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3 years ago

If your chunk of gold weighed 1 N in which case would you have the largest mass of gold?

1 answer:

3 0

Ah ha ! Very interesting question.
Thought-provoking, even.

You have something that weighs 1 Newton, and you want to know
the situation in which the object would have the greatest mass.

Weight = (mass) x (local gravity)

Mass = (weight) / (local gravity)

Mass = (1 Newton) / (local gravity)

"Local gravity" is the denominator of the fraction, so the fraction
has its greatest value when 'local gravity' is smallest. This is the
clue that gives it away.

If somebody offers you 1 chunk of gold that weighs 1 Newton,
you say to him:

"Fine ! Great ! Golly gee, that's sure generous of you.
But before you start weighing the chunk to give me, I want you
to take your gold and your scale to Pluto, and weigh my chunk
there. And if you don't mind, be quick about it."

The local acceleration of gravity on Pluto is 0.62 m/s² ,
but on Earth, it's 9.81 m/s.

So if he weighs 1 Newton of gold for you on Pluto, its mass will be
1.613 kilograms, and it'll weigh 15.82 Newtons here on Earth.

That's almost 3.6 pounds of gold, worth over $57,000 !

It would be even better if you could convince him to weigh it on
Halley's Comet, or on any asteroid. Wherever he's willing to go
that has the smallest gravity. That's the place where the largest
mass weighs 1 Newton.

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Planets are not uniform inside. Normally, they are densest at the center and have decreasing density outward toward the surface.

elena-s [515]

Answer:

$g=13.42\frac{m}{s^2}$

Explanation:

1) Notation and info given

$\rho_{center}=13000 \frac{kg}{m^3}$ represent the density at the center of the planet

$\rho_{surface}=2100 \frac{kg}{m^3}$ represent the densisty at the surface of the planet

r represent the radius

$r_{earth}=6.371x10^{6}m$ represent the radius of the Earth

2) Solution to the problem

So we can use a model to describe the density as function of the radius

$r=0, \rho(0)=\rho_{center}=13000 \frac{kg}{m^3}$

$r=6.371x10^{6}m, \rho(6.371x10^{6}m)=\rho_{surface}=2100 \frac{kg}{m^3}$

So we can create a linear model in the for y=b+mx, where the intercept b= $\rho_{center}=13000 \frac{kg}{m^3}$ and the slope would be given by $m=\frac{y_2-y_1}{x_2-x_1}=\frac{\rho_{surface}-\rho_{center}}{r_{earth}-0}$

So then our linear model would be

$\rho (r)=\rho_{center}+\frac{\rho_{surface}-\rho_{center}}{r_{earth}}r$

Since the goal for the problem is find the gravitational acceleration we need to begin finding the total mass of the planet, and for this we can use a finite element and spherical coordinates. The volume for the differential element would be $dV=r^2 sin\theta d\phi d\theta dr$ .

And the total mass would be given by the following integral

$M=\int \rho (r) dV$

Replacing dV we have the following result:

$M=\int_{0}^{2\pi}d\phi \int_{0}^{\pi}sin\theta d\theta \int_{0}^{r_{earth}}(r^2 \rho_{center}+\frac{\rho_{surface}-\rho_{center}}{r_{earth}}r)$

We can solve the integrals one by one and the final result would be the following

$M=4\pi(\frac{r^3_{earth}\rho_{center}}{3}+\frac{r^4_{earth}}{4} \frac{\rho_{surface}-\rho_{center}}{r_{earth}})$

Simplyfind this last expression we have:

$M=\frac{4\pi\rho_{center}r^3_{earth}}{3}+\pi r^3_{earth}(\rho_{surface}-\rho_{center})$

$M=\pi r^3_{earth}(\frac{4}{3}\rho_{center}+\rho_{surface}-\rho_{center})$

$M=\pi r^3_{earth}[\rho_{surface}+\frac{1}{3}\rho_{center}]$

And replacing the values we got:

$M=\pi (6.371x10^{6}m)^2(\frac{1}{3}13000 \frac{kg}{m^3}+2100 \frac{kg}{m^3})=8.204x10^{24}kg$

And now that for any shape the gravitational acceleration is given by:

$g=\frac{MG}{r^2_{earth}}=\frac{(6.67408x10^{-11}\frac{m^3}{kgs^2})*8.204x10^{24}kg}{(6371000m)^2}=13.48\frac{m}{s^2}$

4 0

3 years ago

Why do most scientists follow a set order of steps when carrying out a scientific investigation?

eimsori [14]

<span>Scientists follow a set order of steps when carrying out a scientific investigation to make sure that the method, interpretation and results that they have obtained are repeatable and reliable. This kind of information can be truly said that their data is true and valid. </span>

6 0

3 years ago

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An example of a high explosive is what?

nordsb [41]

The answer is Dynamite.
Explosive, any substance or device that can be made to produce a volume of rapidly expanding gas in an extremely brief period. Chemical explosives are of two types; detonating, or high explosives and deflagrating, or low, explosives. Detonating explosives, such as TNT and dynamite, are characterized by extremely rapid decomposition and development of high pressure, whereas deflagrating explosives, such as black and smokeless powders, involve merely fast burning and produce relatively low pressures.

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3 years ago

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Only 5 questions plz answer.

Masja [62]

Question 18: a
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What is meant by a colored shadow? If "shadow" means

Natasha_Volkova [10]

Explanation:

Red, green, and blue are therefore called additive primaries of light. ... When you block two lights, you see a shadow of the third color—for example, block the red and green lights and you get a blue shadow. If you block only one of the lights, you get a shadow whose color is a mixture of the other two.

First, your definition of a shadow is incorrect. A shadow is an area that receives less light than its surroundings because a specific source of light is blocked by whatever is "casting" the shadow. Your example of being outside reveals this. The sky and everything around you in the environment (unless you are surrounded by pitch black buildings) is sending more than enough light into your shadow, to reveal the pen to your eyes. The sky itself diffuses the sunlight everywhere, and the clouds reflect plenty of light when they are not directly in front of the Sun.

If you are indoors and have two light bulbs, you can throw two shadows at the same time, possibly of different darknesses, depending on the brightness of the light bulbs.

It can take a lot of work to get a room pitch black. One little hole or crack in some heavy window curtains can be enough to illuminate the room. There are very few perfectly dark shadows.

4 0

3 years ago