Answer:
a
Solid Wire
Stranded Wire ![I_2 = 0.00978 \ A](https://tex.z-dn.net/?f=I_2%20%20%3D%20%20%200.00978%20%5C%20%20A%20)
b
Solid Wire
Stranded Wire
Explanation:
Considering the first question
From the question we are told that
The radius of the first wire is ![r_1 = 1.53 mm = 0.0015 \ m](https://tex.z-dn.net/?f=r_1%20%20%3D%201.53%20mm%20%3D%200.0015%20%5C%20%20m)
The radius of each strand is ![r_0 = 0.306 \ mm = 0.000306 \ m](https://tex.z-dn.net/?f=r_0%20%3D%20%200.306%20%5C%20mm%20%3D%20%200.000306%20%5C%20m)
The current density in both wires is ![J = 1750 \ A/m^2](https://tex.z-dn.net/?f=J%20%20%3D%20%201750%20%5C%20%20A%2Fm%5E2)
Considering the first wire
The cross-sectional area of the first wire is
![A = \pi r^2](https://tex.z-dn.net/?f=A%20%20%20%3D%20%5Cpi%20%20r%5E2)
= >
= >
Generally the current in the first wire is
![I = J*A](https://tex.z-dn.net/?f=I%20%20%3D%20%20J%2AA)
=> ![I = 1750*7.0695 *10^{-6}](https://tex.z-dn.net/?f=I%20%20%3D%20%201750%2A7.0695%20%2A10%5E%7B-6%7D)
=>
Considering the second wire wire
The cross-sectional area of the second wire is
![A_1 = 19 * \pi r^2](https://tex.z-dn.net/?f=A_1%20%20%3D%20%2019%20%2A%20%20%5Cpi%20r%5E2)
=> ![A_1 = 19 *3.142 * (0.000306)^2](https://tex.z-dn.net/?f=A_1%20%20%3D%20%2019%20%2A3.142%20%2A%20%20%280.000306%29%5E2)
=> ![A_1 = 5.5899 *10^{-6} \ m^2](https://tex.z-dn.net/?f=A_1%20%20%3D%20%205.5899%20%2A10%5E%7B-6%7D%20%5C%20%20m%5E2)
Generally the current is
![I_2 = J * A_1](https://tex.z-dn.net/?f=I_2%20%20%3D%20%20J%20%20%2A%20%20A_1)
=> ![I_2 = 1750 * 5.5899 *10^{-6}](https://tex.z-dn.net/?f=I_2%20%20%3D%20%20%201750%20%20%2A%20%205.5899%20%2A10%5E%7B-6%7D%20)
=> ![I_2 = 0.00978 \ A](https://tex.z-dn.net/?f=I_2%20%20%3D%20%20%200.00978%20%5C%20%20A%20)
Considering question two
From the question we are told that
Resistivity is ![\rho = 1.69* 10^{-8} \Omega \cdot m](https://tex.z-dn.net/?f=%5Crho%20%20%3D%20%201.69%2A%2010%5E%7B-8%7D%20%5COmega%20%5Ccdot%20m)
The length of each wire is ![l = 6.25 \ m](https://tex.z-dn.net/?f=l%20%3D%20%206.25%20%5C%20%20m)
Generally the resistance of the first wire is mathematically represented as
=>
=>
Generally the resistance of the first wire is mathematically represented as
=>
=>
Answer:
11.98 N
Explanation:
Normal force = mg = 2.03 * 9.81
coeff of static friction must be overcome for the book to begin moving
.602 = F / (2.03 * 9.81) = 11.98 N
Answer:
s = 20 m
Explanation:
given,
mass of the roller blader = 60 Kg
length = 10 m
inclines at = 30°
coefficient of friction = 0.25
using conservation of energy
u = 9.89 m/s
Using second law of motion
ma =μ mg
a = μ g
a = 0.25 x 9.8
a = 2.45 m/s²
Using third equation of motion ,
v² - u² = 2 a s
0² - 9.89² = 2 x 2.45 x s
s = 20 m
the distance moved before stopping is 20 m
Answer:
<h2>FUNDAMENTAL UNITS INVOLVED ARE : NEWTON AND SECOND .</h2>
<h2>FORMULA OF PRESSURE = </h2>
<h2>P=F/A </h2>
how does the electric force between two charged particles change if the distance between them is increased by a factor of 3?
a. it is reduced by a factor of 3