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grin007 [14]
2 years ago
3

In space, astronauts don’t have gravity to keep them in place. That makes doing even simple tasks difficult. Gene Cernan was the

first astronaut who worked on a task outside a spaceship. He said of the experience, “Every time I’d push or turn a valve, it would turn my entire body at zero gravity. I had nothing to hold on to.” As he worked, Gene Cernan’s heart rate and temperature went so high that his fellow astronauts worried that he wouldn’t survive.
Think about routine tasks that astronauts might need to do inside and outside a spaceship. Choose several tasks, and describe the features the ship and spacesuits should have to account for zero gravity as the astronaut completes the task. Use Newton’s laws of motion in your analysis.

pls help me
Physics
2 answers:
saul85 [17]2 years ago
4 0

Newton's First Law: A body will remain at rest or in motion in a straight line unless acted upon by a force.

<span>Newton's Second Law: </span>Change in motion is proportional to the applied force and parallel to it.

<span>Newton's Third Law: </span>To every action there is an equal and opposite reaction.

<span>By Newton's first law, two things are really cheap: Coasting and sitting still. Out in space, sitting still is a little hard to define - do I mean compared with the Sun, or Mars? Also, most of the time we are in an orbit of some sort, so sitting still doesn't really make sense. What we need to think about is how to change from one kind of orbit to another kind of orbit.</span><span>The sensation of weightlessness that astronauts experience seem to make their tasks almost effortless. However, as Newton's third law of motion suggests, working in space can be physically demanding. 

-turning a valve
-walking around
-trying to shower


</span>
Olegator [25]2 years ago
4 0

Answer:

Let's start with Newton's laws:

First Law: A body will remain at rest or in motion in a straight line unless acted upon by a force.

Second Law: Change in motion is proportional to the applied force and parallel to it.

Third Law: To every action, there is an equal and opposite reaction.

Usually, when you are in the earth, the third law is hard to see, this is because when, for example, you throw a ball, the force that the ball does to you is not comparable with the force that the gravity is doing to you, and maintaining you in place. But in the space, you don't have this force "protecting" you, so if you throw a ball in zero gravity, you will be impulsed in the opposite direction at where the ball is thrown.

Good ideas to have in mind when working in zero gravity are protection strings, that limit the distance that you can go outside the spaceship.

Having magnetized suits also can be very helpfull when working outside the ship, because it allows you to "glued" to it. (and for example, if you are removing out a screw, collecting this screw with some magnetic device would be necessary, because you don't want scraps of metal in the outer space, they can be dangerous)

Nowdays some spacesuits even come with small propulsors, that allow the astronauts to move with some control when they are working outside the spaceship.

For the task inside of it, there are a lot of problems, for example, the effect that the zero gravity has in your body, in how you can eat food and how you use the bathroom, and there are a lot of technological advances that simplify a bit the approach that astronauts need to take when doing these tasks.

Where maybe one of the most important, "non-technological" things you need to have in the spaceship, is a capacitated team that can bring a solution to different problems, like engineering problems or medical problems (and of course, al the equipment these specialist need to do it).

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A. 4 cm behind the mirror
<span>  For any mirror, </span><span><span>so</span><span>si</span>=<span>f^2</span></span>. Therefore, by plugging in the values, you get <span>18<span>si</span>=144. 144/18 = 8, so </span><span><span>si</span>=8</span>
<span>  The focal point is located 12 cm from the mirror, and </span><span>si</span><span> is the distance of the image from the focal point, so the image is 4 cm from the mirror. The mirror is convex, so then the focal point and the image are both behind the mirror.</span>
5 0
2 years ago
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A basketball with mass of 0.8 kg is moving to the right with velocity 6 m/s and hits a volleyball with mass of 0.6 kg that stays
IceJOKER [234]

Answer:

26.67 m/s

Explanation:

From the law of conservation of linear momentum, the initial sum of momentum equals the final sum.

p=mv where p is momentum, m is the mass of object and v is the speed of the object

Initial momentum

The initial momentum will be that of basketball and volleyball, Since basketball is initially at rest, its initial velocity is zero

p_i= m_bv_b+m_vv_v=8*6+0.6*0=48 Kg.m/s

Final momentum

p_f= m_bv_b+m_vv_v=8*4+0.6*v_v=32+0.6v Kg.m/s\\32+0.6v_v=48\\0.6v=16\\v_v=16/0.6=26.66666667\approx 26.67 m/s

4 0
2 years ago
A pump jack scaffold must be fitted with two ______ gripping mechanisms to prevent slippage.
FromTheMoon [43]
A pump jack scaffold must be fitted with two positive gripping mechanisms to prevent slippage. Pump jacks are a uniquely designed scaffold consisting of a platform supported by movable brackets on vertical poles. The brackets are designed to be raised and lowered in a manner similar to an automobile jack. It is important to make sure that pump jack brackets have two positive gripping mechanisms to prevent any failure or slippage. 
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2 years ago
jenny's model train is set up on a circular track. There are six telephone poles evenly spaced around the track. It takes the en
d1i1m1o1n [39]

Answer:

T = 60 s

Explanation:

There are 6 poles on the track which are equally spaced

so the angular separation between the poles is given as

\theta = \frac{2\pi}{6}

\theta = \frac{\pi}{3}

so the angular speed of the train is given as

\omega = \frac{\theta}{t}

\omega = \frac{\pi}{30} rad/s

now we have time period of the train given as

T = \frac{2\pi}{\omega}

T = \frac{2\pi}{\frac{\pi}{30}}

T = 60 s

5 0
2 years ago
A boy throws a water ballon such that it hits his sister standing 10m away. The boy threw the water balloon at an angle of 35 de
Tresset [83]

Answer: 7.734 m/s

Explanation:

We have the following data:

\theta=35\° The angle at which the water ballon was thrown

x=10 m  The horizontal distance of the water ballon

g=-9.8 m/s^{2} The acceleration due gravity

We need to find the initial velocity V_{o} at which the water ballon was thrown, and we can find it by the following equation:

x=V_{o}cos \theta T (1)

Where T=2t is the total time the water ballon is on air

On the other hand, when we talk about parabolic motion (as in this situation) the water ballon reaches its maximum height just in the middle of this parabola, when V=0 and the time t is half the time T it takes the complete parabolic path.

So, if we use the following equation, we will find t:

V=V_{o}+gt=0 (2)

Isolating t:

t=\frac{-V_{o}}{g} (3)

Remembering T=2t:

T=2\frac{-V_{o}}{g} (4)

Substituting (4) in (1):

x=V_{o}cos \theta (2\frac{-V_{o}}{g}) (5)

Isolating V_{o}:

V_{o}=\sqrt{\frac{x g}{-2 cos \theta}} (6)

V_{o}=\sqrt{\frac{(10 m)(9.8 m/s^{2})}{-2 cos(35\°)}} (7)

Finally:

V_{o}=7.734 m/s

4 0
2 years ago
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