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vazorg [7]
3 years ago
11

PLEASE HELP (WILL GIVE BRAINLIEST TO BEST ANSWER!!!!!)

Physics
1 answer:
Luden [163]3 years ago
3 0

Answer:

1) a radio are uses by astronomy

2) 6 bilion waves

3) expert vertified

Explanation:

1) in contrast to an "ordinary" telescope, which receives visible light, a radio telescope "sees" radio waves emitted by radio sources, typically by means of a large parabolic ("dish") antenna, or arrays of them.and Radio telescopes are also the primary means to track space probes, and are used in the SETI project. so must been

radio are almostly

ceiver by astronomy

2) Radio waves have the longest wavelengths in the electromagnetic spectrum.

3)Expert Verified

Radio telescopes are telescopes that are specially designed for observation of long light wavelengths

CARRY ON ✨

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A 30.0-μF capacitor is connected to a 49.0-Ω resistor and a generator whose rms output is 30.0 V at 60.0 Hz. (a) Find the rms
Natali5045456 [20]

Explanation:

Given that,

Capacitor = 30μC

Resistor = 49.0Ω

Voltage = 30.0 V

Frequency = 60.0 Hz

We need to calculate the impedance

Using formula of impedance

Z=\sqrt{R^2+X_{c}^2}.....(I)

We need to calculate the value of X_{c}

Using formula of X_{c}

X_{c}=\dfrac{1}{2\pi f c}

X_{c}=\dfrac{1}{2\times\pi\times60.0\times30\times10^{-6}}

X_{c}=88.42\ \Omega

Put the value of X_{c} into the formula of impedance

Z=\sqrt{(49.0)^2+(88.42)^2}

Z=101.08\ \Omega

(a). We need to calculate the rms current in the circuit

Using formula of rms current

I_{rms}=\dfrac{V}{Z}

I_{rms}=\dfrac{30.0}{101.08}

I_{rms}=0.30\ A

The rms current in the circuit is 0.30 A.

(b). We need to calculate the rms voltage drop across the resistor

Using formula of rms voltage

V_{rms}=I_{rms}\times R

Put the value into the formula

V_{rms}=0.30\times49.0

V_{rms}=14.7\ V

The rms voltage drop across the resistor is 14.7 V

(c). We need to calculate the rms voltage drop across the capacitor

Using formula of rms voltage

V_{rms}=I_{rms}\times X_{C}

V_{rms}=0.30\times88.42

V_{rms}=26.53\ V

The rms voltage drop across the capacitor is 26.53 V.

Hence, This is the required solution.

4 0
3 years ago
The crew of a cargo plane wishes to drop a crate of supplies on a target below. To hit the target, when should the crew drop the
lara [203]
To hit the target the crew drop the crate before the plane is directly over the target. It is because <span>because the cargo has forward velocity and therefore before it reaches the ground it travels some distance. The answer is A. Hope it helps. </span>
5 0
3 years ago
Read 2 more answers
What prevents the pressure from increasing as a cloud contracts due to its gravity?
Vesnalui [34]

Thermal energy is converted to radiative energy via molecular collisions and released as photons.

4 0
3 years ago
In certain cases, using both the momentum principle and energy principle to analyze a system is useful, as they each can reveal
kramer

Explanation:

The gravitational force equation is the following:

F_G = G * \frac{m_1 m_2}{r^2} \\

Where:

G = Gravitational constant = 6.67408 * 10^{-11} m^3 kg^{-1} s^{-2}

m1 & m2 = the mass of two related objects

r = distance between the two related objects

The problem gives you everything you need to plug into the formula, except for the gravitational constant. Let me know if you need further clarification.

8 0
3 years ago
The earth and the moon exert forces on each other which forces is greater? explain
Helen [10]

Thank you for your question, what you say is true, the gravitational force exerted by the Earth on the Moon has to be equal to the centripetal force.

An interesting application of this principle is that it allows you to determine a relation between the period of an orbit and its size. Let us assume for simplicity the Moon's orbit as circular (it is not, but this is a good approximation for our purposes).

The gravitational acceleration that the Moon experience due to the gravitational attraction from the Earth is given by:

ag=G(MEarth+MMoon)/r2

Where G is the gravitational constant, M stands for mass, and r is the radius of the orbit. The centripetal acceleration is given by:

acentr=(4 pi2 r)/T2

Where T is the period. Since the two accelerations have to be equal, we obtain:

(4 pi2 r) /T2=G(MEarth+MMoon)/r2

Which implies:

r3/T2=G(MEarth+MMoon)/4 pi2=const.

This is the so-called third Kepler law, that states that the cube of the radius of the orbit is proportional to the square of the period.

This has interesting applications. In the Solar System, for example, if you know the period and the radius of one planet orbit, by knowing another planet's period you can determine its orbit radius. I hope that this answers your question.


8 0
3 years ago
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