The diameter of the wire is 2.8 * 10^-3 m.
<h3>What is the length?</h3>
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
A = 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?
<span>Force = Work done / distance = 4Nm / 2m = 2N</span>
Answer: Physical changes only change the appearance of a substance, not its chemical composition. Chemical changes cause a substance to change into an entirely substance with a new chemical formula. Chemical changes are also known as chemical reactions.
Explanation:
Answer:
Gravity is the centripetal force when the moon orbits the earth.
Answer:
![g'=63.74\ m/s^2](https://tex.z-dn.net/?f=g%27%3D63.74%5C%20m%2Fs%5E2)
Explanation:
It is given that,
Weight of the person on Earth, W = 818 N
Weight of a person is given by the following formula as :
![W=mg](https://tex.z-dn.net/?f=W%3Dmg)
g is the acceleration due to gravity on earth
![m=\dfrac{W}{g}](https://tex.z-dn.net/?f=m%3D%5Cdfrac%7BW%7D%7Bg%7D)
![m=\dfrac{818\ N}{9.8\ m/s^2}](https://tex.z-dn.net/?f=m%3D%5Cdfrac%7B818%5C%20N%7D%7B9.8%5C%20m%2Fs%5E2%7D)
m = 83.46 kg
The mass of an object is same everywhere. It does not depend on the location.
Let W' is the weight of the person on the surface of a nearby planet, W' = 5320 N
g' is the acceleration due to gravity on that planet. So,
![g'=\dfrac{W'}{m}](https://tex.z-dn.net/?f=g%27%3D%5Cdfrac%7BW%27%7D%7Bm%7D)
![g'=63.74\ m/s^2](https://tex.z-dn.net/?f=g%27%3D63.74%5C%20m%2Fs%5E2)
So, the acceleration due to gravity on that planet is
. Hence, this is the required solution.