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stepan [7]
3 years ago
5

The radius of a circular hat is 3.7 inches. Find the circumference. Use 3.14 for π. (1 point)

Mathematics
2 answers:
miss Akunina [59]3 years ago
5 0

Answer:

The circumference is 23.236\ in

Step-by-step explanation:

we know that

The circumference of a circle is equal to

C=2\pi r

we have

r=3.7\ in

substitute

C=2(3.14)(3.7)=23.236\ in

pashok25 [27]3 years ago
3 0
The answer is <span>23.236

the equation for circumference is C=2(pie)r

So you take 2x3.14 which equals 6.28. 

Then you multiply 6.28 by 3.7 to get your answer.</span>
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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}}

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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)

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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.

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