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Blababa [14]
3 years ago
14

explain each of the following kinds of rockets: Solid-Fuel Rocket, Liquid-Fuel Rocket, Ion Rocket and Plasma Rocket.

Engineering
1 answer:
Rudik [331]3 years ago
5 0

Answer:

ur answer friend

Explanation:

answer

<em>S</em><em>o</em><em>l</em><em>i</em><em>d</em><em>-</em><em>F</em><em>u</em><em>e</em><em>l</em><em> </em><em>R</em><em>o</em><em>c</em><em>k</em><em>e</em><em>t</em><em> </em><em>-</em><em> </em>a solid-propellant rocket or solid rocket is a rocket with a rocket engine that uses solid propellants. The earliest rockets were solid-fuel rockets powered by gunpowder; they were used in warfare by the Chinese, Indians, Mongols and Persians, as early as 13th century.

<em>L</em><em>i</em><em>q</em><em>u</em><em>i</em><em>d</em><em>-</em><em>F</em><em>u</em><em>e</em><em>l</em><em> </em><em>R</em><em>o</em><em>c</em><em>k</em><em>e</em><em>t</em><em> </em><em>-</em><em> </em>a liquid-propellant rocket or liquid rocket utilizes a rocket engine that use liqiud propellants. An inert gas stored in a tank at a high pressure is sometimes used instead of pumps in simpler small engines to force the propellants into the combustion chamber.

<em>I</em><em>o</em><em>n</em><em> </em><em>R</em><em>o</em><em>c</em><em>k</em><em>e</em><em>t</em><em> </em><em>-</em><em> </em>an ion thruster or ion drive is a form of electric propulsion used for spacecraft propulsion. It creates thrust by accelerating ions using electricity. The Deep Space 1 spacecraft, powered by an ion thruster, changed velocity by 4.3 km/s ( 9600 mph ) while consuming less than 74 kg ( 163 lb ) of xenon.

<em>P</em><em>l</em><em>a</em><em>s</em><em>m</em><em>a</em><em> </em><em>R</em><em>o</em><em>c</em><em>k</em><em>e</em><em>t</em><em> </em><em>-</em><em> </em>in this type of rocket, a combination of electric and magnetic fields are used to break down the atoms and molecules of a propellant gas into a collection of particles that have either a positive charge (ions) or a negative charge (electrons). In other words, the propellant gas becomes a plasma.

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nirvana33 [79]

Answer:

You need a 120V to 24V commercial transformer  (transformer 1:5), a 100 ohms resistance, a 1.5 K ohms resistance and a diode with a minimum forward current of 20 mA (could be 1N4148)

Step by step design:

  1. Because you have a 120V AC voltage supply you need an efficient way to reduce that voltage as much as possible before passing to the rectifier, for that I recommend a standard 120V to 24V transformer.  120 Vrms = 85 V and 24 Vrms = 17V = Vin
  2. Because 17V is not 15V you still need a voltage divider to step down that voltage, for that we use R1 = 100Ω and R2 = 1.3KΩ. You need to remember that more than 1 V is going to be in the diode, so for our calculation we need to consider it. Vf = (V*R2)/(R1+R2), V = Vin - 1 = 17-1 = 16V and Vf = 15, Choosing a fix resistance R1 = 100Ω and solving the equation we find R2 = 1.5KΩ
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Our circuit meet the average voltage (Va) specification:

Va = (15)/(pi) = 4.77V considering the diode voltage or 3.77V without considering it

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total time  = 304.21 s

Explanation:

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solution

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y = 1 - e^{(-kt^n)}   ...........1    

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0.5 = 1 - e^{(-k(114)^{1.1})}    

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k = 0.003786  = 37.86 × 10^{4}

so here

y = 1 - e^{(-kt^n)}

1 - y  =  e^{(-kt^n)}

take ln both side

ln(1-y) = -k × t^n  

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t = \sqrt[n]{-\frac{ln(1-y)}{k}}    .............2

now we will put the value of y = 87% in equation  with k and find out t

t = \sqrt[1.1]{-\frac{ln(1-0.87)}{37.86*10^{-4}}}

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