The answer would be 2.63. Your welcome. This has been changed to the correct answer.
Explanation:
The third class lever cannot magnify our force because in third class lever the effort it between the load and the fulcrum. Also, in this type of lever no matter where the force is applied, it is always greater than the force of load. Hence, That type of lever cannot magnify our force.
It will be
E = mgh.
where h and g are constant thus
m can be written as 4/3πr^3*density
E = 4/3πr^3* density
E? = 4/3π(2R)^3* density
= 4/3π8r^3
thus the e will be 4/3π8r^3* density/4/3πr^3*density nd thus you get 8E ..
Answer:
Θ=0.01525 rad
or
Θ=0.87°
Explanation:
Given data
wavelength λ=2.5 µm =2.5×10⁻⁶m
Diameter d=0.20 mm =0.20×10⁻³m
To find
Angle Θ in radians and degree
Solution
Circular apertures have first dark fringe at
Θ=(1.22λ)/d
Substitute the given values
So
Θ=[1.22(2.5×10⁻⁶m)]/0.20×10⁻³m
Θ=0.01525 rad
or
Θ=0.87°
