By calculation, the diameter of the wire is 2.8 * 10^-3 m.
<h3>How do we obtain the length?</h3>
The following data are given in the question;
Mass of the wire = 1.0 g or 1 * 10^-3 Kg
Resistance = 0.5 ohm
Resistivity of copper = 1.7 * 10^-8 ohm meter
Density of copper = 8.92 * 10^3 Kg/m^3
V = m/d
But v = Al
Al = m/d
A = m/ld
Resistance = ρl/A
= ρl/m/ld =
l^2 = Rm/ρd
l = √ Rm/ρd
l = √0.5 * 1 * 10^-3 / 1.7 * 10^-8 * 8.92 * 10^3
l = 1.82 m
A = πr^2
Also;
A = m/ld
A = 1 * 10^-3 Kg / 1.82 m * 8.92 * 10^3 Kg/m^3
Area of the wire = 6.2 * 10^-5 m^2
r^2 = A/ π
r = √A/ π
r = √6.2 * 10^-5 m^2/3.142
r = 1.4 * 10^-3 m
Diameter = 2r = 2( 1.4 * 10^-3 m) = 2.8 * 10^-3 m
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Missing parts;
Suppose you wish to fabricate a uniform wire from 1.00g of copper. If the wire is to have a resistance of R=0.500Ω and all the copper is to be used, what must be (a) the length and (b) the diameter of this wire?
We know, F = m.a
Here, m = 1.5 Kg
a = v/t = 2/0.3 = 6.67 m/s
Substitute their values,
F = 1.5 * 6.67
F = 10 N
In short, Your Answer would be 10 Newtons
Hope this helps!
You first need to convert the hours into minutes, then use the speed equals distance over time formula.
<span>
The length of daylight on the moon is about 29.5 days.</span>
From the earliest days, the Moon has been there in the Solar
System and there has never been a period when we couldn't gaze upward in the
night sky and either observe the Moon hanging there, or realize that it would
be back the precise one night from now (i.e. a New Moon).
A day on the Moon keeps going as long as 29.5 Earth days. We
can say that it would take 29.5 days for the Sun to move the distance over the
sky and come back to its unique position once more.
