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goldfiish [28.3K]
3 years ago
5

Consider a civilization broadcasting a signal with a power of 1.2×104 watts. The Arecibo radio telescope, which is about 300 met

ers in diameter, could detect this signal if it is coming from as far away as 110 light-years. Suppose instead that the signal is being broadcast from the other side of the Milky Way Galaxy, about 70000 light-years away.
Physics
1 answer:
Talja [164]3 years ago
6 0
The same power (1.9×10^4 watts) gets diminished by Inverse square law 
<span>(ratio of distances)² </span>
<span>= (138/70000)² </span>
<span>= 3.886506 X 10^-6 . </span>
<span>This is the diminution factor for the given sensitivity of the telescope. But at 3.8865 millionth below the sensitivity its is far below detection; it needs to be enhanced by collecting the energy over an area so many times more (by an aperture multiplied by same factor) </span>
<span>1/[3.886506 X 10^-6] = 257298.88 </span>
<span>In other words the diameter should have the ratio of square root of this </span>
<span>(70000/138). </span>
<span>This multiplied by 300m dish gives </span>
<span>300 X (70000/138) = 152173.913 m = 152.1739 km. </span>
<span>This should be the aperture of the new telescope for detecting the signal.</span>
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Answer:

a-1 Graph is attached. The relation is linear.

a-2 The corresponding height for 68 kPa Pressure is 7.54 m

a-3 The corresponding weight for 68 kPa Pressure is 1394726kg

b The original height of the column is 5.98 m

Explanation:

Part a

a-1

The graph is attached with the solution. The relation is linear as indicated by the line.

a-2

By the equation

P=\rho \times g \times h

Here

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  • ρ is the density of the oil whose SG is 0.92. It is calculated as

                                       \rho=S.G \times \rho_{water}\\\rho=0.92 \times 1000 kg/m^3\\\rho=920 kg/m^3\\

  • g is the gravitational constant whose value is 9.8 m/s^2
  • h is the height which is to be calculated

                                        P=\rho \times g \times h\\h=\frac{P}{\rho \times g}\\h=\frac{68 \times 10^3}{920 \times 9.8}\\h=7.54m

So the height of column is 7.54m

a-3

By the relation of volume and density

M=\rho \times V

Here

  • ρ is the density of the oil which is 920 kg/m^3
  • V is the volume of cylinder with diameter 16m calculated as follows

                             V=\pi r^2h\\V=3.14\times (8)^2 \times 7.54\\V=1515.23 m^3

Mass is given as

                             M=\rho \times V\\M=920 \times 1515.23\\M=1394726kg

So the mass of oil leading to 68kPa is 1394726kg

Part b

Pressure variation is given as

                            \Delta P=P_{obs}-P_{atm}\\\Delta P=115-101 kPa\\\Delta P=14 kPa\\

Now corrected pressure is as

P_c=P_g-\Delta P\\P_c=68-14 kPa\\P_c=54 kPa

Finding the value of height for this corrected pressure as

P_c=\rho \times g \times h\\h=\frac{P_c}{\rho \times g}\\h=\frac{54 \times 10^3}{920 \times 9.8}\\h=5.98m

The original height of column is 5.98m

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