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: a) The acceletarion is directed to the center on the turntable. b) 5 cm; ac= 0.59 m/s^2; 10 cm, ac=1.20 m/s^2; 14 cm, ac=1.66 m/s^2
Explanation: In order to explain this problem we have to consider teh expression of the centripetal accelartion for a circular movement, which is given by:
ac=ω^2*r where ω and r are the angular speed and teh radios of the circular movement.
w=2*π*f
We know that the turntable is set to 33 1/3 rev/m so
the frequency 33.33/60=0.55 Hz
then w=2*π*0.55=3.45 rad/s
Finally the centripetal acceleration at differents radii results equal:
r= 0.05 m ac=3.45^2*0.05=0.50 m/s^2
r=0.1 ac=3.45^2*0.1=1.20 m/s^2
r=0.14 ac=3.45^2*0.14=1.66 m/s^2
Answer:
1.3 x 10⁻⁴ m
Explanation:
= wavelength of the light = 450 nm = 450 x 10⁻⁹ m
n = order of the bright fringe = 1
θ = angle = 0.2°
d = separation between the slits
For bright fringe, Using the equation
d Sinθ = n
Inserting the values
d Sin0.2° = (1) (450 x 10⁻⁹)
d (0.003491) = (450 x 10⁻⁹)
d = 1.3 x 10⁻⁴ m
No, because he was a philosopher
A simple electromagnet consisting of a coil of insulated wire wrapped around an iron core<span>. A </span>core<span> of ferromagnetic material like </span>iron<span> serves to increase the magnetic field created. The strength of magnetic field </span>generated<span> is proportional to the amount of </span>current<span> through the winding.</span>
