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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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A woman can row a boat at 5.60 km/h in still water. (a) If she is crossing a river where the current is 2.80 km/h, in what direc
katrin2010 [14]

Answer:

a) θ=210°, b) t=1.155hr, c) t=1.333hr, d) t=1.333hr, e) θ=180° (straight across), f) t=1hr.

Explanation:

So, the very first thing we nee to do when solving this problem is draw a diagram that represents it. In the attached picture I show a diagram for each part of this problem.

part a)

So, for her to move in a direction directly opposite her starting point, the x-component of her velocity must be de same as the velocity of the river in the opposite direction. We can use this fact to find the angle we need. If we analize the triangle I drew in the diagram, we can ses that:

cos \theta = \frac {V_{river}}{V_{boat}}

When solving for theta, we get that:

\theta =cos^{-1} ( \frac {V_{river}}{V_{boat}})

so now we can substitute the corresponding values:

\theta =cos^{-1} ( \frac {2.80km/hr}{5.60km/hr}})

Which yields:

\theta = 60^{o}

but we are measuring the angle relative to the line perpendicular to the river, positive if down the river. So we need to subtract the angle from 270° so we get:

θ=270°-60°=210°

part b)

for part b, we need to find what the y-component for the velocity of the boat is for an angle of 210° as shown in the problem, so we get that:

V_{y}=5.60km/hr*cos(210^{o})

V_{y}=-4.85km/hr

The woman will head in a negative 5.60km distance from one side to the other, so we get that the time it takes her to go to the other side of the river is:

t=\frac{y}{V_{y}}

t=\frac{5.60km}{4.85km/hr}=1.155hr

part c)

In order to find the time it takes her to travel 2.80km down and up the river, we need to find the velocities she will have in both directions. First, down stream:

V_{ds}=V_{river}+V{boat}

V_{ds}=2.80km/hr+5.60km/hr=8.40km/hr

and now up stream:

V_{us}=V_{boat}-V{river}

V_{us}=5.60km/hr-2.80km/hr=2.80km/hr

Once we got these two velocities we will now need to find the time to take each trip:

time down stream:

t_{ds}=\frac{x}{v_{ds}}

t_{ds}=\frac{2.80km}{8.40km/hr}=0.333hr

and the time up stream:

t_{us}=\frac{x}{v_{us}}

t_{us}=\frac{2.80km}{2,80km/hr}=1hr

so the total time will be:

t_{ds}+t_{us}=0.333hr+1hr=1.333hr

d) the time it takes the boat to go upstream and then downstream for the same distance is the same as the time we got on part c, since both times will be the same but they will come in different order, but their sum will be just the same:

t=1.333hr

e) For her to cross the river faster, she must row in a 180° direction (this is in a direction straight accross the river) that way she will use all her velocity to move across the river. (Even though she will move a certain distance horizontally and will not reach a point opposite to the starting point.)

f) In order to find the time it takes her to get to the other side, we need to divide the distance into the velocity of the boat.

t=\frac{d}{v_{boat}}

t=\frac{5.60km}{5.60km/hr}

so

t= 1hr

4 0
3 years ago
Read 2 more answers
If 25.0 g of water at 21c is mixed with 45.0 g of water at 75c, what is the final temperature of the mixture?
mrs_skeptik [129]
56C



Hope that helps, Good luck! (:
5 0
3 years ago
What is the relationship between sunspots and geomagnetic storms?
shutvik [7]
Sunspots are spot like structure on the surface of sun, they are created due to geomagnetic flux which inhibits the convection of heat waves thus reducing the temperature of spot with some factor
8 0
3 years ago
The unit of length mist suitable for measuring the thickness of a cell phone is a_________? ​
lbvjy [14]

Answer:

Explanation: Width

4 0
3 years ago
Your friend, who is in a field 100 meters away from you, kicks a ball towards you with an initial velocity of 16 m/s. Assuming t
LekaFEV [45]

Answer:

Time, t = 5.355 seconds

Explanation:

Given the following data;

Distance = 100 m

Initial velocity = 16 m/s

Deceleration = 1 m/s²

To find the time, we would use the second equation of motion;

But since the ball is decelerating, it's acceleration would be negative.

S = ut + ½at²

Where;

S represents the displacement or height measured in meters.

u represents the initial velocity measured in meters per seconds.

t represents the time measured in seconds.

a represents acceleration measured in meters per seconds square.

Substituting into the equation, we have;

100 = 16t - 0.5t²

200 = 32t - t²

t² + 32t - 200 = 0

Solving the quadratic equation using the quadratic formula;

The quadratic equation formula is;

x = \frac {-b \; \pm \sqrt {b^{2} - 4ac}}{2a}

Substituting into the equation, we have;

x = \frac {-32 \; \pm \sqrt {32^{2} - 4*1*(-200)}}{2*1}

x = \frac {-32\pm \sqrt {1024 - (-800)}}{2}

x = \frac {-32 \pm \sqrt {1024 + 800}}{2}

x = \frac {-32 \pm \sqrt {1824}}{2}

x = \frac {-32 \pm 42.71}{2}

x_{1} = \frac {-32 + 42.71}{2}

x_{1} = \frac {10.71}{2}

x1 = 5.355

We do not need the negative value of x, so we proceed.

Therefore, time = 5.355 seconds

3 0
3 years ago
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