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Romashka [77]
3 years ago
8

It is proposed that a spaceship might be propelled in the solar system by radiation pressure, using a large sail made of foil. W

hat must the area (in m2) of the sail be if the radiation force is to be equal in magnitude to the Sun's gravitational attraction? Assume that the mass of the ship + sail is 1500 kg, that the sail is perfectly reflecting, and that the sail is oriented perpendicular to the Sun's rays. G = 6.67 times 10-11 N m2/kg2, power of the sun = 3.9 times 1026 W, c = 3.0 times 108 m/s, and mass of the sun = 1.99 times 1030 kg.
Physics
1 answer:
Anvisha [2.4K]3 years ago
3 0

Answer:

962291.57928 m²

Explanation:

P_r = Pressure = 2\dfrac{I}{c}  (full reflection)

I = Intensity = \dfrac{P}{A}=\dfrac{P}{4\pi r^2}

P = Power = 3.9\times 10^{26}\ W

c = Speed of light = 3\times 10^8\ m/s

M = Mass of Sun = 1.99\times 10^{30}\ kg

m = Mass of ship = 1500 kg

G = Gravitational constant = 6.67 × 10⁻¹¹ m³/kgs²

Force of radiation is given by

F_r=P_rA\\\Rightarrow F_r=2\dfrac{I}{c}\times A\\\Rightarrow F_r=2\dfrac{P}{4\pi r^2c} A

This force will balance the gravitational force as stated in the question

\dfrac{GMm}{r^2}=2\dfrac{P}{4\pi r^2c} A\\\Rightarrow A=\dfrac{4\pi cGMm}{2P}\\\Rightarrow A=\dfrac{4\times \pi\times 3\times 10^8\times 6.67\times 10^{-11}\times 1.99\times 10^{30}\times 1500}{2\times 3.9\times 10^{26}}\\\Rightarrow A=962291.57928\ m^2

The area of the must be 962291.57928 m²

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Answer:

T = 693.147 minutes

Explanation:

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Brine water flows into the tank at a rate of 6 L/min which has a concentration of 0.1 kg/L

Hence, the amount of salt that is added to the tank per minute = (6\times0.1)kg/min=0.6kg/min

Also, there is a continuous outflow from the tank at a rate of 6 L/min.

Hence, amount of salt subtracted from the tank per minute = 6x kg/min

Now, the rate of change of salt concentration in the tank = \frac{dx}{dt}

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3 0
3 years ago
Light of wavelength 600 nm passes though two slits separated by 0.22mm and is observed on a screen 1.1m behind the slits. The lo
Ne4ueva [31]

Answer:

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Explanation:

We have to take into account the expression for the position of the fringes

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y(m=1)=\frac{(1)(600*10^{-9}m)(1.1m)}{0.22m}=3*10^{-6}m\\\\y(m=1)=\frac{(-1)(600*10^{-9}m)(1.1m)}{0.22m}=-3*10^{-6}m

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M = molar mass of the helium gas = 4.0 g/mol

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tino4ka555 [31]

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