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
Learn more about resistivity:brainly.com/question/14547003
#SPJ4
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?
Electricity. Ruler. 69. N.
Answer:
20.96 m/s^2 (or 21)
Explanation:
Using the formula (final velocity - initial velocity)/time = acceleration, we can plug in values and manipulate the problem to give us the answer.
At first, we know a car is going 8 m/s, that is its initial velocity.
Then, we know the acceleration, which is 1.8 m/s/s
We also know the time, 7.2 second.
Plugging all of these values in shows us that we need to solve for final velocity. We can do so by manipulating the formula.
(final velocity - initial velocity) = time * acceleration
final velocity = time*acceleration + initial velocity
After plugging the found values in, we get 20.96 m/s/s, or 21 m/s
This number has 3 sig figs.
Answer:
Option b. Effective nuclear charge increases as we move to the right across a row in the periodic table
Explanation:
The <em>effective nuclear charge </em>is a measure of how strong the protons in the nucleus of an atom attract the outermost electrons of such atom.
The <em>effective nuclear charge</em> is the net positive charge experienced by valence electrons and is calculated (as an approximation) by the equation: Zeff = Z – S, where Z is the atomic number and S is the number of shielding electrons.
The shielding electrons are those electrons in between the interesting electrons and the nucleus of the atom.
Since the shielding electrons are closer to the nucleus, they repel the outermost electrons and so cancel some of the attraction exerted by the positive charge of the nucleus, meaning that the outermost electrons feel less the efect of attraction of the protons. That is why in the equation of Zeff, the shielding electrons (S) subtract the total from the atomic number Z.
The <em>effective nuclear charge</em>, then, is responsible for some properties and trends in the periodic table. Here, you can see how this explains the trend of the atomic radius (size of the atom) accross a row in the periodic table.
- As the<em> effective nuclear charge</em> is larger, in a same row of the periodic table, the shielding effect is lower, the outermost electrons are more strongly attracted by the nucleus, and the size of the atoms decrease. That is why as we move to the right in the periodic table, the size of the atoms decrease.
