Answer:
magnification will be -0.025
Explanation:
We have given the radius of curvature = 12 cm
And object distance = 3 m
So focal length 
Now for mirror we know that
So 

v = 0.750 m
Now magnification of the mirror is 
To solve this problem it is necessary to apply the concepts related to the magnetic field.
According to the information, the magnetic field INSIDE the plates is,

Where,
Permeability constant
Electromotive force
r = Radius
From this deduction we can verify that the distance is proportional to the field

Then the distance relationship would be given by




On the outside, however, it is defined by

Here the magnetic field is inversely proportional to the distance, that is

Then,




Answer:
From the question we are told that
The length of the rod is 
The speed is v
The angle made by the rod is 
Generally the x-component of the rod's length is

Generally the length of the rod along the x-axis as seen by the observer, is mathematically defined by the theory of relativity as

=> ![L_xo = [L_o cos (\theta )] \sqrt{1 - \frac{v^2}{c^2} }](https://tex.z-dn.net/?f=L_xo%20%20%3D%20%20%5BL_o%20cos%20%28%5Ctheta%20%29%5D%20%20%5Csqrt%7B1%20%20-%20%5Cfrac%7Bv%5E2%7D%7Bc%5E2%7D%20%7D)
Generally the y-component of the rods length is mathematically represented as

Generally the length of the rod along the y-axis as seen by the observer, is also equivalent to the actual length of the rod along the y-axis i.e
Generally the resultant length of the rod as seen by the observer is mathematically represented as

=> ![L_r = \sqrt{[ (L_o cos(\theta) [\sqrt{1 - \frac{v^2}{c^2} }\ \ ]^2+ L_o sin(\theta )^2)}](https://tex.z-dn.net/?f=L_r%20%20%3D%20%5Csqrt%7B%5B%20%28L_o%20cos%28%5Ctheta%29%20%5B%5Csqrt%7B1%20-%20%5Cfrac%7Bv%5E2%7D%7Bc%5E2%7D%20%7D%5C%20%5C%20%5D%5E2%2B%20L_o%20sin%28%5Ctheta%20%29%5E2%29%7D)
=> ![L_r= \sqrt{ (L_o cos(\theta)^2 * [ \sqrt{1 - \frac{v^2}{c^2} } ]^2 + (L_o sin(\theta))^2}](https://tex.z-dn.net/?f=L_r%3D%20%5Csqrt%7B%20%28L_o%20cos%28%5Ctheta%29%5E2%20%2A%20%5B%20%5Csqrt%7B1%20-%20%5Cfrac%7Bv%5E2%7D%7Bc%5E2%7D%20%7D%20%5D%5E2%20%2B%20%28L_o%20sin%28%5Ctheta%29%29%5E2%7D)
=> ![L_r = \sqrt{(L_o cos(\theta) ^2 [1 - \frac{v^2}{c^2} ] +(L_o sin(\theta))^2}](https://tex.z-dn.net/?f=L_r%20%20%3D%20%5Csqrt%7B%28L_o%20cos%28%5Ctheta%29%20%5E2%20%5B1%20-%20%5Cfrac%7Bv%5E2%7D%7Bc%5E2%7D%20%5D%20%2B%28L_o%20sin%28%5Ctheta%29%29%5E2%7D)
=> ![L_r = \sqrt{L_o^2 * cos^2(\theta) [1 - \frac{v^2 }{c^2} ]+ L_o^2 * sin(\theta)^2}](https://tex.z-dn.net/?f=L_r%20%3D%20%20%5Csqrt%7BL_o%5E2%20%2A%20cos%5E2%28%5Ctheta%29%20%20%5B1%20-%20%5Cfrac%7Bv%5E2%20%7D%7Bc%5E2%7D%20%5D%2B%20L_o%5E2%20%2A%20sin%28%5Ctheta%29%5E2%7D)
=> ![L_r = \sqrt{ [cos^2\theta +sin^2\theta ]- \frac{v^2 }{c^2}cos^2 \theta }](https://tex.z-dn.net/?f=L_r%20%20%3D%20%20%5Csqrt%7B%20%5Bcos%5E2%5Ctheta%20%2Bsin%5E2%5Ctheta%20%5D-%20%5Cfrac%7Bv%5E2%20%7D%7Bc%5E2%7Dcos%5E2%20%5Ctheta%20%7D)
=> 
Hence the length of the rod as measured by a stationary observer is

Generally the angle made is mathematically represented

=> 
=>
Explanation:
Answer:
When air temperature is higher, the weather is warmer.