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azamat
2 years ago
11

Question:- A planet X moves in an elliptical path as shown below. At which point does the planet moves faster. Assume, Ethiopian

institute of space science and technology, allows to launch a space station with you. While you are in orbit, you gets thirsty. The only thing in the space station were a cup of water. you go to the cup of water and pick it up,But unfortunately your hand is twisted and the cup is turned down.Because the cup is turned down, water pour on to the floor.How would you get water?​
Physics
1 answer:
babymother [125]2 years ago
6 0

We can get water from the sweat and exhaled breath of the people present in the space station.

<h3>How would you get water?​</h3>

We can get water from the sweat and exhaled breath. The water we drink is recycled from the sweat and exhaled breath of the people present in the space station which was collected through  condensation on the Space Station's walls.

So we can conclude we can get water from the sweat and exhaled breath of the people present in the space station.

Learn more about water here: brainly.com/question/1313076

#SPJ1

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3 years ago
A mass of 4.10 kg is suspended from a 1.69 m long string. It revolves in a horizontal circle as shown in the figure.
nikklg [1K]

The horizontal component of the tension in the string is a centripetal force, so by Newton's second law we have

• net horizontal force

F_{\rm tension} \sin(\theta) = \dfrac{mv^2}R

where m=4.10\,\rm kg, v=2.85\frac{\rm m}{\rm s}, and R is the radius of the circular path.

As shown in the diagram, we can see that

\sin(\theta) = \dfrac Rr \implies R = r\sin(\theta)

where r=1.69\,\rm m, so that

F_{\rm tension} \sin(\theta) = \dfrac{mv^2}R \\\\ \implies F_{\rm tension} = \dfrac{mv^2}{r\sin^2(\theta)}

The vertical component of the tension counters the weight of the mass and keeps it in the same plane, so that by Newton's second law we have

• net vertical force

F_{\rm \tension} \cos(\theta) - mg = 0 \\\\ \implies F_{\rm tension} = \dfrac{mg}{\cos(\theta)}

Solve for \theta :

\dfrac{mv^2}{r\sin^2(\theta)} = \dfrac{mg}{\cos(\theta)} \\\\ \implies \dfrac{\sin^2(\theta)}{\cos(\theta)} = \dfrac{v^2}{rg} \\\\ \implies \dfrac{1-\cos^2(\theta)}{\cos(\theta)} = \dfrac{v^2}{rg} \\\\ \implies \cos^2(\theta) + \dfrac{v^2}{rg} \cos(\theta) - 1 = 0

Complete the square:

\cos^2(\theta) + \dfrac{v^2}{rg} \cos(\theta) + \dfrac{v^4}{4r^2g^2} = 1 + \dfrac{v^4}{4r^2g^2} \\\\ \implies \left(\cos(\theta) + \dfrac{v^2}{2rg}\right)^2 = 1 + \dfrac{v^4}{4r^2g^2} \\\\ \implies \cos(\theta) + \dfrac{v^2}{2rg} = \pm \sqrt{1 + \dfrac{v^4}{4r^2g^2}} \\\\ \implies \cos(\theta) = -\dfrac{v^2}{2rg} \pm \sqrt{1 + \dfrac{v^4}{4r^2g^2}}

Plugging in the known quantities, we end up with

\cos(\theta) \approx 0.784 \text{ or } \cos(\theta) \approx -1.27

The second case has no real solution, since -1\le\cos(\theta)\le1 for all \theta. This leaves us with

\cos(\theta) \approx 0.784 \implies \theta \approx \cos^{-1}(0.784) \approx \boxed{38.3^\circ}

7 0
2 years ago
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