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galina1969 [7]
3 years ago
12

A fan connected to a 120-volt electrical system by an extension cord was worn through and exposed the bare, energized conductor,

which made contact with the ladder. The ground wire was not attached to the male end of the cord's plug. When the energized conductor made contact with the ladder, the path to ground included the worker's body, resulting in death. What contributed to the electrocution?
Physics
2 answers:
kari74 [83]3 years ago
8 0

Answer: the ladder

Explanation: since the energized conductor is already in contact with the ladder there by making electric current to flow. The base of the ladder is on the ground there by making the circuit to be complete and causing electrocution.

iragen [17]3 years ago
8 0

Answer:

The worker completed the circuit.

Explanation:

Thinking process:

Electricity is the flow of electrons in a circuit. There is one condition for the electrons to flow - the completion of a circuit. In order for that to be established, there must be a side with low electron concentration or affinity for electrons.

The earth has an infinite affinity for electrons. Thus, earth wires are used to channel an excess amount of electrons there. This prevents the short circuiting injuring a person. Hence, a three-pin plug.

Because there was not insulation, like rubber, an alternative pathway could not be found for electrons. Hence the worker was electrocuted as the electric current passed through him.

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Suppose kinetic energy is 4 joules of a moving object what will happen if we increase the speed twice
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Answer:

The kinetic energy is 16 joules.                        

Explanation:

Kinetic energy (E) is given by:

E = \frac{1}{2}mv^{2}

Where:

m: is the mass

v: is the speed

When the kinetic energy is 4 joules, the speed of the object is:

v_{1}^{2} = \frac{2E_{1}}{m} = \frac{8}{m}                         

Now, if the speed is increased twice then we have:

E_{2} = \frac{1}{2}mv_{2}^{2} = \frac{1}{2}m(2v_{1})^{2} = \frac{1}{2}m[4(\frac{8}{m})] = 16 J

Therefore, the kinetic energy is 4 times the initial value.  

I hope it helps you!                                

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3 years ago
Compare and contrast four types <br> of friction
malfutka [58]
Some of these frictions depend on the Pressure, temperature of atmosphere.

Static Friction: This is the  friction force when two objects in contact  are not moving relative to each other. This friction is higher than kinetic friction.

Kinetic or Dynamic friction: this the friction force opposing the motion of objects, when two objects in contact are in motion relative to each other.  It is less than the static friction. The two surfaces are rubbing against each other as they move.

Rolling friction:  This is the friction when two objects are in contact and one object is rolling over the other - like a wheel on a road. The point of contact appears as stationary. The rolling friction is very less compared to static friction & dynamic friction.

Lubricated friction: this is the friction between two solid surfaces in contact with a layer of lubricant fluid flowing in between them. This friction is the least.

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Internal friction: when an object is compressed and forced to deform, like in a piece of rubber, there is friction between the layers, that opposes this deformation.

Skin friction is the friction that opposes movement of a fluid across a solid surface.  This is also called drag.  When a coin is dropped in water, there is a friction called drag on the coin. Same is the case when a ball is thrown, a drag is experienced by the ball due to the drag of air.

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3 years ago
The op amp in this circuit is ideal. R3 has a maximum value of 100 kΩ and σ is restricted to the range of 0.2 ≤ σ ≤ 1.0. a. Calc
Firlakuza [10]

I have attached the circuit image missing in the question.

Answer:

A) The range of vo is; -6.6V≤ vo ≤-1V

B) σ = 0.1861

Explanation:

A) First of all, Let VΔ be the voltage from the potentiometer contact to the ground.

Thus; [(0 - vg)/(2000)] +[(0 - vΔ)/(50,000)] = 0

So, [(- vg)/(2000)] +[(- vΔ)/(50,000)] = 0

Simplifying further; -25 vg - vΔ = 0

From the question, vg = 40mV = 0.04 V

So - 25(0.04) = vΔ

So: vΔ = - 1 V

Now, [vΔ/(σRΔ)] + [(vΔ - 0)/(50,000)] + [(vΔ - vo)/((1 - σ)RΔ))] = 0

So, multiplying each term by RΔ to get; [vΔ/(σ)] + [(vΔ x RΔ)/(50,000)] + [(vΔ - vo)/((1 - σ))] = 0

So RΔ = 100kΩ or 100,000Ω from the question.

So, substituting for RΔ, we get,

[vΔ/(σ)] + [2vΔ] + [(vΔ - vo)/((1 - σ))] = 0

Let's put the value of - 1 for vΔ as gotten before.

So, ( - 1/σ) - 2 + [(-1 - vo)/(1 - σ)] = 0

Now let's make vo the subject of the equation to get;

-1 - vo = (1 - σ)[2 + (1/σ)]

-1 - vo = 2 - 2σ + (1/σ) - 1

-vo = 1 + 2 - 2σ + (1/σ) - 1

-vo = 2 - 2σ + (1/σ)

vo = - 1 (2 - 2σ + (1/σ))

When σ = 0.2; vo = - 1(2 - 0.4 + 5) =

- 1 x 6.6 = - 6.6V

Also when σ = 1;

vo = - 1(2 - 2 + 1) = - 1V

Therefore, the range of vo is;

- 6.6V ≤ vo ≤ - 1V

B) it will saturate at vo = - 7V

So, from;

vo = - 1 (2 - 2σ + (1/σ))

-7 = - 1 (2 - 2σ + (1/σ))

Divide both sides by (-1)

7 = (2 - 2σ + (1/σ))

Now, subtract 2 from both sides to get; 5 = - 2σ + (1/σ)

Multiply each term by α to get;

5σ = - 2σ^(2) + 1

So 2σ^(2) + 5σ - 1 = 0

Solving simultaneously and picking the positive value , we get σ to be approximately 0.1861

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