Answer:
In the clarification portion elsewhere here, the definition of the concern is mentioned.
Explanation:
So like optical telescopes capture light waves, introduce it to concentrate, enhance it, as well as make it usable through different instruments via study, so radio telescopes accumulate weak signal light waves, introduce that one to focus, enhance it, as well as make this information available during research. To research naturally produced radio illumination from stars, galaxies, dark matter, as well as other natural phenomena, we utilize telescopes.
Optical telescopes detect space-borne visible light. There are some drawbacks of optical telescopes mostly on the surface:
- Mostly at night would they have been seen.
- Unless the weather gets cloudy, bad, or gloomy, they shouldn't be seen.
Although radio telescopes monitor space-coming radio waves. Those other telescopes, when they are already typically very massive as well as costly, have such an improvement surrounded by optical telescopes. They should be included in poor weather and, when they travel through the surrounding air, the radio waves aren't obscured by clouds. Throughout the afternoon and also some at night, radio telescopes are sometimes used.
The equivalent resistance when two resistors are connected in series is
the sum of their individual resistances.
The marking on the resistor that says "1000 W" is the rating that tells
how much power the resistor can safely dissipate, without overheating
or exploding. (The 'W' stands for 'Watts'.) It doesn't tell us anything about
their individual resistances. So we don't have enough information to calculate
their series equivalent.
Answer:
The same as the escape velocity of asteorid A (50m/s)
Explanation:
The escape velocity is described as follows:

where
is the universal gravitational constant,
is the mass of the asteroid and
is the radius
and since the scape velocity is 50m/s:

Now, if the astroid B has twice mass and twice the radius, we have that tha mass is: 
and the radius is: 
inserting these values into the formula for escape velocity:

and we have found that
, so the two asteroids have the same escape velocity.
We found that the expression for escape velocity remains the same as for asteroid A, this because both quantities (radius and mass) doubled, so it does not affect the equation.
The answer is
Asteroid B would have an escape velocity the same as the escape velocity of asteroid A