The frequency of the wheel is given by:

where N is the number of revolutions and t is the time taken. By using N=100 and t=10 s, we find the frequency of the wheel:

And now we can find the angular speed of the wheel, which is related to the frequency by:
Find me on animal jam my username is aparri
It would be rock by erosion bc the rocks erode to form a volcano
Answer:
The focal length is 16.86 cm and the distance of the man if he wants to form an upright image of his chin that is twice the chin's actual size is 8.43 cm.
Explanation:
Given that,
Object distance u=1.54 m =154 cm
Image distance v = 15.2 cm
Magnification = 2
We need to calculate the focal length
Using formula of mirror

Put the value into the formula



We need to calculate the focal length
Using formula of magnification

Put the value into the formula


Using formula of for focal length






Hence, The focal length is 16.86 cm and the distance of the man if he wants to form an upright image of his chin that is twice the chin's actual size is 8.43 cm.