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creativ13 [48]
2 years ago
14

The Mass of the Sun is 1.989 × 10^30 kg. The speed of light is 3 x 10^8 m/s. Assuming the Sun is 75% hydrogen, how much Energy c

ould be produced if all of the hydrogen is converted into energy? Use Einstein's famous equation E = mc^2 where E will be in Joules.
Physics
1 answer:
bogdanovich [222]2 years ago
8 0

Answer:

E = 13.428 \times 10^{46} J

Explanation:

Mass of the Sun = 1.989 \times 10^{30} Kg

Amount of Hydrogen = 75% of Mass of Sun,

thus mass of Hydrogen (m) in the Sun is,

m = 1.989 \times 10^{30} \times \frac{75}{100}

m = 1.492 \times 10^{30}

Speed of Light (c) = 3 \times 10^{8} m/s

Thus, energy(E) produced if whole of hydrogen is converted into energy,

E = mc²

E = 1.492 \times 10^{30} \times (3 \times 10^{8} )^{2}

E = 1.492 \times 9 \times 10^{46}

E = 13.428 \times 10^{46} J

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3 years ago
On a distant planet, a rock falls in 48.4 s from the top of a 1.10e+02 m cliff to the planet surface below. What is the accelera
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this can be solve using the formala of free fall

t = sqrt( 2y/ g)

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8 0
3 years ago
You wish to cool a 1.83 kg block of tin initially at 88.0°C to a temperature of 57.0°C by placing it in a container of kerosene
uranmaximum [27]

Answer:

0.273 liters are needed to accomplish this task without boiling.

Explanation:

The minimum boiling point of kerosene is 150\,^{\circ}C. According to this question, we need to determine the minimum volume of liquid such that heat received is entirely sensible, that is, with no phase change.

If we consider a steady state process and that energy interactions with surrounding are negligible, then we get the following formula by the Principle of Energy Conservation:

\rho_{k}\cdot V_{k}\cdot c_{k}\cdot (T-T_{k,o}) = m_{t}\cdot c_{t}\cdot (T_{t,o}-T) (1)

Where:

\rho_{k} - Density of kerosene, measured in kilograms per cubic meter.

V_{k} - Volume of kerosene, measured in cubic meters.

c_{k}, c_{t} - Specific heats of the kerosene and tin, measured in joule per kilogram-Celsius.

T_{k,o}, T_{t,o} - Initial temperatures of kerosene and tin, measured in degrees Celsius.

T - Final temperatures of the kerosene-tin system, measured in degrees Celsius.

Please notice that the block of tin is cooled at the expense of the temperature of the kerosene until thermal equilibrium is reached.

From (1), we clear the volume of kerosene:

V_{k} = \frac{m_{t}\cdot c_{t}\cdot (T_{t,o}-T)}{\rho_{k}\cdot c_{k}\cdot (T-T_{k,o})}

If we know that m_{t} = 1.83\,kg, c_{t} = 218\,\frac{J}{kg\cdot ^{\circ}C}, T_{t,o} = 88\,^{\circ}C, T_{k,o} = 24.0\,^{\circ}C, T = 57\,^{\circ}C, c_{k} = 2010\,\frac{J}{kg\cdot ^{\circ}C} and \rho_{k} = 820\,\frac{kg}{m^{3}}, then the volume of the liquid needed to accomplish this task without boiling is:

V_{k} = \frac{(1.83\,kg)\cdot \left(218\,\frac{J}{kg\cdot ^{\circ}C} \right)\cdot (88\,^{\circ}C-57\,^{\circ}C)}{\left(820\,\frac{kg}{m^{3}} \right)\cdot \left(2010\,\frac{J}{kg\cdot ^{\circ}C} \right)\cdot (57\,^{\circ}C-24\,^{\circ}C)}

V_{k} = 2.273\times 10^{-4}\,m^{3}

V_{k} = 0.273\,L

0.273 liters are needed to accomplish this task without boiling.

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ANSWER: Both are shiny and are solid at room temperature.
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