Answer:
The total resistance of the wire is = 
Explanation:
Since the wires will both be in contact with the voltage source at the same time and the current flows along in their length-wise direction, the two wires will be considered to be in parallel.
Hence, for resistances in parallel, the total resistance, 

Parameters given:
Length of wire = 1 m
Cross sectional area of copper 
Cross sectional area of aluminium wire
![A_{al}= \pi( R^{2}-r^{2})\\\\ = \pi \times [ (2\times 10^{-3} )^{2}-(1\times 10^{-3} )^{2}] =9.42\times10^{-6} m^{2}\\](https://tex.z-dn.net/?f=A_%7Bal%7D%3D%20%5Cpi%28%20R%5E%7B2%7D-r%5E%7B2%7D%29%5C%5C%5C%5C%20%3D%20%5Cpi%20%5Ctimes%20%5B%20%282%5Ctimes%2010%5E%7B-3%7D%20%20%29%5E%7B2%7D-%281%5Ctimes%2010%5E%7B-3%7D%20%20%29%5E%7B2%7D%5D%20%3D9.42%5Ctimes10%5E%7B-6%7D%20m%5E%7B2%7D%5C%5C)
Resistivity of copper 
Resistivity of Aluminium 
Resistance of copper 
Resistance of aluminium 
The total resistance of the wire can be obtained as follows;


∴ The total resistance of the wire = 
Answer:
heat, energy that is transferred from one body to another as the result of a difference in temperature. If two bodies at different temperatures are brought together, energy is transferred—i.e., heat flows—from the hotter body to the colder. example: stove
Explanation:
hope this helps
Answer:
2.2N
Explanation:
Given parameters:
Work done = 379.5J
Height = 173m
Unknown:
Amount of force exerted on the sled = ?
Solution:
The amount of force she exerted on the sled is the same as her weight.
Work done is the force applied to move a body through a distance.
Work done = mgh
m is the mass
g is the acceleration due to gravity
h is the height
mg = weight;
Work done = weight x h
379.5 = weight x 173
weight =
= 2.2N
Answer:
velocity. height. weight. possition. place. energy. force.
Explanation: 50/50 % chance they are wrong and write.
All the stars, spheres and galaxies that can be perceived nowadays
make up just 5 percent of the universe.
The former 95 percent is prepared of stuff stargazers can't see, notice or even
understand. These secretive substances are called dark energy and
dark matter. Experts determine their presence grounded on their gravitational
influence on what little bits of the universe can be perceived.