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3 years ago

Answer the following questions

Physics

2 answers:

kiruha [24]3 years ago

4 0

I ueueeieueueuekdududieisidudud

ad-work [718]3 years ago

4 0

Number one I think is A

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What is the value of sin r and sin i in physics

taurus [48]

Answer:

★The second law of refraction

The ratio of sine of angle of incidence to the sine of angle of refraction is a constant for a light of given colour and for a given pair of media. This law is also called Snell's law of refraction. If 'i' is the angle of incidence and 'r' is the angle of refraction then, Sin i/Sin r = constant

This constant value is called the refractive index of the second medium with respect to the first.

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3 years ago

Nvhfbvhefvhabjbnvjhhjaqiuv.....

Agata [3.3K]

Answer:

I don’t understand it

Explanation:

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3 years ago

Overcoming an object inertia always requires an

Zinaida [17]

Opposite force in the opposite.

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3 years ago

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:

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

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3 years ago

What is the weight, in pounds, of a 195-kg object on jupiter?

Nostrana [21]

429.901 is the weight of 195 kg on Jupiter

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3 years ago

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