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lord [1]
3 years ago
14

What's the diameter of a dish antenna that will receive 10−20W of power from Voyager at this time? Assume that the radio transmi

tter on Voyager transmits equally in all directions(isotropically). In fact, the antenna on Voyager focuses the signal in a beam aimed at the earth, so this problem over-estimates the size of the receiving dish needed.
Physics
1 answer:
Murrr4er [49]3 years ago
3 0

Complete Question:

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? Assume that the radio transmitter on Voyager transmits equally in all directions(isotropically).  In fact, the antenna on Voyager focuses the signal in a beam aimed at the earth, so this problem over-estimates the size of the receiving dish needed.

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.

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The loaded cab of an elevator has a mass of 3.0 x 10 3 kg and moves 200 m up the shaft in 20 s at constant speed. At what averag
Alexeev081 [22]

The average rate at which the cable does work is 294,000 J/s.

The given parameters:

  • <em>mass, m = 3000 kg</em>
  • <em>height, h = 200 m</em>
  • <em>time of motion, t = 20 s</em>

The average rate at which the cable does work is calculated as follows;

P = \frac{E}{t} \\\\P = \frac{mgh}{t} \\\\P = \frac{3000 \times 9.8 \times 200}{20} \\\\P = 294,000 \ J/s

Thus, the average rate at which the cable does work is 294,000 J/s.

Learn more about energy and power here: brainly.com/question/13387946

4 0
3 years ago
a man crossed a road 8.25m wide at a speed of 2.01m/s,how long does it take to get man to cross the road​
Svetradugi [14.3K]

Answer:

t = 4.1 seconds

Explanation:

It is given that,

Width of road which is to be crossed by a man is 8.25 m, it means it is distance to be covered.

Speed of man is 2.01 m/s

We need to find the time taken by the man to cross the road. It is a concept of speed. Speed of a person is given by total distance covered divided by time taken. So,

v=\dfrac{d}{t}

t is time taken

t=\dfrac{d}{v}\\\\t=\dfrac{8.25}{2.01}\\\\t=4.1\ s

So, the time taken by the man to cross the road is 4.1 seconds.

5 0
4 years ago
What player(s) can take a goal kick?
inysia [295]

Answer:

c) goalie

Explanation:

8 0
4 years ago
Read 2 more answers
A train starting from rest picks up a speed of 20 m/s in 20 s while travelling on a straight path. It continues to move at the s
Andrej [43]

Answer:

1300 m

Explanation:

As the path is straight, so the speed is equivalent to velocity. Now. assuming that the acceleration and deceleration of the train are constant. So, change of velocity with respect to time for acceleration as well as deceleration is constant. Hence, the slope of the speed-time graph is constant for the time of acceleration as well as deceleration. The speed for the time from 20 s to 70 s is constant, so slope for this interval of time is zero. The speed-time graph is shown in the figure.

The total distance covered by the train during the entire journey is the area of the speed-time graph.

Area=\frac{1}{2}(20\times 20)+ 50\times 20+\frac{1}{2}(20\times 10)

=200+1000+100

=1300

As velocity is in m/s and time is in s so the unit of area is m

Hence, the total distance is  1300m.

8 0
4 years ago
What is the period (in hours) of a satellite circling Mars 100 km above the planet's surface? The mass of Mars is 6.42 × 1023 kg
scZoUnD [109]

To solve this problem it is necessary to apply the concepts related to the Centrifugal Force and the Gravitational Force. Since there is balance on the body these two Forces will be equal, mathematically they can be expressed as

F_c = F_g

\frac{mv^2}{r} = \frac{GmM}{r^2}

Where,

m = Mass

G =Gravitational Universal Constant

M = Mass of the Planet

r = Distance/Radius

Re-arrange to find the velocity we have,

v^2 = \frac{GM}{r}

At the same time we know that the period is equivalent in terms of the linear velocity to,

T = \frac{2\pi}{\frac{v}{r}}

v = \frac{2\pi r}{T}

If our values are that the radius of mars is 3400 km and the distance above the planet is 100km more, i.e, 3500km we have,

v^2 = \frac{GM}{r}

( \frac{2\pi r}{T})^2 =  \frac{GM}{r}

T = \sqrt{\frac{4\pi^2 r^3}{GM}}

Replacing we have,

T = \sqrt{\frac{4\pi^2 (3500*10^3)^3}{(6.67430*10^{-11})(6.42*10^23)}}

T = 6285.09s (\frac{1min}{60s})(\frac{1hour}{60min})

T= 1.74hour

Therefore the correct answer is C.

7 0
3 years ago
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