoil.

So

$M g h + \mu M g \cos \theta \times L = \frac{1}{2} M {V}^{2}$

$\implies \cancel{M} g h + \mu \cancel{M} g \cos \theta \times \frac{h}{\sin} \theta = \frac{1}{2} \cancel{M} {V}^{2}$

$\implies g h + \mu g \cot \theta \times h = \frac{1}{2} {V}^{2}$

$\implies 10 h + 0.5 \times 10 \times \cot {45}^{\circ} \times h = \frac{1}{2} \times {5}^{2}$

$\implies 15 h = \frac{25}{2}$

$\implies h = \frac{25}{30} = \frac{5}{6}$ m (OPTION (2)

## Two smaller spherical raindrops collide and merge into one larger spherical raindrop. What is the radius and surface energy of this larger drop, compared to the two smaller ones?

MetaPhysik
Featured 1 month ago

Radius increases by 26% while the surface energy drops 21%

#### Explanation:

For a sphere, its volume and surface A area are:
V=(4pi)/2R^3; A = 4pir^2
${E}_{s}$ = surface energy = gamma

A) Volume consideration.

Let
${r}_{1} =$radius of small droplet 1
${r}_{2} =$radius of small droplet 2
$R =$radius of merged droplet

Because density of water and total mass of the droplets do not change, the volume before and after the merger must be the same.

Volume of two small droplets = Volume of merged droplet

$\frac{4 \pi}{3} {r}_{1}^{3} + \frac{4 \pi}{3} {r}_{2}^{3} = \frac{4 \pi}{3} {R}^{3}$

${R}^{3} = {r}_{1}^{3} + {r}_{2}^{3}$
R = (r_1^3+r_2^3)^(⅓)

If the small droplets have the same size, then ${r}_{1} = {r}_{2} = r$,

R= (2r^3)^(⅓) = root(3)2 r = 1.26r

$\frac{\Delta r}{r} =$(1.26r-r)/r = 0.26 = 26%

The radius of the merged droplet is 26% larger.

(B) Surface area consideration , assuming small droplets have identical radius.

Let
$\sigma =$ surface tension
${E}_{s} =$ surface energy $= \sigma A = 4 \pi {r}^{2} \sigma$
${E}_{1} = {E}_{2} = 4 \pi {r}^{2} \sigma =$ surface energy of small droplets
${E}_{3} = 4 \pi {R}^{2} \sigma$ = surface energy of the merger droplet

The percentage change in surface energy is:

$\frac{\Delta E}{{E}_{1} + {E}_{2}} = \frac{{E}_{3} - \left({E}_{1} + {E}_{2}\right)}{{E}_{1} + {E}_{2}} = \frac{{E}_{3} - 2 {E}_{1}}{2 {E}_{1}} = {E}_{3} / \left(2 {E}_{1}\right) - 1$

$\because R = \sqrt[3]{2} r$
therefore E_3/(2E_1)-1=(4piR^2sigma)/(8pir^2sigma)-1= (4pisigma(root(3)2r)^2)/(8pir^2sigma)-1=-0.21 =- 21% #

The surface energy has reduced by 21%.

## You push a crate full of nooks across the floor at a constant speed of 0.5 meter per second. You then remove some of the books and push exactly the same as you did before. How does the crate’s motion differ, if at all?

Gió
Featured 1 week ago

I would say that the crate now is accelerated.

#### Explanation:

During the first part you exert the exact force $\vec{F}$ to maintain the box in uniform motion, i.e., constant velocity (acceleration$= 0$). In doing so you are winning against kinetic friction ${f}_{k}$ but just! So your force and friction balance (almost) so that there is no acceleration but only constant velocity.

When you remove some books the box becomes lighter so that the normal reaction $N$ decreases as well reducing the contribution of kinetic friction. Now if you push with the same force as before you face less friction so that you get a component of force that gives you acceleration!

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