I need help please ..

De acuerdo con la tercera ley de movimiento planetario de Kepler, la masa de un planeta es directamente proporcional al cubo de

Sunny_sXe [5.5K]

Answer:

La masa del Sol es $2.509\times 10^{31}$ kilogramos.

Step-by-step explanation:

Tras una lectura cuidadosa al enunciado, tenemos que la Tercera Ley de Kepler queda descrita por la siguiente relación:

$M \propto \frac{r^{3}}{T^{2}}$

$M = k\cdot \frac{r^{3}}{T^{2}}$ (Eq. 1)

Donde:

$r$ - Distancia entre los centros del planeta y el satélite, medido en kilómetros.

$T$ - Período oribital del satélite, medido en días.

$k$ - Constante de proporcionalidad, medida en kilogramo-días cuadrados por kilómetro cúbico.

$M$ - Masa del planeta, medida en kilogramos.

Podemos obtener la masa del Sol mediante la siguiente relación:

$\frac{M_{S}}{M_{E}} = \frac{\frac{r_{E}^{3}}{T_{E}^{2}} }{\frac{r_{M}^{3}}{T_{M}^{2}} }$

$\frac{M_{S}}{M_{E}} = \left(\frac{T_{M}}{T_{E}} \right)^{2}\cdot \left(\frac{r_{E}}{r_{M}} \right)^{3}$ (Eq. 2)

Donde:

$T_{M}$ , $T_{E}$ - Períodos orbitales de la Luna y la Tierra, medidos en días.

$r_{E}$ , $r_{M}$ - Distancias entre la Tierra y el Sol, así como entre la Luna y la Tierra, medidas en kilómetros.

$M_{S}$ , $M_{E}$ - Masas del Sol y la Tierra, medidos en kilogramos.

Si $M_{E} = 75.97\times 10^{24}\,kg$ , $T_{E} = 365.3\,d$ , $T_{M} = 27.3\,d$ , $r_{M} = 3.84\times 10^{5}\,km$ y $r_{E} = 1.496\times 10^{8}\,km$ , entonces tenemos que la masa del Sol es:

$M_{S} = \left(\frac{T_{M}}{T_{E}} \right)^{2}\cdot \left(\frac{r_{E}}{r_{M}} \right)^{3}\cdot M_{E}$

$M_{S} = \left(\frac{27.3\,d}{365.3\,d} \right)^{2}\cdot \left(\frac{1.496\times 10^{8}\,km}{3.84\times 10^{5}\,km} \right)^{3}\cdot (75.97\times 10^{24}\,kg)$

$M_{S} = 2.509\times 10^{31}\,kg$

La masa del Sol es $2.509\times 10^{31}$ kilogramos.

7 0

3 years ago

The Center for Medicare and Medical Services reported that there were 295,000 appeals for hospitalization and other Part A Medic

Ymorist [56]

Answer:

(a) 0.00605

(b) 0.0403

(c) 0.9536

(d) 0.98809

Step-by-step explanation:

We are given that 40% of first-round appeals were successful (The Wall Street Journal, October 22, 2012) and suppose ten first-round appeals have just been received by a Medicare appeals office.

This situation can be represented through Binomial distribution as;

$P(X=r)= \binom{n}{r}p^{r}(1-p)^{n-r} ; x = 0,1,2,3,....$

where, n = number of trials (samples) taken = 10

r = number of success

p = probability of success which in our question is % of first-round

appeals that were successful, i.e.; 40%

So, here X ~ $Binom(n=10,p=0.40)$

(a) Probability that none of the appeals will be successful = P(X = 0)

P(X = 0) = $\binom{10}{0}0.40^{0}(1-0.40)^{10-0}$

= $1*0.6^{10}$ = 0.00605

(b) Probability that exactly one of the appeals will be successful = P(X = 1)

P(X = 1) = $\binom{10}{1}0.40^{1}(1-0.40)^{10-1}$

= $10*0.4^{1} *0.6^{10-1}$ = 0.0403

(c) Probability that at least two of the appeals will be successful = P(X>=2)

P(X >= 2) = 1 - P(X = 0) - P(X = 1)

= 1 - $\binom{10}{0}0.40^{0}(1-0.40)^{10-0} - \binom{10}{1}0.40^{1}(1-0.40)^{10-1}$

= 1 - 0.00605 - 0.0403 = 0.9536

(d) Probability that more than half of the appeals will be successful = P(X > 0.5)

For this probability we will convert our distribution into normal such that;

X ~ N( $\mu = n*p=4,\sigma^{2}= n*p*q = 2.4$ )

and standard normal z has distribution as;

Z = $\frac{X-\mu}{\sigma}$ ~ N(0,1)

P(X > 0.5) = P( $\frac{X-\mu}{\sigma}$ > $\frac{0.5-4}{\sqrt{2.4} }$ ) = P(Z > -2.26) = P(Z < 2.26) = 0.98809

3 0

3 years ago

Find the distance between the points (4,-2) and (0,10)

vagabundo [1.1K]

We can solve this problem by using the distance formula. The distance formula is: $\sqrt{(x_{1}-x)^{2} + (y_{1} - y)^{2}}$ We can now put in values and solve.