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n200080 [17]
3 years ago
11

A human being can be electrocuted if a current as small as 55 mA passes near the heart. An electrician working with sweaty hands

makes good contact with the two conductors he is holding. If his resistance is 2300 Ω, what might the fatal voltage in volts be?
Physics
2 answers:
Flura [38]3 years ago
8 0

Answer:

126.5 V

Explanation:

Using Ohm's Law,

V = IR............................. Equation 1

Where V = Voltage, I = current, R = resistance.

Note: The current needed to bring about the fatal voltage is equal to the current that will cause human being to be electrocuted.

Given: I = 55 mA = 55/1000 = 0.055 A, R = 2300 Ω

Substitute into equation 1

V = 0.055×2300

V = 126.5 V.

Hence the fatal voltage = 126.5 V

evablogger [386]3 years ago
4 0

Answer:

The lethal voltage for the electrician under those conditions is 126.5 V.

Explanation:

To discover what is the lethal voltage to the electrician we need to find out what is the voltage that produces 55 mA = 0.055 A when across a resistance of 2300 Ohms (Electrician's body resistancy). For that we'll use Ohm's Law wich is expressed by the following equation:

V = i*R

Where V is the voltage we want to find out, i is the current wich is lethal to the electrician and R is his body resistance. By applying the given values we have:

V = 0.055*2300 = 126.5 V.

The lethal voltage for the electrician under those conditions is 126.5 V.

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3 years ago
A 12.70 g bullet has a muzzle velocity (at the moment it leaves the end of a firearm) of 430 m/s when rifle with a weight of 25.
Norma-Jean [14]

Answer:

2.1844 m/s

Explanation:

The principle of conservation of momentum can be applied here.

when two objects interact, the total momentum remains the same  provided no external forces are acting.

Consider the whole system , gun and bullet. as an isolated system, so the net momentum is constant. In particular before firing the gun, the net momentum is zero. The conservation of momentum,

0=m_{bullet}*v_{bullet}  + m_{rifle}*v_{rifle}  \\

assume the bullet goes to right side and the gravitational acceleration =10 ms^{-2}

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0=m_{bullet}*v_{bullet}  + m_{rifle}*v_{rifle}  \\\\0=(12.70*10^{-3}) *430ms^{-1} +(\frac{25}{10} )*v_{rifle} \\v_{rifle} =-2.1844ms^{-1}

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3 0
3 years ago
Two sources of light of wavelength 700 nm are 9 m away from a pinhole of diameter 1.2 mm. How far apart must the sources be for
Lelechka [254]

Answer:

The distance is  D  =  0.000712 \ m

Explanation:

From the question we are told that

    The wavelength of  the  light source is  \lambda  =  700 \ nm = 700 *10^{-9} \  m

     The distance from a pin hole is  x  =  9\ m

       The  diameter of the pin  hole is  d =  1.2 \ mm  =  0.0012 \ m

     

Generally the distance which the light source need to be in order for their diffraction patterns to be resolved by Rayleigh's criterion is

mathematically represented as

              D  =  \frac{1.22 \lambda }{d }

substituting values

             D  =  \frac{1.22 * 700 *10^{-9} }{ 0.0012 }

             D  =  0.000712 \ m

5 0
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The best logical answer is A
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