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?
Answer:
What is the radius of the table tennis ball?
⇒ 2.1 cm
What is the radius of the golf ball?
2.0 cm
Explanation:
divide the radius and round it to the nearest 10th place..... but hope that help ;)
0.5mv^2
0.5 times 40 times 3^2
The kinetic energy is 180
Answer:
1.25 focal lengths
Explanation:
The lens equation states that:
where
f is the focal length
p is the object distance
q is the image distance
In this problem, the image is 4 times as far from the lens as is the object: this means that
If we substitute this into the lens equation and we rearrange it, we get
so, the object distance measured in focal lengths is
1.25 focal lenghts
Answer:
light and sound are both trasverse waves
Explanation:
