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maksim [4K]
3 years ago
9

During a solar eclipse, the Moon, Earth, and Sun all lie on the same line, with the Moon between the Earth and the Sun. The Moon

has a mass of 7.36 × 1022 kg; the Earth has a mass of 5.98 × 1024 kg; and the Sun has a mass of 1.99 × 1030 kg. The distance between the Moon and the Earth is 3.84 × 108 m; the distance between the Earth and the Sun is 1.496 × 1011 m. Calculate the total gravitational potential energy for this arrangement.
Physics
1 answer:
Andreyy893 years ago
8 0

Answer:

U = 5.37*10^33 J

Explanation:

The gravitational potential energy between two bodies is given by:

U_{1,2}=-G\frac{m_1m_2}{r_{1,2}}

G: Cavendish's constant = 6.67*10^-11 m^3/kg.s

For three bodies the total gravitational potential energy is:

U_{T}=U_{1,2}+U_{1,3}+U_{2,3}\\\\U_{T}=-G[\frac{m_1m_2}{r_{1,2}}+\frac{m_1m_3}{r_{1,3}}+\frac{m_2m_3}{r_{2,3}}]

BY replacing the values of the parameters for 1->earth, 2->moon and 3->sun you obtain:

U_{T}=-(6.67*10^{-11}m^3/kg.s)[\frac{(5.98*10^{24}kg)(7.36*10^{22}kg)}{3.84*10^{8}m}+\\\\\frac{(5.98*10^{24}kg)(1.99*10^{30}kg)}{1.496*10^{11}m}+\frac{(7.36*10^{22}kg)(1.99*10^{30}kg)}{1.496*10^{11}m-3.84*10^8m}]\\\\U_{T}=5.37*10^{33}J

hence, the total gravitational energy is 5.37*10^33 J

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Calculate the wavelength of each frequency of electromagnetic radiation: a. 100.2 MHz (typical frequency for FM radio broadcasti
Natalka [10]

Answer:

a). 100.2 MHz (typical frequency for FM radio broadcasting)

The wavelength of a frequency of 100.2 Mhz is 2.99m.

b. 1070 kHz (typical frequency for AM radio broadcasting) (assume four significant figures)

The wavelength of a frequency of 1070 khz is 280.3 m.

c. 835.6 MHz (common frequency used for cell phone communication)

The wavelength of a frequency of 835.6 Mhz is 0.35m.

Explanation:

The wavelength can be determined by the following equation:

c = \lambda \cdot \nu  (1)

Where c is the speed of light, \lambda is the wavelength and \nu is the frequency.  

Notice that since it is electromagnetic radiation, equation 1 can be used. Remember that light propagates in the form of an electromagnetic wave.

<em>a). 100.2 MHz (typical frequency for FM radio broadcasting)</em>

Then, \lambda can be isolated from equation 1:

\lambda = \frac{c}{\nu} (2)

since the value of c is 3x10^{8}m/s. It is necessary to express the frequency in units of hertz.

\nu = 100.2 MHz . \frac{1x10^{6}Hz}{1MHz} ⇒ 100200000Hz

But 1Hz = s^{-1}

\nu = 100200000s^{-1}

Finally, equation 2 can be used:

\lambda = \frac{3x10^{8}m/s}{100200000s^{-1}}

\lambda = 2.99 m

Hence, the wavelength of a frequency of 100.2 Mhz is 2.99m.

<em>b. 1070 kHz (typical frequency for AM radio broadcasting) (assume four significant figures)</em>

<em> </em>

\nu = 1070kHz . \frac{1000Hz}{1kHz} ⇒ 1070000Hz

But  1Hz = s^{-1}

\nu = 1070000s^{-1}

Finally, equation 2 can be used:

\lambda = \frac{3x10^{8}m/s}{1070000s^{-1}}

\lambda = 280.3 m

Hence, the wavelength of a frequency of 1070 khz is 280.3 m.

<em>c. 835.6 MHz (common frequency used for cell phone communication) </em>

\nu = 835.6MHz . \frac{1x10^{6}Hz}{1MHz} ⇒ 835600000Hz

But  1Hz = s^{-1}

\nu = 835600000s^{-1}

Finally, equation 2 can be used:

\lambda = \frac{3x10^{8}m/s}{835600000s^{-1}}

\lambda = 0.35 m

Hence, the wavelength of a frequency of 835.6 Mhz is 0.35m.

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Ten identical steel wires have equal lengths L and equal "spring constants" k. The wires are connected end to end so that the re
lapo4ka [179]

Answer:

K_{system} = \frac{k}{10}

Explanation:

When the springs are connected end to end, it means they are connected in series. When the springs are connected in series, the stress applied to the system gets applied to each of the springs without any change in magnitude while the strain of the system is the sum total of strains of each spring. The spring constant of the resultant system is given as,

\frac{1}{K_{system}} = (\frac{1}{K_{1}})+(\frac{1}{K_{2}})+(\frac{1}{K_{3}})+ (\frac{1}{K_{4}})+.....+(\frac{1}{K_{n}})

Here, n = 10

Spring constant of each spring = k

Thus,

\frac{1}{K_{system}} = (\frac{1}{K_{1}})+(\frac{1}{K_{2}})+(\frac{1}{K_{3}})+ (\frac{1}{K_{4}})+.....+(\frac{1}{K_{10}})

\frac{1}{K_{system}} = (\frac{1}{k})+(\frac{1}{k})+(\frac{1}{k})+(\frac{1}{k})+(\frac{1}{k})+(\frac{1}{k})+(\frac{1}{k})+(\frac{1}{k})+(\frac{1}{k})+(\frac{1}{k})

\frac{1}{K_{system}} = \frac{10}{k}

K_{system} = \frac{k}{10}

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3 years ago
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