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4 years ago

What are some physical properties of a star

Physics

2 answers:

Juli2301 [7.4K]4 years ago

8 0

Distance, how much energy generated in the star, size, temp, brightness

zloy xaker [14]4 years ago

7 0

1. Apparent magnitude
2. Colur Index
3. Annual parallax/variation period if it is Cepheid variable/Red Shift
4. Radial Velocity & Tangential Velocity
Other quantities are derived from the above basic observational data.

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Supposing the half-life of element x to be 1 year, if you have a 100 gram sample of x, how much of element x would be left after

Vesnalui [34]

6.25 gram of element x would be left after 4 years.

<h3>What's half life of an element?</h3>

Half - life is the time period of decay of an element to half of its initial number of nuclei.
After n no. of half lives, no. of nuclei left in the sample is initial number of nuclei / 2^n

<h3>What's the amount of 100 gm of element x left after 4 half lives ?</h3>

Leftover amount of the element x = 100/2^4

= 100/16 = 6.25 gram

Thus, we can conclude that 6.25 gram of element x would be left after 4 years.

Learn more about the half life here:

brainly.com/question/15086552

#SPJ1

3 0

2 years ago

Show that on a hypothetical planet having half the diameter of the Earth but twice its density, the acceleration of free fall is

sergiy2304 [10]

Answer:

Steps include:

Find the ratio between the mass of this hypothetical planet and the mass of planet earth.
Calculate the ratio between the acceleration of free fall (gravitational field strength) near the surface of that hypothetical planet and near the surface of the earth.

Assumption: both planet earth and this hypothetical planet are spheres of uniform density.

Explanation:

<h3>Mass of that hypothetical planet</h3>

Let $M_0$ , $r_0$ , and $\rho_0$ denote the mass, radius, and density of planet earth.

The radius and density of this hypothetical planet would be $(1/2)\, r_0$ and $2\, \rho_0$ , respectively.

The volume of a sphere is $\displaystyle V = \frac{4}{3}\, \pi\, r^3$ .

Therefore:

The volume of planet earth would be $\displaystyle V_0 = \frac{4}{3}\, \pi\, {r_0}^{3}$ .

The volume of this hypothetical planet would be: $\displaystyle V = \frac{4}{3}\, \pi\, \left(\frac{r_0}{2}\right)^3 = \frac{1}{6}\, \pi\, {r_0}^3$ .

The mass of an object is the product of its volume and density. Hence, the mass of the earth could also be represented as:

$\displaystyle M_0 = \rho_0\, V_0 = \frac{4}{3}\, \pi\, {r_0}^{3}\, \rho_0$ .

In comparison, the mass of this hypothetical planet would be:

$\begin{aligned}M &= \rho\, V\\ &= (2\, \rho_0) \cdot \left(\frac{1}{6}\, \pi\, {r_0}^3\right) \\ &= \frac{1}{3}\, \pi\, {r_0}^{3}\, \rho_0 \end{aligned}$ .

Compare these two expressions. Notice that $\displaystyle M = \frac{1}{4}\, M_0$ . In other words, the mass of the hypothetical planet is one-fourth the mass of planet earth.

<h3>Gravitational field strength near the surface of that hypothetical planet</h3>

The acceleration of free fall near the surface of a planet is equal to the gravitational field strength at that very position.

Consider a sphere of uniform density. Let the mass and radius of that sphere be $M$ and $r$ , respectively. Let $G$ denote the constant of universal gravitation. Right next to the surface of that sphere, the strength of the gravitational field of that sphere would be:

$\displaystyle g = \frac{G \cdot M}{r^2}$ .

(That's the same as if all the mass of that sphere were concentrated at a point at the center of that sphere.)

Assume that both planet earth and this hypothetical planet are spheres of uniform density.

Using this equation, the gravitational field strength near the surface the earth would be:

$\displaystyle g_0 &= \frac{G \cdot M_0}{{r_0}^2}$ .

On the other hand, the gravitational field strength near the surface of that hypothetical planet would be:

$\begin{aligned}g &= \frac{G \cdot (M_0/4)}{{(r_0/2)}^2} = \frac{G \cdot M_0}{{r_0}^2}\end{aligned}$ .

Notice that these two expressions are equal. Therefore, the gravitational field strength (and hence the acceleration of free fall) would be the same near the surface of the earth and near the surface of that hypothetical planet.

As a side note, both $g$ and $g_0$ could be expressed in terms of $\rho_0$ and $r_0$ alongside the constants $\pi$ and $G$ :

$\begin{aligned}g &= \frac{G \cdot M}{r^2}\\ &= \frac{G \cdot (1/3)\, \pi\, {r_0}^3 \, \rho_0}{(r_0/2)^2} = \frac{4}{3}\,\pi\, G\, r_0\, \rho_0\end{aligned}$ .

$\begin{aligned}g_0 &= \frac{G \cdot M_0}{{r_0}^2}\\ &= \frac{G \cdot (4/3)\, \pi\, {r_0}^3 \, \rho_0}{{r_0}^2} = \frac{4}{3}\,\pi\, G\, r_0\, \rho_0\end{aligned}$ .