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PilotLPTM [1.2K]
3 years ago
13

The height reached by the rocket depends on the amount of fuel

Mathematics
2 answers:
emmasim [6.3K]3 years ago
4 0
Initial displacement = dboost = v0t + 1/2at^2 
<span>hboost = dboost sin theta </span>
<span>xboost = dboost cos theta </span>
<span>vboost = v0 + at </span>
<span>vxboost = vboost cos theta </span>
<span>vyboost = vboost sin theta </span>

<span>To get the maximum altitude, use conservation of energy </span>
<span>initial PE + vertical KE </span>
<span>= mg (hboost) + 1/2 m vyboost^2 </span>
<span>= final PE = mg (hfinal) </span>

<span>hfinal = hboost + vyboost^2 / 2g </span>
<span>= hboost + ((v0 + at) * sin (theta))^2 / 2g </span>

<span>b) flight time = </span>
<span>boost time + time to rise ballistically from hboost to hfinal </span>
<span>+ time to fall from hfinal </span>

<span>The time to fall a distance h can be gotten by: </span>
<span>h = 1/2 gt^2, so t = sqrt (h/2g) </span>

<span>So flight time = t + sqrt ((hfinal - hboost)/2g)+ sqrt (hfinal/2g) </span>

<span>Horizontal range = xboost + vxboost * time of ballistic flight </span>

<span>= (v0t + 1/2at^2) cos theta </span>
<span>+ (v0 + at) cos theta * (sqrt ((hfinal - hboost)/2g)+ sqrt (hfinal/2g)) </span>

<span>Lots of plugging and chugging to do now. </span>

<span>EDIT--the problem said acceleration was constant, so we don't have to worry about loss of mass during the boost. Realistic? No. But that's what the problem says. Once the thing is ballistic, it's a moot point anyway

</span>
gavmur [86]3 years ago
3 0
True as u cant have a rocket flying at a high height if its low on fuel as if the fuel runs out mid flight they will just crash

brainliest:))))))))


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The following functions give the populations of four towns with time tt in years.
expeople1 [14]

Solution :

Transforming given equations :

1)\ P = 600( 1 + \dfrac{12}{100})^t\\\\2)\ P = 1000( 1 + \dfrac{3}{100})^t\\\\3)\ P = 200( 1 + \dfrac{8}{100})^t\\\\4)\ P = 900( 1 - \dfrac{10}{100})^t

From above equations we can see that equation 1) has largest percentage growth rate and the percent growth rate is 12% .

Hence, this is the required solution.

6 0
3 years ago
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If vector c = vector b + vector a, under what circumstance does c = b+a ?
Vitek1552 [10]
The sum of the magnitudes will be the magnitude of the sum when the vectors have the same direction.
6 0
3 years ago
Which is greater, 0.28 or 0.205? Explain how you know.
LuckyWell [14K]

Answer:

Step-by-step explanation:

0.28 is roughly the same as 0.280, which has 3 significant digits (same as 0.205 has 3 significant digits).  It's easier when we compare numbers having the same number of decimal places.

Comparing 0.280 to 0.205, we see that the hundredths place of 0.280 (which is 8) is greater than the hundredths place of 0.205 (which is 0).  Thus 0.28 is greateer than 0.205.

6 0
2 years ago
If the volume of a box is 2x3 + 4x2 − 30xwhich of the dimensions are possible with the given x-value?
Kipish [7]

The possible value of x = 4, dimensions 8 by 9 by 1 (option D), if the volume of a box is 2 x^{3} + 4 x^{2} -30x.

Step-by-step explanation:

The given is,

                        2 x^{3} + 4 x^{2} -30x................................(1)

Step:1

    Check for option A,

             x = 1, dimensions 8 by 9 by 1  

            From the equation (1),

                      Volume = 2 (1^{3}) + 4 (1^{2} )-30(1)

                                    =2+4-30 = -24...................(2)

            From the dimensions,

                      Volume = ( 8 × 9 × 1 )

                                     = 72............................................(3)

            From equation (2) and (3)

                                -24 ≠ 72

            So, X=1; dimensions 8 by 9 by 1 is not possible.

   Check for option B,

             x = 1, dimensions 2 by 5 by 3

            From the equation (1),

                      Volume = 2 (1^{3}) + 4 (1^{2} )-30(1)

                                    =2+4-30 = -24...................(4)

            From the dimensions,

                      Volume = ( 2 × 5 × 3 )

                                     = 30.........................................(5)

            From equation (4) and (5)

                                -24 ≠ 30

            So, X=1; dimensions 2 by 5 by 3 is not possible.

   Check for option C,

            x = 4, dimensions 2 by 5 by 3

            From the equation (1),

                      Volume = 2 (4^{3}) + 4 (4^{2} )-30(4)

                                    =2(64)+4(16)-30(4)

                                    = 128+64-120

                                    = 72.............................................(6)

            From the dimensions,

                      Volume = ( 2 × 5 × 3 )

                                     = 30............................................(7)

            From equation (6) and (7)

                               72 ≠ 30

            So, X=4; dimensions 2 by 5 by 3 is not possible.

    Check for option C,

            x = 4, dimensions 8 by 9 by 1

            From the equation (1),

                      Volume = 2 (4^{3}) + 4 (4^{2} )-30(4)

                                    =2(64)+4(16)-30(4)

                                    = 128+64-120

                                    = 72............................................(8)

            From the dimensions,

                      Volume = ( 8 × 9 × 1 )

                                    = 72............................................(9)

            From equation (8) and (9)

                               72 = 72

            So, X=4; dimensions 8 by 9 by 3 is possible.

Result:

           The possible value of x = 4, dimensions 8 by 9 by 1 (option D), if the volume of a box is 2 x^{3} + 4 x^{2} -30x.

         

4 0
3 years ago
What are the values of x and y?
Olenka [21]
X=5.4 and Y=10.5.
Used the law of cosine equation, then SOHCAHTOA, and then Pythagorean theorem.
6 0
2 years ago
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