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?
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I believe this is right but double check to make sure :)
Answer:
Q = 5 L/s
Explanation:
To find the flow you use the following formula (para calcular el caudal usted utiliza la siguiente formula):

V: Volume (volumen) = 200L
t: time (tiempo) = 40 s
you replace the values of the parameters to calculate Q (usted reemplaza los valores de los parámteros V y t para calcular el caudal):

Hence, the flow is 5 L/s (por lo tanto, el caudal es de 5L/s)
The angular velocity of the wheel at the bottom of the incline is 4.429 rad/sec
The angular velocity (ω) of an object is the rate at which the object's angle position is changing in relation to time.
For a wheel attached to an incline angle, the angular velocity can be computed by considering the conservation of energy theorem.
As such the total kinetic energy (K.E) and rotational kinetic energy (R.K.E) at a point is equal to the total potential energy (P.E) at the other point.
i.e.
P.E = K.E + R.K.E







Therefore, we can conclude that the angular velocity of the wheel at the bottom of the incline is 4.429 rad/sec
Learn more about angular velocity here:
brainly.com/question/1452612