The fatal current is 51 mA = 0.051 Ampere.
The resistance is 2,050Ω .
Voltage = (current) x (resistance)
= (0.051 Ampere) x (2,050 Ω) = 104.6 volts .
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This is what the arithmetic says IF the information in the question
is correct.
I don't know how true this is, and I certainly don't plan to test it,
but I have read that a current as small as 15 mA through the
heart can be fatal, not 51 mA .
If 15 mA can do it, and the sweaty electrician's resistance is
really 2,050 Ω, then the fatal voltage could be as little as 31 volts !
The voltage at the wall-outlets in your house is 120 volts in the USA !
THAT's why you don't want to stick paper clips or a screwdriver into
outlets, and why you want to cover unused outlets with plastic plugs
if there are babies crawling around.
Answer:
(b) Water
Explanation:
The density of the liquid is found by dividing its mass by its volume.
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<h3>mass</h3>
The mass of the liquid is the difference in mass between the full and empty graduated cylinder:
liquid mass = (145 g) -(45 g) = 100 g
<h3>volume</h3>
The cylinder is said to contain 100 mL of the liquid, so that is the volume of interest.
<h3>density</h3>
The ratio of mass to volume is the liquid's density:
ρ = (100 g)/(100 mL) = 1 g/mL
This density identifies the most likely liquid as water.
By the work-energy theorem, the total work done on the car is equal to the change in its kinetic energy:
<em>W</em> = ∆<em>K</em>
<em>W</em> = 1/2 (0.34 kg) (22.9 m/s)² - 1/2 (0.34 kg) (6.5 m/s)²
<em>W</em> ≈ 82 J
The mass of the object does not change by moving it to another place. ... At the center of the earth the net gravitational force is zero, so the weight will be zero, but its masses will remain same. Hence the mass at the centre of earth will be equl to 50 kg.
Answer:
a) always. b) electric field lines are defined by the path positive test charges travel.
Explanation:
By convention, field lines always follow the direction that it would take a positive test charge (small enough so it can´t disrupt the field created by a charge distribution), under the influence of an electric field, at the same point where the test charge is located.
So any positive charge, subject to an electric field influence, moves along the field line that passes through its current position, in the same way that a positive test charge would.
We could say also that the electric force on a positively charged particle is in the same direction as the electric field that produces that force (due to some charge distribution) , which is true, but it doesn´t explain why.
