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cestrela7 [59]
2 years ago
3

The Voyager 1 spacecraft is now beyond the outer reaches of our solar system, but earthbound scientists still receive data from

the spacecraft's 20-W radio transmitter. Voyager is expected to continue transmitting until about 2025, when it will be some 25 billion km from Earth. What's the diameter of a dish antenna that will receive 10−20W of power from Voyager at this time?
Note it is not 3.7m or 2.24m.
Physics
1 answer:
KATRIN_1 [288]2 years ago
7 0

Answer:

d = 2,236 m.

Explanation:

The received power on Earth, can be calculated as the product of the intensity (or power density) times the area that intercepts the power radiated.

As we assume that the transmitter antenna is ominidirectional, power is spreading out over a sphere with a radius equal to the distance to the source.

So, we can get the power density as follows:

I = P /A = P / 4*π*r², where P = 20 W, and r= 25 billion km = 25*10¹² m.

⇒ I = 20 W / 4*π* (25*10¹²)² m²

The received power, is just the product of this value times the area of the receiver antenna, which we assumed be a circle of diameter d:

Pr = I. Ar =( 20W / 4*π*(25*10¹²)² m²) * π * (d²/4) = 10⁻²⁰ W

Simplifying common terms, we can solve for d:

d= √(16*(25)²*10⁴/20) = 2,236 m.

Actually, the receiver antenna will have some gain (which it means that will concentrate more power in a given direction) so the real diameter will be less than calculated for an isotropic antenna.

Typical gain values for ESA (Earth Station Antennas) are in the range of 45-50 dBi.

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Answer:

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First we need to calculate the initial speed V_{0}

x=V_{0} *t+\frac{1}{2} *a*(t^{2} )\\62.5m=V_{0} *4.15s+\frac{1}{2} *-5.25\frac{m}{s^{2} } *(4.15^{2} )\\V_{0}=25.953\frac{m}{s}

Once we have the initial speed, we can isolate the final speed from following equation:

V_{f} =V_{0} +a*t  V_{f}= 4.165 \frac{m}{s}  

Then we can calculate the aceleration where the car stops 0.5 m before striking the tree.

To do that, we replace 62 m in the first formula, as follows:

x=V_{0} *t+\frac{1}{2} *a*(t^{2} )\\62m=25.953\frac{m}{s}*4.15s+\frac{1}{2} *-a\frac{m}{s^{2} } *(4.15^{2} )\\a=-5.19\frac{m}{s^{2} }

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A cylindrical tank of methanol has a mass of 40 kgand a volume of 51 L. Determine the methanol’s weight, density,and specific gr
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Answer:

Weight  W = 392.4 N

Density  \rho = 784.31 \frac{kg}{m^{3} }

Specific gravity S = 0.78431

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Explanation:

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Volume (V) = 0.051 m^{3}

Weight W = m × g

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Density \rho = \frac{mass }{volume}

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Specific gravity (S) = \frac{\rho}{\rho_{water} }

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This is the specific gravity of the methanol.

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