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Maksim231197 [3]
3 years ago
12

select the correct answer from each drop-down menu. study this model of a gravitational slingshot then answer the questions. In

this model, the velocity of the spacecraft of the position 2 is (equal to - greater than- less than) the velocity of the craft at position 4. At point 1, the direction of the spacecraft changes because of ( the gravitational force between earth and the spacecraft - the momentum of the space craft - secondary fuel tanks). Position 3 represents the (final destination of the spacecraft - gravitational force of the spacecraft on earth - orbital path or a of earth).​

Geography
2 answers:
MArishka [77]3 years ago
6 0
<h2>1. Right answer: the velocity of the spacecraft at position 2 is <u>greater than</u> the velocity of the craft at position 4. </h2>

This is due the gravity field of the Earth is used to accelerate the craft. This is true when in a specific point the direction of the movement of the craft is the same direction of the movement of the planet.

In this case the craft will be “catched” by the Earth’s gravitational field, making the craft  to enter a circular orbit.

<h2>2. Right answer: At point 1, the direction of the spacecraft changes because of <u>the gravitational force between earth and the spacecraft. </u></h2>

As explained in the first answer, this is the exact point where the trajectory of the spacecraft enters into a circular orbit because of the attraction due gravity of the Earth and therefore changes its direction.

<h2>3. Right answer: Position 3 represents the <u>orbital path of Earth </u></h2>

Being this the orbital path of the Earth and considering the trajectory of the craft, the condition of accelerating the craft is accomplished. If the orbital path of the Earth were the opposite from the shown in the figure, the effect on the craft would be <u>braking. </u>

Note all of these is related to the gravitational assistance, this consists in a maneuver in which the energy of the gravitational field of a planet or satellite is used to obtain an acceleration or braking of the probe or craft, changing its trajectory.

This maneuver is also called <em>slingshot effect</em>, swing-by or gravity assist.

NeTakaya3 years ago
3 0

1. Greater than

2. the gravitational force between earth and the spacecraft

3. orbital path or velocity of earth

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You have a wire that is 23 cm long. You wish to cut it into two pieces. One piece will be bent into the shape of a square. The o
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Answer:

In explanation you see the answer but it I see it send it  different  so I just added a file so you can see clearly the way how this exercise works. Please, see the file attached

Explanation:

Let x = circumference of the circle

then

(53-x) = the perimeter of the square

and \frac{53-\times}{4} = the side of the square

and

\left(\frac{53-\times}{4}\right)^2 = the area of the square

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Find the area of the circle using the circumference

find the radius (r)

2*pi*r = x

r = \frac{\times}{2*\mathbit{\pi}}

r = \frac{\times}{6.28}

Find the area of the circle

\mathbit{A}=\pi r^2

Replace r with \frac{\times}{6.28}

A = π * \left(\frac{\times}{6.28}\right)^2=  * \left(\frac{\times^2}{39.48}\right)      

cancel pi into 39.48

A =  \left(\frac{\times^2}{12.566}\right)

Total area of circle and square

A =  \left(\frac{\times^2}{12.566}\right)+\left(\frac{53-\times}{4}\right)^2  = \left(\frac{\times^2}{12.566}\right)  + \left(\frac{{2809-106\times+\times}^2}{16}\right)

:

convert these fractions to decimal coefficients

A(x) = .0796x^2 + .0625x^2 - 6.625x + 175.5625

A(x) = .1421x^2 - 6.625x + 175.5625

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Find the axis of symmetry of this quadratic equation (min area)

x =   \frac{-\left(-6.625\right)}{2*.1421}

x =   \frac{6.625}{.2842}

x = 23.31 cm is the circumference when they have min area

Download docx
7 0
3 years ago
If AC = 48 meters, what is the perimeter of the field?
denis23 [38]
I think it would be 
6+6+8+8 bc 6x8=48 if that makes sense
so the answer would be 28 m

or 12+12+4+4 bc 12x4=48
so the answer is 32 m

it really depends on how you break 48 up

have a blessed day! 
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