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?
Explanation:
Half-life is the time taken for a radioactive material to decay to half its original composition:
Original mass = 48g
Half- life = 2hr
After four half lives;
Initially: 48g
First halving 24
Second halving 12
Third halving 6
Fourth halving 3
After second half life, we would have 12g
At fourth halving, we would have 3g
Answer:
4.7 is 10 as much as the number 0.47.
If you multiply 0.47 x 10 it will equal 4.7
Explanation:
Answer:
b. able to travel through a vacuum.
Explanation:
The most distinguishing factor of an electromagnetic waves is that they are able to travel through a vacuum.
These waves do not require materials in a medium for propagation.
- Electromagnetic waves are formed by the propagation of the electric and magnetic fields.
- They vibrate at an angle of 90° .
- They are unlike like mechanical waves that requires that requires materials in medium for their propagation.
