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Nutka1998 [239]
4 years ago
10

Help please its due today will mark you brainliest

Engineering
1 answer:
Tems11 [23]4 years ago
6 0

Answer:

launch- The first stage is ignited at launch and burns through the powered ascent until its propellants are exhausted. The first stage engine is then extinguished, the second stage separates from the first stage, and the second stage engine is ignited. The payload is carried atop the second stage into orbit

powered ascent-The first stage is ignited at launch and burns through the powered ascent until its propellants are exhausted. The first stage engine is then extinguished, the second stage separates from the first stage, and the second stage engine is ignited. The payload is carried atop the second stage into orbit

coasting flight-

When the rocket runs out of fuel, it enters a coasting flight. The vehicle slows down under the action of the weight and drag since there is no longer any thrust present. The rocket eventually reaches some maximum altitude which you can measure using some simple length and angle measurements and trigonometry.

ejection charge-At the end of the delay charge, an ejection charge is ignited which pressurizes the body tube, blows the nose cap off, and deploys the parachute. The rocket then begins a slow descent under parachute to a recovery. The forces at work here are the weight of the vehicle and the drag of the parachute.

slow decent- slow downs (i guess)

recovery-A recovery period is typically characterized by abnormally high levels of growth in real gross domestic product, employment, corporate profits, and other indicators. This is a turning point from contraction to expansion and often results in an increase in consumer confidence

Explanation:

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An incandescent light bulb can be regarded as a resistor. If its power output is 100W, calculate the resistance of the light bul
stira [4]

Answer:121\ \Omega

0.909\ A

Explanation:

Given

Power P=100\ W

Voltage applied V=110\ V

Resistance of the bulb is given by

P=\frac{V^2}{R}

100=\frac{110^2}{R}

R=\frac{12100}{100}

R=121\ \Omega

Current drawn by the Power source is given by

P=V\cdot I

I=\frac{P}{V}

I=\frac{100}{110}

I=0.909\ A

8 0
4 years ago
Breh/bro <br><br>what is this, finally I can learn programming​
LenKa [72]

Answer:

It's an intoduction to hacking and systematic programming.

Explanation:

Yes, you might be able to grasp a few things from it, but it also may be a way hackers could hack you, by luring you to click it.

8 0
3 years ago
A storage tank, used in a fermentation process, is to be rotationally molded from polyethylene plastic. This tank will have a co
NNADVOKAT [17]

Answer:

The volume up to cylindrical portion is approx  32355 liters.

Explanation:

The tank is shown in the attached figure below

The volume of the whole tank is is sum of the following volumes

1) Hemisphere top

Volume of hemispherical top of radius 'r' is

V_{hem}=\frac{2}{3}\pi r^3

2) Cylindrical Middle section

Volume of cylindrical middle portion of radius 'r' and height 'h'

V_{cyl}=\pi r^2\cdot h

3) Conical bottom

Volume of conical bottom of radius'r' and angle \theta is

V_{cone}=\frac{1}{3}\pi r^3\times \frac{1}{tan(\frac{\theta }{2})}

Applying the given values we obtain the volume of the container up to cylinder is

V=\pi 1.5^2\times 4.0+\frac{1}{3}\times \frac{\pi 1.5^{3}}{tan30}=32.355m^{3}

Hence the capacity in liters is V=32.355\times 1000=32355Liters

3 0
4 years ago
NAME JUICE WRLDS SONG THAT HE BLEW UP ON
creativ13 [48]

Answer:

lucid dreams :)

Explanation:

5 0
3 years ago
Read 2 more answers
Series and parrarel circuts combination a. Find the currents I1, I2, I3, I4, I5, and I6. a 5 k R1 = 1 k R7 = 2 k I1 I428 V 6 k R
satela [25.4K]

Answer:

Explanation:

R1 = 1 k; R2 = 1k; R3 = 3k; R4 = 6k; R5 = 5k; R6 = 6k and R7 = 2k

Vt = 428V

The series and parallel circuit combination is as follow:

(R6║R7 + R5) + R4 + R3║R2 + R1

(6*2/6 + 2) + 5 = 13/ k

(13/2*6/13/6 + 6) = 78/31k

78/3 + 3 = 171/3 = 57k

57k║R2 = 57k║1k = 57/58k + R1 = (57/58 + 1)k = 115/58k = 2k

It = Vt/2k = 428/2000 = 0.2A

∴ I1 = 0.2A

I1 = I2 + I3

Using current divider rules to obtain I2 and I3

∴ I2 = I1 X (I2/I2 + I3) = 0.2 X ( 1/4) = 0.05A

and I3 = I1 X (I3/I2 + I3) = 0.2 X (3/4) = 0.15A

I3 = I4 + I5, using current divider

I4 = I3 X (I4/I4 + I5) = 0.15 X (6/6 + 5) = 0.08

I5 = 0.15 - 0.08 = 0.07A

I5 = I6 + I7, using current divider

I6 = I5 X (I6/I6 + I7) = 0.07 X (6/6 + 2) = 0.05A

I7 = 0.07 - 0.05 = 0.02A

8 0
3 years ago
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