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Sav [38]
3 years ago
9

What was the only option for getting the Apollo 13 astronauts back to Earth alive?

Physics
2 answers:
Vera_Pavlovna [14]3 years ago
4 0
They got back in the Lunar Explorer Module
Cerrena [4.2K]3 years ago
3 0

Answer:

Using the Lunar Module to travel back to Earth

Explanation:

During the Apollo 13 Mission (Launched on April 11, 1970), two days in the mission on the way to the Moon, an explosion occurred in the oxygen cylinder of the Service Module due to which Lunar landing was aborted and a quick plan was devised to bring back the Astronauts alive on Earth.

The oxygen tank of service module was used to generate electrical power and provide breathable air in the command module. The command module became uninhabitable because of quick loss of oxygen in the service module. Due to this the only option was to use the Lunar Module to get back on Earth.

Lunar module was designed to support 2 Astronauts for 2 days on Moon. A lot of procedures were drafted by Mission Control, Houston so that it could support 3 astronauts for 4 days. Although the crew faced a lot of challenges and hardships during the return journey, they came back on Earth safely after going around Moon and making a record of farthest space travel by any human being. The capsule carrying the crew , splashed in South Pacific ocean on April 17, 1970.

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17,874,000 what is the value of 1
kotegsom [21]
Salutations!

17,874,000 what is the value of 1?

The value of 1 is 10 million. The place value would be 10,000,000.

Hope I helped.
3 0
3 years ago
Read 2 more answers
One end of a 7-cm-long spring is attached to the ceiling. When a 5.4 kg mass is hung from the other end, the spring is stretched
mash [69]

Answer:

2.63 cm

Explanation:

Hooke's law gives that the force F is equal to cy where c is spring constant and x is extension

Making c the subject of the formula then

c=\frac {F}{y}

Since F is gm but taking the given mass to be F

c=\frac {5.4 kg}{4.3 cm}=1.2558139534883720930232558139534883720930

By substitution now considering F to be 3.3 kg

y=\frac {3.3 kg}{1.2558139534883720930232558139534883720930}=2.6277777777777 cm\approx 2.63 cm

8 0
4 years ago
What net force would be required for a 50.5 kg wolf to accelerate at 5.0 m/s^2
Basile [38]

Answer:

\boxed {\boxed {\sf 252.5 \ Newtons }}

Explanation:

Force is the push or pull on object that can cause different things, like acceleration. According to Newton's 2nd Law of Motion, it is the product of mass and acceleration.

F=m*a

The mass of the wolf is 50.5 kilograms and the acceleration is 5.0 meters per square second. Therefore:

  • m= 50.5 \ kg
  • a= 5.0 \ m/s^2

Substitute the values into the formula.

F= 50.5 \ kg * 5.0 \ m/s^2

Multiply.

F=252.5 \ kg*m/s^2

  • 1 kilogram meter per square second is equal to 1 Newton.
  • Our answer of 252.5 kg*m/s² is equal to 252.5 Newtons.

F= 252.5 \ N

The force required is <u>252.5 Newtons</u>.

4 0
3 years ago
What is the period of a simple pendulum is its frequency is 20 Hz?
JulijaS [17]

Period  =  (1) / (frequency)

             =  (1) / (20/sec)

             =        1/20 sec 
5 0
4 years ago
Homework B5
nydimaria [60]

Answer:

I hope this helps a little bit.

7 0
3 years ago
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