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Paladinen [302]
3 years ago
9

In a hospital setting, which of the following presents the possibility of being the source of an arc flash? A. Electrocautery De

vice B. Defibrillator C. Muscle Stimulation Electrodes D. Cauterization Tool
Physics
2 answers:
Umnica [9.8K]3 years ago
3 0
The answer to this question would be <span>Electrocautery Device

In a hospital, the device that can cause arc flash would be electrocautery device and circuit breaker. In fact, the electrocautery device uses the arc flash principle to work. The arc flash causing burns that surgeon use to clot bleeding in surgery or to destroy a tissue.</span>
Kamila [148]3 years ago
3 0

Answer:

B. Defibrillator

Explanation:

The electric arc occurs when there is a dielectric break in which it produces an electrical discharge. The term electric arc is the most common, but it can also be called an arc flash or short circuit as it is popularly called. This phenomenon is so strong that it can break the insulation made by air by conducting electrons from one electrode to another through a current flow.

A defibrillator is a device designed to discharge electrical charges to the chest wall (if external) or to the heart muscle fibers (if internal) of a patient with cardiac arrhythmia. Because of its ability to provide an electrical discharge, we can conclude that the defibrillator is a source of an electric arc in the hospital environment.

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B. Direction

Explanation:

Speed is a scalar quantity and doesn't keep track of direction : Velocity is a vector quantity and is direction aware.

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A U-shaped tube open to the air at both ends contains some mercury. A quantity of water is carefully poured into the left arm of
Gekata [30.6K]

Answer:

a) P=2450\ Pa

b) \delta h=23.162\ cm

Explanation:

Given:

height of water in one arm of the u-tube, h_w=25\ cm=0.25\ m

a)

Gauge pressure at the water-mercury interface,:

P=\rho_w.g.h_w

we've the density of the water =1000\ kg.m^{-3}

P=1000\times 9.8\times 0.25

P=2450\ Pa

b)

Now the same pressure is balanced by the mercury column in the other arm of the tube:

\rho_w.g.h_w=\rho_m.g.h_m

1000\times 9.8\times 0.25=13600\times 9.8\times h_m

h_m=0.01838\ m=1.838\ cm

<u>Now the difference in the column is :</u>

\delta h=h_w-h_m

\delta h=25-1.838

\delta h=23.162\ cm

7 0
3 years ago
A car is moving at 19 m/s along a curve on a horizontal plane with radius of curvature 49m.
JulsSmile [24]

Answer:

\mu =0.75

Explanation:

<u>Frictional Force </u>

When the car is moving along the curve, it receives a force that tries to take it from the road. It's called centripetal force and the formula to compute it is:

F_c=m.a_c

The centripetal acceleration a_c is computed as

\displaystyle a_c=\frac{v^2}{r}

Where v is the tangent speed of the car and r is the radius of curvature. Replacing the formula into the first one

F_c=m.\frac{v^2}{r}

For the car to keep on the track, the friction must have the exact same value of the centripetal force and balance the forces. The friction force is computed as

F_r=\mu N

The normal force N is equal to the weight of the car, thus

F_r=\mu .m.g

Equating both forces

\displaystyle \mu .m.g=m.\frac{v^2}{r}

Simplifying

\displaystyle \mu =\frac{v^2}{rg}

Substituting the values

\displaystyle \mu =\frac{19^2}{(49)(9.8)}

\boxed{\mu =0.75}

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