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2 years ago

Discuss how oxygen is used in spacecraft's air supplies in at least 3 paragraphs.

1 answer:

galina1969 [7]2 years ago

3 0

Short-duration spacecraft typically have one backup system and carry their own supply of oxygen. A large portion of the required oxygen is produced on long-duration missions, such as the International Space Station (ISS), which has been in orbit since 1998. Different sources provide the oxygen utilized on the ISS. The water electrolyzer is the primary source of metabolic oxygen. As an alternative to the electrolyzer, oxygen candles (also known as SFOGs) can produce metabolic oxygen. Additionally, oxygen is carried up whenever a cargo ship docks and stored in two tanks on the ISS Airlock. The electrolyzer electrolyzes water to create oxygen by running an electric current through it. Since water is a poor electrical conductor by itself, a little quantity of common salt is dissolved in the water to improve its electrical conductivity. Water is split into hydrogen and oxygen throughout the process.

We must keep in mind that oxygen by itself cannot be inhaled; it must be combined in the proper ratio with nitrogen to make it breathable. Two tanks aboard the ISS are used to store nitrogen, and the cargo ships that travel by from time to time also transport nitrogen cylinders. Through the electrical grid of the station, the solar panels on the station supply the necessary electricity for the oxygen generators. The majority of the required water is transported to the station by cargo supply ships. Condensers, which draw water vapor even from the station's air, ensure that not a drop of water is wasted. Using the proper equipment, water is also recycled from the astronauts' urine.

Through a suitable vent, the hydrogen gas produced during the electrolysis process is released into space. Pressurized tanks at the airlock nodes at the space station are pumped with oxygen when the cargo vehicles arrive there. Pressurized tanks there are also pumped with nitrogen. It goes without saying that the station's atmospheric controls combine the gases in the right amounts for the atmosphere of Earth and then distribute the combination throughout the cabin. The production of oxygen in space is impossible.

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4. Which of the following statements about the state of matter is correct? A. Molecules in a solid are disorderly arranged B. So

Zolol [24]

Answer:

B. Solids are highly compressible

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3 years ago

A swimming pool heater has to be able to raise the temperature of the 40 000 gallons of water in the pool by 10.0 C°.

alexandr402 [8]

Answer:

b. 1 770 kWh

Explanation:

The heat needed to change the temperature of a certain amount of a substance is given by:

$Q=mC\Delta T$

Here m is the mass of the susbtance, C is the specific heat of the substance and $\Delta T$ is the temperature change

$Q=(40000*3.8kg)(4186\frac{J}{kg\cdot ^\circ C})(10^\circ C)\\Q=6.36*10^9J$

Recall that one watt hour is equivalent to 1 watt (1 W) of power sustained for 1 hour. One watt is equal to 1 J/s. So, one watt hour is equal to 3600 J and one kilowatt hour is equal to $3600*10^3 J$

$Q=6.36*10^9J*\frac{1kW\cdot h}{3600*10^3J}\\Q=1766.66kW\cdot h$

6 0

3 years ago

A toy cannon uses a spring to project a 5.38-g soft rubber ball. The spring is originally compressed by 5.08 cm and has a force

yawa3891 [41]

(a) 1.43 m/s

We can solve this problem by using the law of conservation of energy.

The initial total energy stored in the spring-mass system is

$E=U=\frac{1}{2}kx^2$

where

k = 7.91 N/m is the spring constant

$x=5.08 cm = 0.0508 m$

Substituting,

$E=\frac{1}{2}(7.91)(0.0508)^2=0.0102 J$

The final kinetic energy of the ball is equal to the energy released by the spring + the work done by friction:

$E+W_f=K$

where

$K_f=\frac{1}{2}mv^2$ is the kinetic energy of the ball, with

$m=5.38 g = 5.38\cdot 10^{-3} kg$ being the mass of the ball

v being the final speed

$W_f = -F_f d$ is the work done by friction (which is negative since the force of friction is opposite to the motion), where

$F_f = 0.0323 N$ is force of friction

d = 14.5 cm = 0.145 m is the displacement

Substituting,

$W_f = -(0.0323)(0.145)=-4.68\cdot 10^{-3} J$

So, the kinetic energy of the ball as it leaves the cannon is

$K_f = E+W_f = 0.0102 - 4.68\cdot 10^{-3}=0.00552 J$

And so the final speed is

$v=\sqrt{\frac{2K_f}{m}}=\sqrt{\frac{2(0.00552)}{0.00538}}=1.43 m/s$

(b) +5.08 cm

The speed of the ball is maximum at the instant when all the elastic potential energy stored in the spring has been released: in fact, after that moment, the spring does no longer release any more energy, so the kinetic energy of the ball from that moment will start to decrease, due to the effect of the work done by friction.

The elastic potential energy of the spring is

$U=\frac{1}{2}kx^2$

And this has all been released when it becomes zero, so when x = 0 (equilibrium position of the spring). However, the spring was initially compressed by 5.08 cm, so the ball has maximum speed when

x = +5.08 cm

with respect to the initial point.

(c) 1.78 m/s

The maximum speed is the speed of the ball at the moment when the kinetic energy is maximum, i.e. when all the elastic potential energy has been released.

As we calculated in part (a), the total energy released by the spring is

E = 0.0102 J

The work done by friction here is just the work done to cover the distance of

d = 5.08 cm = 0.0508 m

Therefore

$W_f = -(0.0323)(0.0508)=-1.64\cdot 10^{-3} J$

So, the kinetic energy of the ball at the point of maximum speed is

$K_f = E+W_f = 0.0102 - 1.64\cdot 10^{-3}=0.00856 J$

And so the final speed is

$v=\sqrt{\frac{2K_f}{m}}=\sqrt{\frac{2(0.00856)}{0.00538}}=1.78 m/s$

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3 years ago

Do you agree or disagree that folk dance forms can be outside their original formmeaningful s

My name is Ann [436]

I disagree that folk dance forms can be outside their original meaningful form in this scenario.

<h3>What is Folk dance?</h3>

This is the type of dance which is unique to a particular set of people and helps celebrate their culture.

This dance is a ritualistic entertainment which is done in different type of gatherings such as ceremonies etc which is why it isn't outside their original meaningful form.

Read more about Folk dance here brainly.com/question/18269425

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3 years ago

A .71-kg billiard ball moving at 2.5 m/s in the x-direction strikes a stationary ball of the same mass. After the collision, the

Finger [1]

After the collision, the momentum didn't change, so the total momentum in x and y are the same as the initial.

The x component was calculated by subtracting the initial momentum (total) minus the momentum of the first ball after the collision

In the y component, as at the beginning, the total momentum was 0 in this axis, the sum of both the first and struck ball has to be the same in opposite directions. In other words, both have the same magnitude but in opposite directions

$\begin{gathered} Py=0.71kg\cdot2.17\cdot sin(30) \\ Py=0.77kg\cdot m/s \end{gathered}$

This is for both balls after the collision, but one goes in a positive and the other in a negative direction.

7 0

1 year ago