Answer:
High speed optical communication technology
To be able to communicate from the space to the earth and from earth to space is one of the most essential features required during space exploration.
Explanation:
Space exploration involves going into the space, beyond the earth's atmosphere. Landing on other planets and studying their details, going into deeper space beyond the planets to discover new cosmic events or structures is all a part of space exploration.
The key to analyse the studies and observations is being able to communicate the data collected, photos taken etc to the launch centers or space centers on earth. The space centers on earth should also be able to communicate with the persons or the satellites in space.
This is made possible using the optical communication technology which involves the use of optical fibers, lasers etc, since high speeds are more efficient during communication
Answer:
7 m/s
Explanation:
To solve this problem you must use the conservation of energy.

That math speak for, initial kinetic energy plus initial potential energy equals final kinetic energy plus final potential energy.
The initial PE (potential energy) is 0 because it hasn't been raised in the air yet. The final KE (kinetic energy) is 0 because it isn't moving. This gives the following:


K1=U2

Solve for v

Input known values and you get 7 m/s.
Efficiency = (useful output) / (input)
Efficiency = (35 J) / (125 J) = 0.28 = 28%
Answer:
= 4.38 × 10³⁴kgm²/s
Explanation:
Given that,
mass of moon m = 9.5 × 10²²kg
Orbital radius r = 4.28 × 10⁵km
Orbital period T = 28.9days
T = 28.9 × 24 × 60 × 60
= 2,496,960s
Angular momentum of the moon about the planet
L = mvr
L = mr²w
