Answer:
p = 6.64 cm
Explanation:
For this exercise we use the equation of the constructor

where f is the focal length, p and q are the distance to the object and the image, respectively
They tell us the focal length f = 2.2 cm and that the image as far as it can go is q = 3.29 cm, let's find the position of the object that creates this image
1 / p = 1 / 2.2 - 1/3.29
1 / p = 0.15059
p = 6.64 cm
therefore the farthest distance from the object is 6.64 c
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:
31.55 m/s
Explanation:
Let the initial velocity of the arrow is u metre per second.
Angle of projection, θ = 40 degree
range = 100 m
Use the formula for the range.

100 = u^2 Sin(2 x 40) / 9.8
100 x 9.8 = u^2 Sin 80
u^2 = 995.11
u = 31.55 m/s
Answer:
shown in the attachment
Explanation:
The detailed step by step and necessary mathematical application is as shown in the attachment.
Like a seesaw, it shows that the forces aren’t equal because if it was the seesaw would stay put