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PilotLPTM [1.2K]
2 years ago
9

On isolated ground receptacles, the metal yoke ____ integrally bonded to the equipment grounding terminal of the receptacle.

Physics
1 answer:
Ede4ka [16]2 years ago
4 0

On isolated ground receptacles, the metal yoke is not allowed to be integrally bonded to the equipment grounding terminal of the receptacle.

Any device with two distinct switches or receptacles is a duplex device. It can be shaped to fit a Decora opening or a typical duplex plate opening. It should be noted that they can be combination devices with a switch/outlet, switch/pilot light, etc.

Because of grounding connection removal and receptacle, it is utterly undesirable to connect the two bare equipment grounding conductors together directly.

The equipment grounding conductor associated with those circuits must be connected to the box when circuit conductors are terminated on equipment inside a metal box to prevent unneeded current discharge.

Learn more about  grounding conductors here brainly.com/question/14886979

#SPJ4.

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Answer: 107.8\ J

Explanation:

Given

Initial position of object is (4.4 i+5 j)

Final position of object is (11.6 i -2 j)

Force acting (4i-9j)

Work done is given by

\Rightarrow W=F\cdot dx\\\Rightarrow W=(4i-9j)\cdot (11.6i-4.4i-2j-5j)\\\Rightarrow W=(4i-9j)\cdot (7.2i-7j)\\\Rightarrow W=28.8+63\\\Rightarrow W=91.8\ J

Initial kinetic energy

K_i=\dfrac{1}{2}\times 2\times 4^2\\\\K_i=16\ J

Change in kinetic energy is equal to work done by object

\Rightarrow K_f=K_i+W\\\Rightarrow K_f=16+91.8\\\Rightarrow K_f=107.8\ J

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Do you think it is possible to determine an object is moving towards you, or moving away from you just by how it sounds? If
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Answer:

I think so

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Answer:

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<em>In our case</em> our initial velocity is v_i=100ft/s and our final velocity is v_f=0ft/s since it comes to rest, this process while traveling a distance d=1000ft, so we have:

a=\frac{v_f^2-v_i^2}{2d}=\frac{(0m/s)^2-(100ft/s)^2}{2(1000ft)}=-5ft/s^2

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