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mr_godi [17]
3 years ago
6

Satelite communication is only possible using

Physics
2 answers:
Paha777 [63]3 years ago
7 0

the correct answer is B) digital transmission


zimovet [89]3 years ago
6 0
None of the choices is correct.

Satellite communication is quite effective using either analog OR digital transmission.

Sound can't be used for satellite communication at all. (There's no medium to carry the sound to or from the satellite, and if there were, then the round-trip up and down through a geostationary satellite would take at least a minute at the sea-level speed of sound.)

Cellular devices can work but aren't required.
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The Earth's place in space can be described as _________.
Olin [163]
The answer is B because if you use process of elimination, you find that A is invalid because Venus is the second planet. C is out because Mars is the 4th planet. D is out because we are nowhere  near the Andromeda Galaxy. We are millions of light years away.
4 0
3 years ago
Read 2 more answers
A bug is 12 cm from the center of a turntable that is rotating with a frequency of 45 rev/min . What minimum coefficient frictio
Agata [3.3K]

Answer:

The minimum coefficient of friction is 0.27.

Explanation:

To solve this problem, start with identifying the forces at play here. First, the bug staying on the rotating turntable will be subject to the centripetal force constantly acting toward the center of the turntable (in absence of which the bug would leave the turntable in a straight line). Second, there is the force of friction due to which the bug can stick to the table. The friction force acts as an intermediary to enable the centripetal acceleration to happen.

Centripetal force is written as

F_c = m\frac{v^2}{r}

with v the linear velocity and r the radius of the turntable. We are not given v, but we can write it as

v = r\omega

with ω denoting the angular velocity, which we are given. With that, the above becomes:

F_c = m\frac{v^2}{r}=m\omega^2 r

Now, the friction force must be at least as much (in magnitude) as Fc. The coefficient (static) of friction μ must be large enough. How large?

F_r=\mu mg \geq m\omega^2 r = F_c\implies\\\mu \geq \frac{\omega^2 r}{g}

Let's plug in the numbers. The angular velocity should be in radians per second. We are given rev/min, which can be easily transformed by a factor 2pi/60:

\frac{1 rev}{1 min}\cdot\frac{\frac{2\pi rad}{rev}}{\frac{60s}{1 min}}=\frac{2\pi}{60}\frac{rad}{s}

and so 45 rev/min = 4.71 rad/s.

\mu \geq \frac{\omega^2 r}{g}=\frac{4.71^2\frac{1}{s^2}\cdot 0.12m}{9.8\frac{m}{s^2}}=0.27

A static coefficient of friction of at least be 0.27 must be present for the bug to continue enjoying the ride on the turntable.



3 0
3 years ago
1. Synthesize Information You push your
RSB [31]

Answer:separate

Explanation:

6 0
3 years ago
What is the equivalent resistance in this circuit?<br><br> What is the current in this circuit?
kramer

All these resistors are in series so we can take the sum of them by:

Rtotal = R1 + R2 + R3......

So...

Rtotal = 2 + 3 + 4 + 6

Rtotal = 15

So now the total resistance in the circuit is 15 ohms and the potential difference applied to the circuit is 45 volts

Now we can use:

V = IR

Isolate for I

V/R = I

45/15 = I

I = 3 amps (A)

3 0
3 years ago
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If the force on a spring is 2 N and it stretched 0.5 m, what is the spring constant?
andriy [413]
<h2>It solved by the Hooke's law states F=kx</h2>

answer is

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6 0
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