If you accidentally touch the "hot" wire connected to the 120 V line, how much current will pass through your body?
1 answer:
<h2>Complete Question:</h2>
You are on an aluminum ladder that is standing on the ground, trying to fix an electrical connection with a metal screwdriver having a metal handle. Your body is wet because you are sweating from the exertion; therefore, it has a resistance of 1.60 kΩ .
(a) If you accidentally touch the "hot" wire connected to the 120 V line, how much current will pass through your body?
(b) How much electrical power is delivered to your body?
<h2>Answer:</h2>(a) 0.075A
(b) 9W
<h2>Explanation:</h2>The voltage (V) passing across or supplied to a body is directly proportional to the current (I) flowing through the body as stated by Ohm's law. i.e
V ∝ I
=> V = I x R ----------------------(i)
Where;
R = constant of proportionality called resistance of the body
(a) As stated in the question;
The body is wet and thus will conduct electricity and has the following;
V = voltage supplied = 120V
R = resistance of the wet body = 1.60kΩ = 1.6 x 1000Ω = 1600Ω
Substitute these values into equation(i) as follows;
120 = I x 1600
Solve for I;
I =
I = 0.075A
Therefore the amount of current that will pass through your body is 0.075A
(b) Electrical power(P), which is commonly measured in Watts(W), delivered to a body is the product of the current(I) and voltage (V) supplied to the body. i.e
P = I x V ---------------------(ii)
Where;
I = 0.075A [as calculated above]
V = 120V [given in the question]
Substitute these values into equation (ii) as follows;
P = 0.075 x 120
P = 9W
Therefore, the electric power delivered to your body is 9W
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Answer:
<em>0.45 mm</em>
Explanation:
The complete question is
a certain fuse "blows" if the current in it exceeds 1.0 A, at which instant the fuse melts with a current density of 620 A/ cm^2. What is the diameter of the wire in the fuse?
A) 0.45 mm
B) 0.63 mm
C.) 0.68 mm
D) 0.91 mm
Current in the fuse is 1.0 A
Current density of the fuse when it melts is 620 A/cm^2
Area of the wire in the fuse = I/ρ
Where I is the current through the fuse
ρ is the current density of the fuse
Area = 1/620 = 1.613 x 10^-3 cm^2
We know that 10000 cm^2 = 1 m^2, therefore,
1.613 x 10^-3 cm^2 = 1.613 x 10^-7 m^2
Recall that this area of this wire is gotten as
A =
where d is the diameter of the wire
1.613 x 10^-7 =
6.448 x 10^-7 = 3.142 x
=
d = 4.5 x 10^-4 m = <em>0.45 mm</em>
Answer:
I = 2 kgm^2
Explanation:
In order to calculate the moment of inertia of the door, about the hinges, you use the following formula:
(1)
I: moment of inertia of the door
α: angular acceleration of the door = 2.00 rad/s^2
τ: torque exerted on the door
You can calculate the torque by using the information about the Force exerted on the door, and the distance to the hinges. You use the following formula:
(2)
F: force = 5.00 N
d: distance to the hinges = 0.800 m
You replace the equation (2) into the equation (1), and you solve for α:
Finally, you replace the values of all parameters in the previous equation for I:
The moment of inertia of the door around the hinges is 2 kgm^2