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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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What radius of a circle is required to inscribe a regular hexagon with an area of 210.44 cm2 and an apothem of 7.794 cm
alekssr [168]
We know that

[area of a regular hexagon]=6*[area of one <span>equilateral triangle]
</span>210.44=6*[area of one equilateral triangle]
[area of one equilateral triangle]=210.44/6-----> 35.07 cm²

[area of one equilateral triangle]=b*h/2
h=7.794 cm
b=2*area/h------> b=2*35.07/7.794------>b= 9 cm

the length side of a regular hexagon is 9 cm
<span>applying the Pythagorean theorem
</span>r²=h²+(b/2)²------>r²=7.794²+(4.5)²------> r²=81--------> r=9 cm

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</span>
the answer is
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8 0
3 years ago
The feet of the average adult woman are 24.6 cm long, and foot lengths are normally distributed. If 16% of adult women have feet
nevsk [136]

Answer:

Approximately 16% of adult women have feet longer than 27.2 cm.

Step-by-step explanation:

Normal Probability Distribution

Problems of normal distributions can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the z-score of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

The feet of the average adult woman are 24.6 cm long

This means that \mu = 24.6

16% of adult women have feet that are shorter than 22 cm

This means that when X = 22, Z has a p-value of 0.16, so when X = 22, Z = -1. We use this to find \sigma

Z = \frac{X - \mu}{\sigma}

-1 = \frac{22 - 24.6}{\sigma}

-\sigma = -2.6

\sigma = 2.6

Approximately what percent of adult women have feet longer than 27.2 cm?

The proportion is 1 subtracted by the p-value of Z when X = 27.2. So

Z = \frac{X - \mu}{\sigma}

Z = \frac{27.2 - 24.6}{2.6}

Z = 1

Z = 1 has a p-value of 0.84.

1 - 0.84 = 0.16

0.16*100% = 16%.

Approximately 16% of adult women have feet longer than 27.2 cm.

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Rewrite the decimal in the sentence below as a percentage.
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Answer:

0.0066%

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Answer:

m<HGI=21°

Step-by-step explanation:

we know that

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The measure of angle HGI is equal to

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