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xxTIMURxx [149]
3 years ago
11

When is a handrail required for stairs?

Engineering
1 answer:
Leviafan [203]3 years ago
5 0

Answer:

after 8 stepshddnffuddbnggkbdbkloyr

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For a gearbox power and efficiency test apparatus that accommodates interchangeable gearboxes, drive up to 0.5kW. How to select
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Answer:

First we must know the efficiency of the gearbox in which energy is lost by friction between its components, we divide this efficiency by the nominal power and find the real power.

With this value knowing the speed of rotation of the gearbox we select an engine that has a power just above.

Also keep in mind that the motor torque must be greater than the resistance offered by the gearbox and the 0.5Kw machine

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4 years ago
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What careers could you potential do if you
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Engineering careers. If you want to stay in engineering, your job opportunities are very much linked to your degree type, and you probably know what many of them are already. ...

Consulting. ...

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Explanation:

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Is a street the same as a avenue
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On a date when the earth was 147.4x106 km from the sun, a spacecraft parked in a 200 km altitude circular earth orbit was launch
rewona [7]

Answer:

ΔV =  =v_{p} -v_{c} = 3.337 \frac{km}{s}

Explanation:

Distance of earth from sun = R_{2} = 147.4 \times 106 Km

Spacecraft perihelion = R_{2} = 120\times106Km

gravitational parameters are now given as

\mu_{sun} = 132.7\times 10^{9}

\mu_{earth} = 398600

radius of earth  = 6378 Km

Heliocentric spacecraft velocity at earth sphere of influence =

   V_{D}^{v} =\sqrt{2\mu_{sun}} \sqrt{\frac{R_{2} }{R_{1}(R_{1} +R_{2} ) } }

V_{D}^{v} =28.43\frac{km}{s}

Heliocentric velocity of earth = v_{earth} = 30.06\frac{km}{sec}

V_{infinity}= v_{earth}-V_{D}^{v} =30.06-28.43=1.57g\frac{km}{s}

assume

r_{p} =r_{earth} +r_{altitude} =6378 + 200 = 6578Km

Geometric spacecraft velocity of spacecraft at perigee of departure hyperbola

v_{p}=\sqrt{v^{2} _{infinity}+\frac{2\mu_{earth} }{r_{p} } } = 11.12\frac{km}{s}

geometric space craft velocity in its circular parking orbit

v_{c}=\sqrt{\frac{\mu_{earth} }{r_{p} } }  = 7.784 \frac{km}{s}

              ΔV =  =v_{p} -v_{c} = 3.337 \frac{km}{s}

7 0
4 years ago
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