A parallel-plate capacitor with a 4.9 mm plate separation is charged to 57 V . Part A With what kinetic energy, in eV, must a pr
1 answer:
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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:
5 mg,
Explanation:
First of all, let's rewrite the mass in grams using scientific notation.
we have:
m = 0.005 g
To rewrite it in scientific notation, we must count by how many digits we have to move the dot on the right - in this case three. So in scientific notation is
If we want to convert into milligrams, we must remind that
1 g = 1000 mg
So we can use the proportion
and we find