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3 years ago

What is a static load

Engineering

1 answer:

dusya [7]3 years ago

7 0

Answer:

static loading is the load on an actuator whis it is in a fixed pr stationery condition.

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In plumbing what is a video snake used for

aleksley [76]

Answer:

How to stop toilets

Explanation:

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3 years ago

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Technician A says that after replacing a power steering hose, the system should be flushed, refilled, and bled. Technician B say

dedylja [7]

The. Answer will be D

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4 years ago

To protect their employees from caught-in and -between hazards, what must employers do?

marusya05 [52]

Answer:

Employers must comply with OSHA's regulations to safeguard workers from caught-in or -between dangers, which include, but are not limited to, the following:

• Protect power tools and other equipment with moving parts with guards.

• Support, secure, or otherwise make safe any equipment that has pieces that workers could become entangled in.

· Take precautions to avoid workers being crushed by tipped-over heavy machinery.

• Take precautions to avoid pinning workers between equipment and a solid object.

• Provide workers with protection when trenching and excavating.

• Provide ways to prevent constructions, such as scaffolds, from collapsing.

Explanation:

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3 years ago

Historical thinking skills can be divded into there main processes

morpeh [17]

Answer:

asking questions, drawing conclusions, and gathering information.

Explanation:

hope this helps!

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4 years ago

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