Answer:0.1759 v
Explanation:
Intensity of wave at receiver end is
I=
I=
I=
Amplitude of electric field at receiver end
Amplitude of induced emf
=
=
=
The answer is B tell me if I am wrong.
B
V= f x lambda
V= 5m/s
F = 10hz
Lambda = ?
5 = 10 x lamba
5 /10 = lambda
Wavelength =0.5
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
Learn more about resistivity:brainly.com/question/14547003
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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: 8*10^-15 N
Explanation: In order to calculate the force applied on an electron in the middle of the two planes at 500 V we know that, F=q*E
The electric field between the plates is given by:
E = ΔV/d = 500 V/0.01 m=5*10^3 N/C
the force applied to the electron is: F=e*E=8*10^-15 N
